Merge Sort Implementation using C++ - c++

I am having trouble with trying to implement the merge sort algorithm. I would appreciate it if someone can help me out. Here is what I have.
#include <iostream>
#include <deque>
using size_type = std::deque<int>::size_type;
void print(std::deque<int> &v)
{
for(const auto &ref:v)
std::cout << ref << " ";
std::cout << std::endl;
}
void merge(std::deque<int> &vec, size_type p, size_type q, size_type r)
{
int n_1 = q - p;
int n_2 = r - q;
std::deque<int> left, right;
for(auto i = 0; i != n_1; i++)
left.push_back(vec[p + i]);
for(auto j = 0; j != n_2; j++)
right.push_back(vec[q + j]);
int i = 0, j = 0;
std::cout << "left = ";
print(left);
std::cout << "right = ";
print(right);
for(auto k = p; k != r; k++) {
if(i < n_1 && left[i] <= right[j]) {
vec[k] = left[i];
i++;
}
else if(j < n_2){
vec[k] = right[j];
j++;
}
}
}
void merge_sort(std::deque<int> &A, size_type p, size_type r)
{
int q;
if(p < r) {
q = (p + r)/2;
merge_sort(A, p, q);
merge_sort(A, q + 1, r);
merge(A, p, q, r);
}
}
int main()
{
std::deque<int> small_vec = {1, 6, 2, 10, 5, 2, 12, 6};
std::deque<int> samp_vec = {2, 9, 482, 72, 42, 3, 4, 9, 8, 73, 8, 0, 98, 72, 473, 72, 3, 4, 9, 7, 6, 5, 6953, 583};
print(small_vec);
merge_sort(small_vec, 0, small_vec.size());
print(small_vec);
return 0;
}
The output from the program is:
left =
right = 1
left = 1
right = 6
left =
right = 10
left = 1 6
right = 2 10
left =
right = 2
left =
right = 6
left = 2
right = 12 6
left = 1 2 6 10
right = 5 2 12 6
1 2 5 2 6 10 12 6

There are a few issues with your sort. Firstly the merge step is wrong. Second how you call merge is wrong. Ill suggest a few steps to improve the implementation to a correct solution, and maybe itll help you.
First my code for merge:
void merge(std::deque<int> &vec, size_type p, size_type q, size_type r)
{
std::deque<int> left, right;
int i = p, j = q+1;
while(i <= q) //change 1: to a while loop. expresses it a little simpler but
//you weren't inserting the correct left elements here
left.push_back(vec[i++]);
while(j <= r) //change 2: same thing, lets get the correct right values
right.push_back(vec[j++]);
i = 0; j = 0;
for(auto k = p; k <= r; k++) {
//change 3: alter this case to include the left over left elements! this is the main error
if(i < left.size() && left[i] <= right[j] || j >= right.size())
vec[k] = left[i++];
else if(j < right.size())
vec[k] = right[j++];
}
}
Then to change how you call merge_sort to:
merge_sort(small_vec, 0, small_vec.size()-1); //change 4: initialized r wrong
This made the sort work for me. As a review of the problems I found: 1) not grabbing the correct subarrays of left and right. 2) didn't handle merge correctly - forgot to grab all the left elements after right is gone. 3) didn't call merg_sort correctly, initializing the r parameter incorrectly.

Related

Issue with getting to some data in a matrix [duplicate]

How do I print a 5×5 two-dimensional array in spiral order?
Is there any formula so that I can print an array of any size in spiral order?
The idea is to treat the matrix as a series of layers, top-right layers and bottom-left layers. To print the matrix spirally we can peel layers from these matrix, print the peeled part and recursively call the print on the left over part. The recursion terminates when we don't have any more layers to print.
Input matrix:
1 2 3 4
5 6 7 8
9 0 1 2
3 4 5 6
7 8 9 1
After peeling top-right layer:
1 2 3 4
8
5 6 7 2
9 0 1 6
3 4 5 1
7 8 9
After peeling bottom-left layer from sub-matrix:
6 7
5 0 1
9 4 5
3
7 8 9
After peeling top-right layer from sub-matrix:
6 7
1
0 5
4
After peeling bottom-left layer from sub-matrix:
0
4
Recursion terminates.
C functions:
// function to print the top-right peel of the matrix and
// recursively call the print bottom-left on the submatrix.
void printTopRight(int a[][COL], int x1, int y1, int x2, int y2) {
int i = 0, j = 0;
// print values in the row.
for(i = x1; i<=x2; i++) {
printf("%d ", a[y1][i]);
}
// print values in the column.
for(j = y1 + 1; j <= y2; j++) {
printf("%d ", a[j][x2]);
}
// see if more layers need to be printed.
if(x2-x1 > 0) {
// if yes recursively call the function to
// print the bottom left of the sub matrix.
printBottomLeft(a, x1, y1 + 1, x2-1, y2);
}
}
// function to print the bottom-left peel of the matrix and
// recursively call the print top-right on the submatrix.
void printBottomLeft(int a[][COL], int x1, int y1, int x2, int y2) {
int i = 0, j = 0;
// print the values in the row in reverse order.
for(i = x2; i>=x1; i--) {
printf("%d ", a[y2][i]);
}
// print the values in the col in reverse order.
for(j = y2 - 1; j >= y1; j--) {
printf("%d ", a[j][x1]);
}
// see if more layers need to be printed.
if(x2-x1 > 0) {
// if yes recursively call the function to
// print the top right of the sub matrix.
printTopRight(a, x1+1, y1, x2, y2-1);
}
}
void printSpiral(int arr[][COL]) {
printTopRight(arr,0,0,COL-1,ROW-1);
printf("\n");
}
Pop top row
Transpose and flip upside-down (same as rotate 90 degrees counter-clockwise)
Go to 1
Python 2 code:
import itertools
arr = [[1,2,3,4],
[12,13,14,5],
[11,16,15,6],
[10,9,8,7]]
def transpose_and_yield_top(arr):
while arr:
yield arr[0]
arr = list(reversed(zip(*arr[1:])))
print list(itertools.chain(*transpose_and_yield_top(arr)))
For python 3x:
import itertools
arr = [[1,2,3,4],
[12,13,14,5],
[11,16,15,6],
[10,9,8,7]]
def transpose_and_yield_top(arr):
while arr:
yield arr[0]
arr = list(reversed(list(zip(*arr[1:]))))
print(list(itertools.chain(*transpose_and_yield_top(arr))))
I see that no one has use only one for loop and without recursion in the code, and so I want to contribute.
The idea is like this:
Imagine there is a turtle standing at point (0,0), that is, top-left corner, facing east (to the right)
It will keep going forward and each time it sees a sign, the turtle will turn right
So if we put the turtle at point (0,0) facing right-ward, and if we place the signs at appropriate places, the turtle will traverse the array in spiral way.
Now the problem is: "Where to put the signs?"
Let's see where we should put the signs (marked by #, and numbers by O):
For a grid that looks like this:
O O O O
O O O O
O O O O
O O O O
We put the signs like this:
O O O #
# O # O
O # # O
# O O #
For a grid that looks like this:
O O O
O O O
O O O
O O O
We put the signs like this:
O O #
# # O
O # O
# O #
And for a grid that looks like this:
O O O O O O O
O O O O O O O
O O O O O O O
O O O O O O O
O O O O O O O
We put the signs like this:
O O O O O O #
# O O O O # O
O # O O # O O
O # O O O # O
# O O O O O #
We can see that, unless the point is at the top-left part, the signs are places at points where the distances to the closest horizontal border and the closest vertical border are the same, while for the top-left part, the distance to the top border is one more than the distance to the left border, with priority given to top-right in case the point is horizontally centered, and to top-left in case the point is vertically centered.
This can be realized in a simple function quite easily, by taking the minimum of (curRow and height-1-curRow), then the minimum of (curCol and width-1-curCol) and compare if they are the same. But we need to account for the upper-left case, that is, when the minimum is curRow and curCol themselves. In that case we reduce the vertical distance accordingly.
Here is the C code:
#include <stdio.h>
int shouldTurn(int row, int col, int height, int width){
int same = 1;
if(row > height-1-row) row = height-1-row, same = 0; // Give precedence to top-left over bottom-left
if(col >= width-1-col) col = width-1-col, same = 0; // Give precedence to top-right over top-left
row -= same; // When the row and col doesn't change, this will reduce row by 1
if(row==col) return 1;
return 0;
}
int directions[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
void printSpiral(int arr[4][4], int height, int width){
int directionIdx=0, i=0;
int curRow=0, curCol=0;
for(i=0; i<height*width; i++){
printf("%d ",arr[curRow][curCol]);
if(shouldTurn(curRow, curCol, height, width)){
directionIdx = (directionIdx+1)%4;
}
curRow += directions[directionIdx][0];
curCol += directions[directionIdx][1];
}
printf("\n");
}
int main(){
int arr[4][4]= {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
printSpiral(arr, 4, 4);
printSpiral(arr, 3, 4);
}
Which outputs:
1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10
1 2 3 4 8 12 11 10 9 5 6 7
Here are the three interesting ways
Reading in spiral way can be treated like a snake moving towards boundary and turning on hitting the boundary or itself (I find it elegant and most efficient being a single loop of N iterations)
ar = [
[ 0, 1, 2, 3, 4],
[15, 16, 17, 18, 5],
[14, 23, 24, 19, 6],
[13, 22, 21, 20, 7],
[12, 11, 10, 9, 8]]
def print_spiral(ar):
"""
assuming a rect array
"""
rows, cols = len(ar), len(ar[0])
r, c = 0, -1 # start here
nextturn = stepsx = cols # move so many steps
stepsy = rows-1
inc_c, inc_r = 1, 0 # at each step move this much
turns = 0 # how many times our snake had turned
for i in range(rows*cols):
c += inc_c
r += inc_r
print ar[r][c],
if i == nextturn-1:
turns += 1
# at each turn reduce how many steps we go next
if turns%2==0:
nextturn += stepsx
stepsy -= 1
else:
nextturn += stepsy
stepsx -= 1
# change directions
inc_c, inc_r = -inc_r, inc_c
print_spiral(ar)
output:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
A recursive approach would be to print outer layer and call same function for inner rectangle e.g.
def print_spiral(ar, sr=0, sc=0, er=None, ec=None):
er = er or len(ar)-1
ec = ec or len(ar[0])-1
if sr > er or sc > ec:
print
return
# print the outer layer
top, bottom, left, right = [], [], [], []
for c in range(sc,ec+1):
top.append(ar[sr][c])
if sr != er:
bottom.append(ar[er][ec-(c-sc)])
for r in range(sr+1,er):
right.append(ar[r][ec])
if ec != sc:
left.append(ar[er-(r-sr)][sc])
print " ".join([str(a) for a in top + right + bottom + left]),
# peel next layer of onion
print_spiral(ar, sr+1, sc+1, er-1, ec-1)
Finally here is a small snippet to do it, not efficient but fun :), basically it prints top row, and rotates whole rectangle anti-clockwise and repeats
def print_spiral(ar):
if not ar: return
print " ".join(str(a) for a in ar[0]),
ar = zip(*[ reversed(row) for row in ar[1:]])
print_spiral(ar)
This program works for any n*n matrix..
public class circ {
public void get_circ_arr (int n,int [][] a)
{
int z=n;
{
for (int i=0;i<n;i++)
{
for (int l=z-1-i;l>=i;l--)
{
int k=i;
System.out.printf("%d",a[k][l]);
}
for (int j=i+1;j<=z-1-i;j++)
{
int k=i;
{
System.out.printf("%d",a[j][k]);
}
}
for (int j=i+1;j<=z-i-1;j++)
{
int k=z-1-i;
{
System.out.printf("%d",a[k][j]);
}
}
for (int j=z-2-i;j>=i+1;j--)
{
int k=z-i-1;
{
System.out.printf("%d",a[j][k]);
}
}
}
}
}
}
Hope it helps
I was obsessed with this problem when I was learning Ruby. This was the best I could do:
def spiral(matrix)
matrix.empty? ? [] : matrix.shift + spiral(matrix.transpose.reverse)
end
You can check out some of my other solutions by stepping back through the revisions in this gist. Also, if you follow the link back to whom I forked the gist from, you'll find some other clever solutions. Really interesting problem that can be solved in multiple elegant ways — especially in Ruby.
JavaScript solution:
var printSpiral = function (matrix) {
var i;
var top = 0;
var left = 0;
var bottom = matrix.length;
var right = matrix[0].length;
while (top < bottom && left < right) {
//print top
for (i = left; i < right; i += 1) {
console.log(matrix[top][i]);
}
top++;
//print right column
for (i = top; i < bottom; i += 1) {
console.log(matrix[i][right - 1]);
}
right--;
if (top < bottom) {
//print bottom
for (i = right - 1; i >= left; i -= 1) {
console.log(matrix[bottom - 1][i]);
}
bottom--;
}
if (left < right) {
//print left column
for (i = bottom - 1; i >= top; i -= 1) {
console.log(matrix[i][left]);
}
left++;
}
}
};
One solution involves directions right, left, up, down, and their corresponding limits (indices). Once the first row is printed, and direction changes (from right) to down, the row is discarded by incrementing the upper limit. Once the last column is printed, and direction changes to left, the column is discarded by decrementing the right hand limit... Details can be seen in the self-explanatory C code.
#include <stdio.h>
#define N_ROWS 5
#define N_COLS 3
void print_spiral(int a[N_ROWS][N_COLS])
{
enum {up, down, left, right} direction = right;
int up_limit = 0,
down_limit = N_ROWS - 1,
left_limit = 0,
right_limit = N_COLS - 1,
downcount = N_ROWS * N_COLS,
row = 0,
col = 0;
while(printf("%d ", a[row][col]) && --downcount)
if(direction == right)
{
if(++col > right_limit)
{
--col;
direction = down;
++up_limit;
++row;
}
}
else if(direction == down)
{
if(++row > down_limit)
{
--row;
direction = left;
--right_limit;
--col;
}
}
else if(direction == left)
{
if(--col < left_limit)
{
++col;
direction = up;
--down_limit;
--row;
}
}
else /* direction == up */
if(--row < up_limit)
{
++row;
direction = right;
++left_limit;
++col;
}
}
void main()
{
int a[N_ROWS][N_COLS] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
print_spiral(a);
}
Link for Testing and Download.
Given a matrix of chars, implement a method that prints all characters in the following order: first the outer circle,
then the next one and so on.
public static void printMatrixInSpiral(int[][] mat){
if(mat.length == 0|| mat[0].length == 0){
/* empty matrix */
return;
}
StringBuffer str = new StringBuffer();
int counter = mat.length * mat[0].length;
int startRow = 0;
int endRow = mat.length-1;
int startCol = 0;
int endCol = mat[0].length-1;
boolean moveCol = true;
boolean leftToRight = true;
boolean upDown = true;
while(counter>0){
if(moveCol){
if(leftToRight){
/* printing entire row left to right */
for(int i = startCol; i <= endCol ; i++){
str.append(mat[startRow][i]);
counter--;
}
leftToRight = false;
moveCol = false;
startRow++;
}
else{
/* printing entire row right to left */
for(int i = endCol ; i >= startCol ; i--){
str.append(mat[endRow][i]);
counter--;
}
leftToRight = true;
moveCol = false;
endRow--;
}
}
else
{
if(upDown){
/* printing column up down */
for(int i = startRow ; i <= endRow ; i++){
str.append(mat[i][endCol]);
counter--;
}
upDown = false;
moveCol = true;
endCol--;
}
else
{
/* printing entire col down up */
for(int i = endRow ; i >= startRow ; i--){
str.append(mat[i][startCol]);
counter--;
}
upDown = true;
moveCol = true;
startCol++;
}
}
}
System.out.println(str.toString());
}
Two dimensional N*N Matrix is Square matrix
Idea:
We have to traverse in four different directions to traverse like spiral.
We have to traverse inside matrix once one layer of spiral is over.
So total, we need 5 loops, 4 loops to traverse like spiral and 1 loop to traverse through the layers.
public void printSpiralForm(int[][] a, int length)
{
for( int i = 0 , j = length-1 ; i < j ; i++ , j-- )
{
for( int k = i ; k < j ; k++ )
{
System.out.print( a[i][k] + " " ) ;
}
for( int k = i ; k < j ; k++ )
{
System.out.print(a[k][j] + " ");
}
for( int k = j ; k > i ; k-- )
{
System.out.print(a[j][k] + " ") ;
}
for( int k = j ; k > i ; k-- )
{
System.out.print( a[k][i] + " " ) ;
}
}
if ( length % 2 == 1 )
{
System.out.println( a[ length/2 ][ length/2 ] ) ;
}
}
Just keep it simple -->
public class spiralMatrix {
public static void printMatrix(int[][] matrix, int rows, int col)
{
int rowStart=0;
int rowEnd=rows-1;
int colStart=0;
int colEnd=col-1;
while(colStart<=colEnd && rowStart<=rowEnd)
{
for(int i=colStart;i<colEnd;i++)
System.out.println(matrix[rowStart][i]);
for(int i=rowStart;i<rowEnd;i++)
System.out.println(matrix[i][colEnd]);
for(int i=colEnd;i>colStart;i--)
System.out.println(matrix[rowEnd][i]);
for(int i=rowEnd;i>rowStart;i--)
System.out.println(matrix[i][colStart]);
rowStart++;
colEnd--;
rowEnd--;
colStart++;
}
}
public static void main(String[] args){
int[][] array={{1,2,3,4},{5,6,7,8}};
printMatrix(array,2,4);
}
}
This is my implementation:
public static void printMatrix(int matrix[][], int M, int N){
int level = 0;
int min = (M < N) ? M:N;
System.out.println();
while(level <= min/2){
for(int j = level; j < N - level - 1; j++){
System.out.print(matrix[level][j] + "\t");
}
for(int i = level; i < M - level - 1; i++) {
System.out.print(matrix[i][N - level - 1] + "\t");
}
for(int j = N - level - 1; j > level; j--){
System.out.print(matrix[M - level - 1][j] + "\t");
}
for(int i = M - level - 1; i > level; i-- ){
System.out.print(matrix[i][level] + "\t");
}
level++;
}
}
Here is my solution. Please correct if I'm wrong.
class Spiral:
def spiralOrder(self, A):
result = []
c = []
c.append(A[0])
b = A[1:]
while len(b) > 0:
b = self.rotate(b)
c.append(b[0])
b = b[1:]
for item in c:
for fitem in item:
print fitem,
result.append(fitem)
return result
def rotate(self,a):
b = []
l = zip(*a)
for i in xrange(len(l)-1,-1,-1):
b.append(list(l[i]))
return b
if __name__ == '__main__':
a = [[1, 2, 3,3], [4, 5, 6,6], [7, 8, 9,10]]
s = Spiral()
s.spiralOrder(a)
Slash Top Row -> Transpose -> Flip -> Repeat.
void slashTransposeFlip(int[][] m){
if( m.length * m[0].length == 1){ //only one element left
System.out.print(m[0][0]);
}else{
//print the top row
for(int a:m[0]){System.out.print(a+" ");}
//slash the top row from the matrix.
int[][] n = Arrays.copyOfRange(m,1,m.length);
int[][] temp = n;
int rows = temp.length;
int columns = temp[0].length;
//invert rows and columns and create new array
n = new int[columns][rows];
//transpose
for(int x=0;x<rows;x++)
for(int y=0;y<columns;y++)
n[y][x] = temp[x][y];
//flipping time
for (int i = 0; i < n.length / 2; i++) {
int[] t = n[i];
n[i] = n[n.length - 1 - i];
n[n.length - 1 - i] = t;
}
//recursively call again the reduced matrix.
slashTransposeFlip(n);
}
}
Complexity: Single traverse O(n)
Please let me add my single loop answer with complexity O(n). I have observed that during left-right and right-left traverse of the matrix, there is an increase and decrease by one respectively in the row-major index. Similarly, for the top-bottom and bottom-top traverse there is increase and decrease by n_cols. Thus I made an algorithm for that. For example, given a (3x5) matrix with entries the row-major indexes the print output is: 1,2,3,4,5,10,15,14,13,12,11,6,7,8,9.
------->(+1)
^ 1 2 3 4 5 |
(+n_cols) | 6 7 8 9 10 | (-n_cols)
| 11 12 13 14 15
(-1)<-------
Code solution:
#include <iostream>
using namespace std;
int main() {
// your code goes here
bool leftToRight=true, topToBottom=false, rightToLeft=false, bottomToTop=false;
int idx=0;
int n_rows = 3;
int n_cols = 5;
int cnt_h = n_cols, cnt_v = n_rows, cnt=0;
int iter=1;
for (int i=0; i <= n_rows*n_cols + (n_rows - 1)*(n_cols - 1)/2; i++){
iter++;
if(leftToRight){
if(cnt >= cnt_h){
cnt_h--; cnt=0;
leftToRight = false; topToBottom = true;
//cout << "Iter: "<< iter << " break_leftToRight"<<endl;
}else{
cnt++;
idx++;
//cout << "Iter: "<< iter <<" idx: " << idx << " cnt: "<< cnt << " cnt_h: "<< cnt_h<< endl;
cout<< idx << endl;
}
}else if(topToBottom){
if(cnt >= cnt_v-1){
cnt_v--; cnt=0;
leftToRight = false; topToBottom = false; rightToLeft=true;
//cout << "Iter: "<< iter << " break_topToBottom"<<endl;
}else{
cnt++;
idx+=n_cols;
//cout << "Iter: "<< iter << " idx: " << idx << " cnt: "<< cnt << " cnt_v: "<< cnt_h<< endl;
cout << idx <<endl;
}
}else if(rightToLeft){
if(cnt >= cnt_h){
cnt_h--; cnt=0;
leftToRight = false; topToBottom = false; rightToLeft=false; bottomToTop=true;
//cout << "Iter: "<< iter << " break_rightToLeft"<<endl;
//cout<< idx << endl;
}else{
cnt++;
idx--;
//cout << "Iter: "<< iter << " idx: " << idx << " cnt: "<< cnt << " cnt_h: "<< cnt_h<< endl;
cout << idx <<endl;
}
}else if(bottomToTop){
if(cnt >= cnt_v-1){
cnt_v--; cnt=0;
leftToRight = true; topToBottom = false; rightToLeft=false; bottomToTop=false;
//cout << "Iter: "<< iter << " break_bottomToTop"<<endl;
}else{
cnt++;
idx-=n_cols;
//cout << "Iter: "<< iter << " idx: " << idx << " cnt: "<< cnt << " cnt_v: "<< cnt_h<< endl;
cout<< idx << endl;
}
}
//cout << i << endl;
}
return 0;
}
function spiral(a) {
var s = [];
while (a.length) {
// concat 1st row, push last cols, rotate 180 (reverse inner/outer)...
s = s.concat(a.shift());
a = a
.map(function(v) {
s.push(v.pop());
return v.reverse();
})
.reverse();
}
return s;
}
var arr = [
[1, 2, 3, 4],
[12, 13, 14, 5],
[11, 16, 15, 6],
[10, 9, 8, 7]
];
console.log(spiral(arr));// -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
arr = [
[0, 1, 2, 3, 4],
[15, 16, 17, 18, 5],
[14, 23, 24, 19, 6],
[13, 22, 21, 20, 7],
[12, 11, 10, 9, 8]
];
console.log(spiral(arr));// -> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]
For printing a 2-D matrix consider matrix as a composition of rectangles and/or line where smaller rectangle is fitted into larger one, take boundary of matrix which forms a rectangle to be printed, starting with up-left element each time in each layer; once done with this go inside for next layer of smaller rectangle, in case i don't have a rectangle then it should be line to be printed, a horizontal or vertical. I have pasted the code with an example matrix, HTH.
#include <stdio.h>
int a[2][4] = { 1, 2 ,3, 44,
8, 9 ,4, 55 };
void print(int, int, int, int);
int main() {
int row1, col1, row2, col2;
row1=0;
col1=0;
row2=1;
col2=3;
while(row2>=row1 && col2>=col1)
{
print(row1, col1, row2, col2);
row1++;
col1++;
row2--;
col2--;
}
return 0;
}
void print(int row1, int col1, int row2, int col2) {
int i=row1;
int j=col1;
/* This is when single horizontal line needs to be printed */
if( row1==row2 && col1!=col2) {
for(j=col1; j<=col2; j++)
printf("%d ", a[i][j]);
return;
}
/* This is when single vertical line needs to be printed */
if( col1==col2 && row1!=row2) {
for(i=row1; j<=row2; i++)
printf("%d ", a[i][j]);
return;
}
/* This is reached when there is a rectangle to be printed */
for(j=col1; j<=col2; j++)
printf("%d ", a[i][j]);
for(j=col2,i=row1+1; i<=row2; i++)
printf("%d ", a[i][j]);
for(i=row2,j=col2-1; j>=col1; j--)
printf("%d ", a[i][j]);
for(j=col1,i=row2-1; i>row1; i--)
printf("%d ", a[i][j]);
}
Here is my implementation in Java:
public class SpiralPrint {
static void spiral(int a[][],int x,int y){
//If the x and y co-ordinate collide, break off from the function
if(x==y)
return;
int i;
//Top-left to top-right
for(i=x;i<y;i++)
System.out.println(a[x][i]);
//Top-right to bottom-right
for(i=x+1;i<y;i++)
System.out.println(a[i][y-1]);
//Bottom-right to bottom-left
for(i=y-2;i>=x;i--)
System.out.println(a[y-1][i]);
//Bottom left to top-left
for(i=y-2;i>x;i--)
System.out.println(a[i][x]);
//Recursively call spiral
spiral(a,x+1,y-1);
}
public static void main(String[] args) {
int a[][]={{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};
spiral(a,0,4);
/*Might be implemented without the 0 on an afterthought, all arrays will start at 0 anyways. The second parameter will be the dimension of the array*/
}
}
//shivi..coding is adictive!!
#include<shiviheaders.h>
#define R 3
#define C 6
using namespace std;
void PrintSpiral(int er,int ec,int arr[R][C])
{
int sr=0,sc=0,i=0;
while(sr<=er && sc<=ec)
{
for(int i=sc;i<=ec;++i)
cout<<arr[sr][i]<<" ";
++sr;
for(int i=sr;i<=er;++i)
cout<<arr[i][ec]<<" ";
ec--;
if(sr<=er)
{
for(int i=ec;i>=sc;--i)
cout<<arr[er][i]<<" ";
er--;
}
if(sc<=ec)
{
for(int i=er;i>=sr;--i)
cout<<arr[i][sc]<<" ";
++sc;
}
}
}
int main()
{
int a[R][C] = { {1, 2, 3, 4, 5, 6},
{7, 8, 9, 10, 11, 12},
{13, 14, 15, 16, 17, 18}
};
PrintSpiral(R-1, C-1, a);
}
int N = Integer.parseInt(args[0]);
// create N-by-N array of integers 1 through N
int[][] a = new int[N][N];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
a[i][j] = 1 + N*i + j;
// spiral
for (int i = N-1, j = 0; i > 0; i--, j++) {
for (int k = j; k < i; k++) System.out.println(a[j][k]);
for (int k = j; k < i; k++) System.out.println(a[k][i]);
for (int k = i; k > j; k--) System.out.println(a[i][k]);
for (int k = i; k > j; k--) System.out.println(a[k][j]);
}
// special case for middle element if N is odd
if (N % 2 == 1) System.out.println(a[(N-1)/2][(N-1)/2]);
}
}
Java code if anybody is interested.
Input:
4
1 2 3 4
5 6 7 8
9 1 2 3
4 5 6 7
Output: 1 2 3 4 8 3 7 6 5 4 9 5 6 7 2 1
public class ArraySpiralPrinter {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(); //marrix size
//read array
int[][] ar = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
ar[i][j] = sc.nextInt();
}
}
printTopRight(0, 0, n - 1, n - 1, ar);
}
//prints top and right layers.
//(x1,y1) to (x1, y2) - top layer & (x1,y2) to (x2, y2)
private static void printTopRight(int x1, int y1, int x2, int y2, int[][] ar) {
//print row values - top
for (int y = y1; y <= y2; y++) {
System.out.printf("%d ", ar[x1][y]);
}
//print column value - right
for (int x = x1 + 1; x <= x2; x++) {
System.out.printf("%d ", ar[x][y2]);
}
//are there any remaining layers
if (x2 - x1 > 0) {
//call printBottemLeft
printBottomLeft(x1 + 1, y1, x2, y2 - 1, ar);
}
}
//prints bottom and left layers in reverse order
//(x2,y2) to (x2, y1) - bottom layer & (x2,y1) to (x1, y1)
private static void printBottomLeft(int x1, int y1, int x2, int y2, int[][] ar) {
//print row values in reverse order - bottom
for (int y = y2; y >= y1; y--) {
System.out.printf("%d ", ar[x2][y]);
}
//print column value in reverse order - left
for (int x = x2-1; x >= x1; x--) {
System.out.printf("%d ", ar[x][y1]);
}
//are there any remaining layers
if (x2 - x1 > 0) {
printTopRight(x1, y1 + 1, x2 - 1, y2, ar);
}
}
}
This is a recursive version in C that I could think of:-
void printspiral (int[][100],int, int, int, int);
int main()
{
int r,c, i, j;
printf ("Enter the dimensions of the matrix");
scanf("%d %d", &r, &c);
int arr[r][100];
int min = (r<c?r:c);
if (min%2 != 0) min = min/2 +1;
for (i = 0;i<r; i++)
for (j = 0; j<c; j++)
scanf ("%d",&arr[i][j]);
printspiral(arr,0,r,c,min );
}
void printspiral (int arr[][100], int i, int j, int k, int min)
{
int a;
for (a = i; a<k;a++)
printf("%d\n", arr[i][a]);
for (a=i+1;a<j;a++)
printf ("%d\n", arr[a][k-1]);
for (a=k-2; a>i-1;a--)
printf("%d\n", arr[j-1][a]);
for (a=j-2; a>i; a--)
printf("%d\n", arr[a][i]);
if (i < min)
printspiral(arr,i+1, j-1,k-1, min);
}
http://www.technicalinterviewquestions.net/2009/03/print-2d-array-matrix-spiral-order.html
here is the best explanation for the above answer :) along with diagram :)
public static void printSpiral1(int array[][],int row,int col){
int rowStart=0,colStart=0,rowEnd=row-1,colEnd=col-1;
int i;
while(rowStart<=rowEnd && colStart<= colEnd){
for(i=colStart;i<=colEnd;i++)
System.out.print(" "+array[rowStart][i]);
for(i=rowStart+1;i<=rowEnd;i++)
System.out.print(" "+array[i][colEnd]);
for(i=colEnd-1;i>=colStart;i--)
System.out.print(" "+array[rowEnd][i]);
for(i=rowEnd-1;i>=rowStart+1;i--)
System.out.print(" "+array[i][colStart]);
rowStart++;
colStart++;
rowEnd--;
colEnd--;
}
}
public class SpiralPrint{
//print the elements of matrix in the spiral order.
//my idea is to use recursive, for each outer loop
public static void printSpiral(int[][] mat, int layer){
int up = layer;
int buttom = mat.length - layer - 1;
int left = layer;
int right = mat[0].length - layer - 1;
if(up > buttom+1 || left > right + 1)
return; // termination condition
//traverse the other frame,
//print up
for(int i = left; i <= right; i ++){
System.out.print( mat[up][i]+ " " );
}
//print right
for(int i = up + 1; i <=buttom; i ++){
System.out.print(mat[i][right] + " ");
}
//print buttom
for(int i = right - 1; i >= left; i --){
System.out.print(mat[buttom][i] + " ");
}
//print left
for(int i = buttom - 1; i > up; i --){
System.out.print(mat[i][left] + " ");
}
//recursive call for the next level
printSpiral(mat, layer + 1);
}
public static void main(String[] args){
int[][] mat = {{1,2,3,4}, {5,6,7,8}, {9,10,11,12}, {13,14,15,16}};
int[][] mat2 = {{1,2,3}, {4,5,6}, {7,8,9}, {10,11,12}};
SpiralPrint.printSpiral(mat2,0);
return;
}
}
Here is my solution in C#:
public static void PrintSpiral(int[][] matrix, int n)
{
if (matrix == null)
{
return;
}
for (int layer = 0; layer < Math.Ceiling(n / 2.0); layer++)
{
var start = layer;
var end = n - layer - 1;
var offset = end - 1;
Console.Write("Layer " + layer + ": ");
// Center case
if (start == end)
{
Console.Write(matrix[start][start]);
}
// Top
for (int i = start; i <= offset; i++)
{
Console.Write(matrix[start][i] + " ");
}
// Right
for (int i = start; i <= offset; i++)
{
Console.Write(matrix[i][end] + " ");
}
// Bottom
for (int i = end; i > start; i--)
{
Console.Write(matrix[end][i] + " ");
}
// Left
for (int i = end; i > start; i--)
{
Console.Write(matrix[i][start] + " ");
}
Console.WriteLine();
}
}
Here's my approach using an Iterator . Note this solves almost the same problem..
Complete code here : https://github.com/rdsr/algorithms/blob/master/src/jvm/misc/FillMatrix.java
import java.util.Iterator;
class Pair {
final int i;
final int j;
Pair(int i, int j) {
this.i = i;
this.j = j;
}
#Override
public String toString() {
return "Pair [i=" + i + ", j=" + j + "]";
}
}
enum Direction {
N, E, S, W;
}
class SpiralIterator implements Iterator<Pair> {
private final int r, c;
int ri, ci;
int cnt;
Direction d; // current direction
int level; // spiral level;
public SpiralIterator(int r, int c) {
this.r = r;
this.c = c;
d = Direction.E;
level = 1;
}
#Override
public boolean hasNext() {
return cnt < r * c;
}
#Override
public Pair next() {
final Pair p = new Pair(ri, ci);
switch (d) {
case E:
if (ci == c - level) {
ri += 1;
d = changeDirection(d);
} else {
ci += 1;
}
break;
case S:
if (ri == r - level) {
ci -= 1;
d = changeDirection(d);
} else {
ri += 1;
}
break;
case W:
if (ci == level - 1) {
ri -= 1;
d = changeDirection(d);
} else {
ci -= 1;
}
break;
case N:
if (ri == level) {
ci += 1;
level += 1;
d = changeDirection(d);
} else {
ri -= 1;
}
break;
}
cnt += 1;
return p;
}
private static Direction changeDirection(Direction d) {
switch (d) {
case E:
return Direction.S;
case S:
return Direction.W;
case W:
return Direction.N;
case N:
return Direction.E;
default:
throw new IllegalStateException();
}
}
#Override
public void remove() {
throw new UnsupportedOperationException();
}
}
public class FillMatrix {
static int[][] fill(int r, int c) {
final int[][] m = new int[r][c];
int i = 1;
final Iterator<Pair> iter = new SpiralIterator(r, c);
while (iter.hasNext()) {
final Pair p = iter.next();
m[p.i][p.j] = i;
i += 1;
}
return m;
}
public static void main(String[] args) {
final int r = 19, c = 19;
final int[][] m = FillMatrix.fill(r, c);
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
System.out.print(m[i][j] + " ");
}
System.out.println();
}
}
}
Complete pure C program for any 2D array matrix with given row x column.
#include <stdio.h>
void printspiral(int *p,int r, int c) {
int i=0,j=0,m=1,n=0;
static int firstrun=1,gCol;
if (!p||r<=0||c<=0)
return ;
if(firstrun) {
gCol=c;
firstrun=0;
}
for(i=0,j=0;(0<=i && i<c)&&(0<=j && j<r);i+=m,j+=n) {
printf(" %d",p[i+j*gCol]);
if (i==0 && j==1 && (i+1)!=c) break;
else if (i+1==c && !j) {m=0;n=1;}
else if (i+1==c && j+1==r && j) {n=0;m=-1;}
else if (i==0 && j+1==r && j) {m=0;n=-1;}
}
printspiral(&p[i+j*gCol+1],r-2,c-2);
firstrun=1;
printf("\n");
}
int main() {
int a[3][3]={{0,1,2},{3,4,5},{6,7,8}};
int b[3][4]={{0,1,2,3},{4,5,6,7},{8,9,10,11}};
int c[4][3]={{0,1,2},{3,4,5},{6,7,8},{9,10,11}};
int d[3][1]={{0},{1},{2}};
int e[1][3]={{0,1,2}};
int f[1][1]={{0}};
int g[5][5]={{0,1,2,3,4},{5,6,7,8,9},{10,11,12,13,14},{15,16,17,18,19},{20,21,22,23,24}};
printspiral(a,3,3);
printspiral(b,3,4);
printspiral(c,4,3);
printspiral(d,3,1);
printspiral(e,1,3);
printspiral(f,1,1);
printspiral(g,5,5);
return 0;
}
This question is related to this one: Matrix arrangement issues in php
The answers presented seem to work but are complicated to understand. A very simple way to solve this is divide and conquer i.e., after reading the edge, remove it and the next read will be much simpler. Check out a complete solution in PHP below:
#The source number matrix
$source[0] = array(1, 2, 3, 4);
$source[1] = array(5, 6, 7, 8);
$source[2] = array(9, 10, 11, 12);
$source[3] = array(13, 14, 15, 16);
$source[4] = array(17, 18, 19, 20);
#Get the spiralled numbers
$final_spiral_list = get_spiral_form($source);
print_r($final_spiral_list);
function get_spiral_form($matrix)
{
#Array to hold the final number list
$spiralList = array();
$result = $matrix;
while(count($result) > 0)
{
$resultsFromRead = get_next_number_circle($result, $spiralList);
$result = $resultsFromRead['new_source'];
$spiralList = $resultsFromRead['read_list'];
}
return $spiralList;
}
function get_next_number_circle($matrix, $read)
{
$unreadMatrix = $matrix;
$rowNumber = count($matrix);
$colNumber = count($matrix[0]);
#Check if the array has one row or column
if($rowNumber == 1) $read = array_merge($read, $matrix[0]);
if($colNumber == 1) for($i=0; $i<$rowNumber; $i++) array_push($read, $matrix[$i][0]);
#Check if array has 2 rows or columns
if($rowNumber == 2 || ($rowNumber == 2 && $colNumber == 2))
{
$read = array_merge($read, $matrix[0], array_reverse($matrix[1]));
}
if($colNumber == 2 && $rowNumber != 2)
{
#First read left to right for the first row
$read = array_merge($read, $matrix[0]);
#Then read down on right column
for($i=1; $i<$rowNumber; $i++) array_push($read, $matrix[$i][1]);
#..and up on left column
for($i=($rowNumber-1); $i>0; $i--) array_push($read, $matrix[$i][0]);
}
#If more than 2 rows or columns, pick up all the edge values by spiraling around the matrix
if($rowNumber > 2 && $colNumber > 2)
{
#Move left to right
for($i=0; $i<$colNumber; $i++) array_push($read, $matrix[0][$i]);
#Move top to bottom
for($i=1; $i<$rowNumber; $i++) array_push($read, $matrix[$i][$colNumber-1]);
#Move right to left
for($i=($colNumber-2); $i>-1; $i--) array_push($read, $matrix[$rowNumber-1][$i]);
#Move bottom to top
for($i=($rowNumber-2); $i>0; $i--) array_push($read, $matrix[$i][0]);
}
#Now remove these edge read values to create a new reduced matrix for the next read
$unreadMatrix = remove_top_row($unreadMatrix);
$unreadMatrix = remove_right_column($unreadMatrix);
$unreadMatrix = remove_bottom_row($unreadMatrix);
$unreadMatrix = remove_left_column($unreadMatrix);
return array('new_source'=>$unreadMatrix, 'read_list'=>$read);
}
function remove_top_row($matrix)
{
$removedRow = array_shift($matrix);
return $matrix;
}
function remove_right_column($matrix)
{
$neededCols = count($matrix[0]) - 1;
$finalMatrix = array();
for($i=0; $i<count($matrix); $i++) $finalMatrix[$i] = array_slice($matrix[$i], 0, $neededCols);
return $finalMatrix;
}
function remove_bottom_row($matrix)
{
unset($matrix[count($matrix)-1]);
return $matrix;
}
function remove_left_column($matrix)
{
$neededCols = count($matrix[0]) - 1;
$finalMatrix = array();
for($i=0; $i<count($matrix); $i++) $finalMatrix[$i] = array_slice($matrix[$i], 1, $neededCols);
return $finalMatrix;
}
// Program to print a matrix in spiral order
#include <stdio.h>
int main(void) {
// your code goes here
int m,n,i,j,k=1,c1,c2,r1,r2;;
scanf("%d %d",&m,&n);
int a[m][n];
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
}
}
r1=0;
r2=m-1;
c1=0;
c2=n-1;
while(k<=m*n)
{
for(i=c1;i<=c2;i++)
{
k++;
printf("%d ",a[r1][i]);
}
for(j=r1+1;j<=r2;j++)
{
k++;
printf("%d ",a[j][c2]);
}
for(i=c2-1;i>=c1;i--)
{
k++;
printf("%d ",a[r2][i]);
}
for(j=r2-1;j>=r1+1;j--)
{
k++;
printf("%d ",a[j][c1]);
}
c1++;
c2--;
r1++;
r2--;
}
return 0;
}

Problem: Shortest path in a grid between multiple point with a constraint

Problem description:
I'm trying to solve a problem on the internet and I wasn't able to pass all testcases, well, because my logic is flawed and incorrect. The flaw: I assumed starting to the closest 'F' point will get me to the shortest paths always, at all cases.
Thinks I thought of:
Turning this into a graph problem and solve it based on it. > don't think this would work because of the constraint?
Try to obtain all possible solution combinations > does not scale, if !8 combination exist.
#include <iostream>
#include <utility>
#include <string>
#include <vector>
#include <queue>
using namespace std;
#define N 4
#define M 4
int SearchingChallenge(string strArr[], int arrLength) {
int n = arrLength, m = n, steps = 0, food = 0;
// initial position of charlie
int init_j = 0;
int init_i = 0;
queue<pair<int,int>> q;
// directions
vector<int> offsets = {0,-1,0,1,0};
vector<pair<int,int>> food_nodes;
//store visited nodes, no need for extra work to be done.
int visited_nodes[4][4] = {{0}};
// get number of food pieces
for(int i = 0; i < m; i++){
for(int j = 0; j < n ; j++){
if(strArr[i][j] == 'F')
{
food++;
}
if(strArr[i][j] == 'C')
{
strArr[i][j] = 'O';
food_nodes.push_back({i,j});
}
}
}
while(food_nodes.size()>0){
food_nodes.erase(food_nodes.begin());
int break_flag=0;
q.push(food_nodes[0]);
while(!q.empty()){
int size = q.size();
while(size-->0){
pair<int,int> p = q.front();
q.pop();
for(int k = 0; k < 4; k++){
int ii = p.first + offsets[k], jj = p.second + offsets[k+1];
/* if(ii == 0 && jj == 3)
printf("HI"); */
if(jj >= 0 && jj < 4 && ii < 4 && ii >=0){
if(strArr[ii][jj] == 'F'){
strArr[ii][jj] = 'O';
while(!q.empty())
q.pop();
break_flag=1;
food--;
food_nodes.push_back({ii,jj});
break;
}
if(strArr[ii][jj] == 'O')
q.push({ii,jj});
if(strArr[ii][jj] == 'H' && food == 0)
return ++steps;
}
}
if(break_flag==1)
break;
}
steps++;
if(break_flag==1)
break;
}
}
return 0;
}
int main(void) {
// keep this function call here
/* Note: In C++ you first have to initialize an array and set
it equal to the stdin to test your code with arrays. */
//passing testcase
//string A[4] = {"OOOO", "OOFF", "OCHO", "OFOO"};
//failing testcase
string A[4] = {"FOOF", "OCOO", "OOOH", "FOOO"}
int arrLength = sizeof(A) / sizeof(*A);
cout << SearchingChallenge(A, arrLength);
return 0;
}
Your help is appreciated.
I have wrote the javascript solution for the mentioned problem..
function SearchingChallenge(strArr) {
// create coordinate array
const matrix = [
[0, 0], [0, 1], [0, 2], [0, 3],
[1, 0], [1, 1], [1, 2], [1, 3],
[2, 0], [2, 1], [2, 2], [2, 3],
[3, 0], [3, 1], [3, 2], [3, 3]
]
// flatten the strArr
const flattenArray = flatten(strArr)
// segreagate and map flattenArray with matrix to get coordinate of food,charlie and home
const segregatedCoordinates = flattenArray.reduce((obj, char, index) => {
if (char === 'F') obj['food'].push(matrix[index])
else if (char === 'C') obj['dog'] = matrix[index]
else if (char === 'H') obj['home'] = matrix[index]
return obj
}, { "food": [], dog: null, home: null })
// construct possible routes by permutating food coordinates
let possibleRoutes = permuate(segregatedCoordinates['food'])
// push dog and home in possibleRoutes at start and end positions
possibleRoutes = possibleRoutes.map((route) => {
return [segregatedCoordinates['dog'], ...route, segregatedCoordinates['home']]
})
// Calculate distances from every possible route
const distances = possibleRoutes.reduce((distances, route) => {
let moveLength = 0
for (let i = 0; i < route.length - 1; i++) {
let current = route[i], next = route[i + 1]
let xCoordinatePath = current[0] > next[0] ? (current[0] - next[0]) : (next[0] - current[0])
let yCoordinatePath = current[1] > next[1] ? (current[1] - next[1]) : (next[1] - current[1])
moveLength += xCoordinatePath + yCoordinatePath
}
distances.push(moveLength)
return distances
}, [])
return Math.min(...distances);
}
function permuate(arr) {
if (arr.length <= 2) return (arr.length === 2 ? [arr, [arr[1], arr[0]]] : arr)
return arr.reduce((res, ele, index) => {
res = [...res, ...permuate([...arr.slice(0, index), ...arr.slice(index + 1)]).map(val => [ele, ...val])]
return res
}, [])
}
function flatten(inputtedArr) {
return inputtedArr.reduce((arr, row) => {
arr = [...arr, ...row]
return arr
}, [])
}
console.log(SearchingChallenge(['FOOF', 'OCOO', 'OOOH', 'FOOO']));
You can write a DP solution where you have a 4x4x8 grid. The first two axis represent the x and y coordinate. The third one represent the binary encoding of which food item you picked already.
Each cell in the grid stores the best number of moves to get at this cell having eaten the specified foods. So for example, grid[2][2][2] is the cost of getting to cell (2,2) after having eaten the second piece of food only.
Then you set the value of the start cell, at third index 0 to 0, all the other cells to -1. You keep a list of the cells to propagate (sorted by least cost), and you add the start cell to it.
Then you repeatedly take the next cell to propagate, remove it and push the neighboring cell with cost +1 and updated food consume. Once you reach the destination cell with all food consumed, you're done.
That should take no more than 4x4x8 updates, with about the same order of priority queue insertion. O(n log(n)) where n is xy2^f. As long as you have few food items this will be almost instant.
C++ solution
I used both dfs and bfs for this problem
TIME COMPLEXITY - (4^(N×M))+NO_OF_FOODS×N×M
#include <bits/stdc++.h>
using namespace std;
//It is a dfs function it will find and store all the possible steps to eat all food in toHome map
void distAfterEatingAllFood(vector<vector<char>> &m, int countOfFood, int i, int j, int steps, map<pair<int,int>,int>&toHome){
if(i<0 || j<0 || i>=4 || j>=4 || m[i][j]=='*') return;
if(m[i][j]=='F') countOfFood--;
if(countOfFood==0){
if(!toHome.count({i,j}))
toHome[{i,j}] = steps;
else if(toHome[{i,j}]>steps)
toHome[{i,j}] = steps;
return;
}
char temp = m[i][j];
m[i][j] = '*';
distAfterEatingAllFood(m, countOfFood, i+1, j, steps+1, toHome);
distAfterEatingAllFood(m, countOfFood, i-1, j, steps+1, toHome);
distAfterEatingAllFood(m, countOfFood, i, j+1, steps+1, toHome);
distAfterEatingAllFood(m, countOfFood, i, j-1, steps+1, toHome);
m[i][j] = temp;
return;
}
//It is a bfs function it will iterate over the toHome map and find the shortest distance between the last food position to home
int lastFoodToHome(vector<vector<char>> &m, int i, int j, int steps){
queue<pair<pair<int, int>,int>>q;
vector<vector<int>> vis(4, vector<int>(4, 0));
q.push({{i, j}, steps});
vis[i][j] = 1;
int dirX[] = {0, 1, 0, -1};
int dirY[] = {1, 0, -1, 0};
while (!q.empty())
{
int x = q.front().first.first;
int y = q.front().first.second;
int steps = q.front().second;
q.pop();
if (m[x][y] == 'H')
return steps;
for (int k = 0; k < 4; k++)
{
int ni = x + dirX[k];
int nj = y + dirY[k];
if (ni >= 0 && nj >= 0 && ni < 4 && nj < 4 && !vis[ni][nj])
{
if(m[ni][nj] == 'H') return steps + 1;
q.push({{ni, nj}, steps + 1});
vis[i][j] = 1;
}
}
}
return INT_MAX;
}
int main()
{
vector<vector<char>> m(4, vector<char>(4));
int countOfFood = 0, x, y;
for (int i = 0; i < 4; i++){
for (int j = 0; j < 4; j++){
cin >> m[i][j];
if (m[i][j] == 'C'){
x = i;
y = j;
}
if (m[i][j] == 'F')
countOfFood++;
}
}
map<pair<int,int>,int>toHome;
distAfterEatingAllFood(m, countOfFood, x, y, 0, toHome);
int ans = INT_MAX;
for(auto &i:toHome){
ans = min(ans, lastFoodToHome(m, i.first.first, i.first.second, i.second));
}
cout<<ans;
return 0;
}

Is there a way to generate a sequence that includes squares which sum up to an integer square?

decompose(11) must return [1,2,4,10].
Note that there are actually two ways to decompose 11², 11² = 121 = 1 + 4 + 16 + 100 = 1² + 2² + 4² + 10².
For decompose(50) don't return [1, 1, 4, 9, 49] but [1, 3, 5, 8, 49] since [1, 1, 4, 9, 49] doesn't form a strictly increasing sequence.
I created a function but in only some cases provides a strictly increasing sequence all of my solutions add up to the correct number, what changes do i have to make to enable the return of a strictly increasing sequence?
vector<ll> Decomp::decompose(ll n){
ll square = n * n, j = 1, nextterm = n - 1, remainder, sum = 0;
float root;
vector<ll> sequence;
do
{
sequence.push_back(nextterm);
sum = sum + (nextterm * nextterm);
remainder = square - sum;
root = sqrt(remainder - 1);
if (root - (int)root > 0)
{
root = (int)root;
}
j = 1;
nextterm = (int)root;
if (remainder == 1)
{
sequence.push_back(1);
}
} while (root > 0);
reverse(sequence.begin(),sequence.end());
for (int i=0; i < sequence.size(); i++)
{
cout << sequence[i] << endl;
}
}
Here is a simple recursive approach, basically exploring all the possibilities.
It stops once a solution is found.
Output:
11 : 1 2 4 10
50 : 1 3 5 8 49
And the code:
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
bool decompose_dp (long long int sum, long long int k, std::vector<long long int> &seq) {
while (k > 0) {
long long int sump = sum - k*k;
if (sump == 0) {
seq.push_back(k);
return true;
}
if (sump < 0) {
k--;
continue;
}
long long int kp = k-1;
while (kp > 0) {
if (decompose_dp(sump, kp, seq)) {
seq.push_back(k);
return true;
}
kp --;
}
k--;
}
return false;
}
std::vector<long long int> decompose(long long int n){
long long int square = n * n, j = 1, nextterm = n - 1, remainder, sum = 0;
float root;
std::vector<long long int> sequence;
auto check = decompose_dp (n*n, n-1, sequence);
return sequence;
}
void pr (long long int n, const std::vector<long long int> &vec) {
std::cout << n << " : ";
for (auto k: vec) {
std::cout << k << " ";
}
std::cout << "\n";
}
int main() {
long long int n = 11;
auto sequence = decompose (n);
pr (n, sequence);
n = 50;
sequence = decompose (n);
pr (n, sequence);
}
Here's BFS, DFS and brute force in Python. BFS seems slow for input 50. Brute force yielded 91020 different combinations for input 50.
from collections import deque
def bfs(n):
target = n * n
queue = deque([(target, [], 1)])
while queue:
t, seq, i = queue.popleft()
if t == 0:
return seq
if (t == target and i*i < t) or (t != target and i*i <= t):
queue.append((t - i*i, seq[:] + [i], i + 1))
queue.append((t, seq, i + 1))
def dfs(n):
target = n * n
stack = [(target, [], 1)]
while stack:
t, seq, i = stack.pop()
if t == 0:
return seq
if (t == target and i*i < t) or (t != target and i*i <= t):
stack.append((t - i*i, seq[:] + [i], i + 1))
stack.append((t, seq, i + 1))
def brute(n):
target = n * n
stack = [(target, [], 1)]
result = []
while stack:
t, seq, i = stack.pop()
if t == 0:
result.append(seq)
if (t == target and i*i < t) or (t != target and i*i <= t):
stack.append((t - i*i, seq[:] + [i], i + 1))
stack.append((t, seq, i + 1))
return result
print bfs(50) # [1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20]
print dfs(50) # [30, 40]
#print brute(50)

Find the minimum sum of 'P' elements in an array of N elements such that no more than 'k' consecutive elements are selected together

Suppose the array is 1 2 3 4 5
Here N = 5 and we have to select 3 elements and we cannot select more than 2 consecutive elements, so P = 3 and k = 2. So the output here will be 1 + 2 + 4 = 7.
I came up with a recursive solution, but it has an exponential time complexity. Here is the code.
#include<iostream>
using namespace std;
void mincost_hoarding (int *arr, int max_size, int P, int k, int iter, int& min_val, int sum_sofar, int orig_k)
{
if (P == 0)
{
if (sum_sofar < min_val)
min_val = sum_sofar;
return;
}
if (iter == max_size)
return;
if (k!=0)
{
mincost_hoarding (arr, max_size, P - 1, k - 1, iter + 1, min_val, sum_sofar + arr[iter], orig_k);
mincost_hoarding (arr, max_size, P, orig_k, iter + 1, min_val, sum_sofar, orig_k);
}
else
{
mincost_hoarding (arr, max_size, P, orig_k, iter + 1, min_val, sum_sofar, orig_k);
}
}
int main()
{
int a[] = {10, 5, 13, 8, 2, 11, 6, 4};
int N = sizeof(a)/sizeof(a[0]);
int P = 2;
int k = 1;
int min_val = INT_MAX;
mincost_hoarding (a, N, P, k, 0, min_val, 0, k);
cout<<min_val;
}
Also, if supposedly P elements cannot be selected following the constraint, then we return INT_MAX.
I was asked this question in an interview. After proposing this solution, the interviewer was expecting something faster. Maybe, a DP approach towards the problem. Can someone propose a DP algorithm if there exists one, or a faster algorithm.
I have tried various tests cases and got correct answers. If you find some test cases that are giving incorrect response, please point that out too.
Below is a Java Dynamic Programming algorithm.
(the C++ version should look very similar)
It basically works as follows:
Have a 3D array of [pos][consecutive length][length]
Here length index = actual length - 1), so [0] would be length 1, similarly for consecutive length. This was done since there's no point to having length 0 anywhere.
At every position:
If at length 0 and consecutive length 0, just use the value at pos.
Otherwise, if consecutive length 0, look around for the minimum in all the previous positions (except pos - 1) with length - 1 and use that plus the value at pos.
For everything else, if pos > 0 && consecutive length > 0 && length > 0,
use [pos-1][consecutive length-1][length-1] plus the value at pos.
If one of those are 0, initialize it to an invalid value.
Initially it felt like one only needs 2 dimensions for this problem, however, as soon as I tried to figure it out, I realized I needed a 3rd.
Code:
int[] arr = {1, 2, 3, 4, 5};
int k = 2, P = 3;
int[][][] A = new int[arr.length][P][k];
for (int pos = 0; pos < arr.length; pos++)
for (int len = 0; len < P; len++)
{
int min = 1000000;
if (len > 0)
{
for (int pos2 = 0; pos2 < pos-1; pos2++)
for (int con = 0; con < k; con++)
min = Math.min(min, A[pos2][len-1][con]);
A[pos][len][0] = min + arr[pos];
}
else
A[pos][0][0] = arr[pos];
for (int con = 1; con < k; con++)
if (pos > 0 && len > 0)
A[pos][len][con] = A[pos-1][len-1][con-1] + arr[pos];
else
A[pos][len][con] = 1000000;
}
// Determine the minimum sum
int min = 100000;
for (int pos = 0; pos < arr.length; pos++)
for (int con = 0; con < k; con++)
min = Math.min(A[pos][P-1][con], min);
System.out.println(min);
Here we get 7 as output, as expected.
Running time: O(N2k + NPk)

Merge Sort Algorithm Assistance

I'm trying to implement the merge-sort algorithm. I started with pseudocode that was available in an algorithms book. The pseudocode indicates the first position in the array as 1 and not 0. I am having a very difficult time trying to implement the code.
Here is what I have. I've tried stepping through the recursion by printing out the results at each step but it is very convoluted at this point.
#include <iostream>
#include <deque>
using size_type = std::deque<int>::size_type;
void print(std::deque<int> &v)
{
for(const auto &ref:v)
std::cout << ref << " ";
std::cout << std::endl;
}
void merge(std::deque<int> &vec, size_type p, size_type q, size_type r)
{
int n_1 = q - p;
int n_2 = r - q;
std::deque<int> left, right;
for(auto i = 0; i != n_1; i++)
left.push_back(vec[p + i]);
for(auto j = 0; j != n_2; j++)
right.push_back(vec[q + j]);
int i = 0, j = 0;
std::cout << "left = ";
print(left);
std::cout << "right = ";
print(right);
for(auto k = p; k != r; k++) {
if((i != n_1 && j != n_2) && left[i] <= right[j]) {
vec[k] = left[i];
i++;
}
else if(j != n_2){
vec[k] = right[j];
j++;
}
}
}
void merge_sort(std::deque<int> &A, size_type p, size_type r)
{
int q;
if(p < r - 1) {
q = (p + r)/2;
merge_sort(A, p, q);
merge_sort(A, q + 1, r);
merge(A, p, q, r);
}
}
int main()
{
std::deque<int> small_vec = {1, 6, 2, 10, 5, 2, 12, 6};
std::deque<int> samp_vec = {2, 9, 482, 72, 42, 3, 4, 9, 8, 73, 8, 0, 98, 72, 473, 72, 3, 4, 9, 7, 6, 5, 6953, 583};
print(small_vec);
merge_sort(small_vec, 0, small_vec.size());
print(small_vec);
return 0;
}
I get the following output when I run the program:
left = 1
right = 6
left = 1 6
right = 2 10
left = 2
right = 12 6
left = 1 2 6 10
right = 5 2 12 6
1 2 5 2 6 10 12 6
The error is here: (i != n_1 && j != n_2) && left[i] <= right[j]) when i != n_1 evaluates to false vec[k] = right[j]; is executed - correct.
But if i != n_1 evaluates to true and j != n_2 to false i.e. j = n_2 your program trys to do this vec[k] = right[j]; again i.e. accesing over the bounds of your deque.
Rewrite your for loop as follows:
if (i<n_1 && (j>=n_2 || left[i] <= right[j])
This loop works only due to C++'s short circuiting of the conditions i.e. when j>=n_2 evaluates to true, left[i] <= right[j] is never checked again and you don't access the deque over bounds.
left[i] <= right[j] is being checked only if both i<n_1 is true and j>=n_2 false otherwise the 2nd branch is executed.
After spending a lot of time and getting some valuable help on another post was able to get the algorithm to run correctly.
CORRECT CODE:
#include <iostream>
#include <deque>
using size_type = std::deque<int>::size_type;
void print(std::deque<int> &v)
{
for(const auto &ref:v)
std::cout << ref << " ";
std::cout << std::endl;
}
void print(int arr[], int size)
{
for(int i = 0; i != size; i++)
std::cout << arr[i] << " ";
std::cout << std::endl;
}
void merge(std::deque<int> &vec, size_type p, size_type q, size_type r)
{
int n_1 = q - p + 1;
int n_2 = r - q;
std::deque<int> left, right;
int i = 0, j = 0;
while(i < n_1)
left.push_back(vec[p + i++]);
while(j < n_2)
right.push_back(vec[j++ + q + 1]);
i = 0; j = 0;
//std::cout << "left = ";
//print(left);
//std::cout << "right = ";
//print(right);
for(auto k = p; k <= r; k++) {
if((i < n_1 && left[i] <= right[j]) || j >= n_2) {
vec[k] = left[i++];
}
else if(j < n_2){
vec[k] = right[j++];
}
}
}
void merge_sort(std::deque<int> &A, size_type p, size_type r)
{
int q;
if(p < r) {
q = (r + p) / 2;
std::cout << "q = " << q << std::endl;
//std::cout << "p = " << p << std::endl;
merge_sort(A, p, q);
merge_sort(A, q + 1, r);
merge(A, p, q, r);
}
}
int main()
{
std::deque<int> small_vec = {10, 3, 6, 4, 1, 5, 3, 9, 7, 2, 8};
std::deque<int> samp_vec = {2, 9, 482, 42, 3, 4, 9, 8, 73, 8, 0, 98, 72, 473, 72, 3, 4, 9, 7, 6, 5, 6953, 583};
print(samp_vec);
merge_sort(samp_vec, 0, samp_vec.size() - 1);
print(samp_vec);
return 0;
}