I have two dimensional array of chars, where all numbers, excluding one * (as given in picture (two examples)
My task is to sum up all neighbour integers ( in example 1, neighbours of * are 4,2,5,8 and sum is 4+2+5+8=19)
But in example 2, * doesn't have top neighbour.
My initial code was like:
arr[i-1][j] + arr[i+1][j] + arr[i][j-1] + arr[i][j+1]
But then I understood that in case like a[0][-1] doesn't exist. So can you help me to to solve my problem
You need to explicitly check each one. The following should work:
bool inRange(int i, int j) {
const auto n = 4; // you need to set this somewhere, or pass it in
return (i >= 0) && (i < n) && (j >= 0) && (j < n);
}
auto sum = (inRange(i-1, j) ? arr[i-1][j] : 0)
+ (inRange(i+1, j) ? arr[i+1][j] : 0)
+ (inRange(i, j-1) ? arr[i][j-1] : 0)
+ (inRange(i, j+1) ? arr[i][j+1] : 0);
You can probably write this a little cleaner, but you need to check not only for the -1, but also for when you go over 3.
There can be multiple solutions to this problem, but if you want to avoid checking bound each time you can extend the matrix dimension by 1 than needed. That means if you have an array:
1 * 4 7
8 9 2 3
5 1 2 4
4 3 6 5
Implement it as:
0 0 0 0 0 0
0 1 * 4 7 0
0 8 9 2 3 0
0 5 1 2 4 0
0 4 3 6 5 0
0 0 0 0 0 0
Doing this won't even affect your sum at the end.
Related
I represent a n*m matrix like chessboard.
1 0 2 0
0 3 0 4
5 0 6 0
0 7 0 8
I don't need to store the zeros in my 1d vector.
vector v = {1, 2, 3, 4.. etc}
I ask the user for a row and column number.
How can i return with i. row j. column element?
if (i+j) % 2 != 0
I return with 0, but i don't know what i need to do when
(i+j) % 2 == 0
Can you help me? (sorry for my bad English)
With regular matrices stored as 1D-vector, coordinate to index would be:
(i + j * width) (or i * height + j depending on convention).
with half case to 0, you just have to divide by 2:
if ((i + j) % 2 != 0) return 0;
else return data[(i + j * width) / 2];
I have been stuck with this problem for two days and I still can't get it right.
Basically, I have a 2D array with relations between certain numbers (in given range):
0 = the order doesn't matter
1 = the first number (number in left column) should be first
2 = the second number (number in upper row) should be first
So, I have some 2D array, for example this:
0 1 2 3 4 5 6
0 0 0 1 0 0 0 2
1 0 0 2 0 0 0 0
2 2 1 0 0 1 0 0
3 0 0 0 0 0 0 0
4 0 0 2 0 0 0 0
5 0 0 0 0 0 0 0
6 1 0 0 0 0 0 0
And my goal is to create a new array of given numbers (0 - 6) in such a way that it is following the rules from the 2D array (e.g. 0 is before 2 but it is after 6). I probably also have to check if such array exists and then create the array. And get something like this:
6 0 2 1 4 5
My Code
(It doesn't really matter, but I prefer c++)
So far I tried to start with ordered array 0123456 and then swap elements according to the table (but that obviously can't work). I also tried inserting the number in front of the other number according to the table, but it doesn't seem to work either.
// My code example
// I have:
// relArr[n][n] - array of relations
// resArr = {1, 2, ... , n} - result array
for (int i = 0; i < n; i++) {
for (int x = 0; x < n; x++) {
if (relArr[i][x] == 1) {
// Finding indexes of first (i) and second (x) number
int iI = 0;
int iX = 0;
while (resArr[iX] != x)
iX++;
while (resArr[iI] != i)
iI++;
// Placing the (i) before (x) and shifting array
int tmp, insert = iX+1;
if (iX < iI) {
tmp = resArr[iX];
resArr[iX] = resArr[iI];
while (insert < iI+1) {
int tt = resArr[insert];
resArr[insert] = tmp;
tmp = tt;
insert++;
}
}
} else if (relArr[i][x] == 2) {
int iI = 0;
int iX = 0;
while (resArr[iX] != x)
iX++;
while (resArr[iI] != i)
iI++;
int tmp, insert = iX-1;
if (iX > iI) {
tmp = resArr[iX];
resArr[iX] = resArr[iI];
while (insert > iI-1) {
int tt = resArr[insert];
resArr[insert] = tmp;
tmp = tt;
insert--;
}
}
}
}
}
I probably miss correct way how to check whether or not it is possible to create the array. Feel free to use vectors if you prefer them.
Thanks in advance for your help.
You seem to be re-ordering the output at the same time as you're reading the input. I think you should parse the input into a set of rules, process the rules a bit, then re-order the output at the end.
What are the constraints of the problem? If the input says that 0 goes before 1:
| 0 1
--+----
0 | 1
1 |
does it also guarantee that it will say that 1 comes after 0?
| 0 1
--+----
0 |
1 | 2
If so you can forget about the 2s and look only at the 1s:
| 0 1 2 3 4 5 6
--+--------------
0 | 1
1 |
2 | 1 1
3 |
4 |
5 |
6 | 1
From reading the input I would store a list of rules. I'd use std::vector<std::pair<int,int>> for this. It has the nice feature that yourPair.first comes before yourPair.second :)
0 before 2
2 before 1
2 before 4
6 before 0
You can discard any rules where the second value is never the first value of a different rule.
0 before 2
6 before 0
This list would then need to be sorted so that "... before x" and "x before ..." are guaranteed to be in that order.
6 before 0
0 before 2
Then move 6, 0, and 2 to the front of the list 0123456, giving you 6021345.
Does that help?
Thanks for the suggestion.
As suggested, only ones 1 are important in 2D array. I used them to create vector of directed edges and then I implemented Topological Sort. I decide to use this Topological Sorting Algorithm. It is basically Topological Sort, but it also checks for the cycle.
This successfully solved my problem.
Description:
Given matrix [x] [y], with x- rows and y- number of columns . Filled random numbers from 0 to 5 inclusive .
Description of finding a solution : the solution is considered to be a set of matrix elements that are adjacent to each other ( diagonal neighborhood is not taken into account ) and the sum of the number are 10. Each element of the matrix can be used 1 time for a decision . The solution may have any number of digits. The decision must end any number other than zero .
Example:
given
0 1 2 3 4 5
1 2 3 4 5 0
2 3 4 5 1 2
Solution 1 : (1 - 2 - 3 - 4)
0 **1** 2 3 4 5
1 **2** 3 4 5 0
2 **3** **4** 5 1 2
i tried to do smth like this, but it is wrong, i dont know when i must stop,
Solution it is a class which contains mair of indexes, pls help me.
void xxx(int colCount, int rowCount, int currentRow, int currentCol, int** matrix, int sum, Solution *solution, int solCount) {
sum += matrix[currentRow][currentCol];
matrix[currentRow][currentCol] = -1;
if(sum > 10){
sum - = matrix[currentRow][currentCol];
return;
} else if(sum == 10){
solution[solCount].additem(currentRow, currentCol);
return xxx(5,5,currentRow - 1, currentCol, matrix, sum, solution, solCount+1);
} else {
//UP
if( currentRow > 0 && matrix [currentRow - 1][currentCol] != -1){
xxx(5,5,currentRow - 1, currentCol, matrix, sum, solution,solCount);
}
//LEFT
if(currentCol > 0 && matrix [currentRow][currentCol-1] != -1){
xxx(5,5,currentRow, currentCol - 1, matrix, sum, solution,solCount);
}
//DOWN
if(currentRow + 1 < colCount && matrix[currentRow + 1][currentCol] != -1){
xxx(5,5,currentRow + 1, currentCol, matrix, sum, solution,solCount);
}
//RIGHT
if(currentCol + 1 < rowCount && matrix[currentRow][currentCol + 1] != -1){
xxx(5,5,currentRow, currentCol + 1, matrix, sum, solution,solCount);
}
}
}
I would really appreciate it if someone can point me to the right direction with this problem. I am trying to find all the different combinations of various numbers each with a different number of columns (in C++). for example consider the number 2:
two columns:
2 = { 2 , 0 }
{ 0 , 2 }
{ 1 , 1 }
three columns :
2 = { 0 , 0 , 2 }
{ 0 , 2 , 0 }
{ 2 , 0 , 0 }
{ 1 , 1 , 0 }
{ 0 , 1 , 1 }
{ 1 , 0 , 1 }
four columns:
2 = { 0 , 0 , 0 , 2 }
{ 0 , 0 , 2 , 0 }
{ 0 , 2 , 0 , 0 }
{ 2 , 0 , 0 , 0 }
{ 1 , 1 , 0 , 0 }
{ 0 , 0 , 1 , 1 }
{ 0 , 1 , 1 , 0 }
{ 1 , 0 , 0 , 1 }
{ 1 , 0 , 1 , 0 }
{ 0 , 1 , 0 , 1 }
thanks in advance!
Here's my attempt:
void combinations(int n, int columns, std::vector<int>& soFar)
{
if (columns == 1)
{
for (auto e : soFar)
std::cout << e << " ";
std::cout << n << '\n';
return;
}
for (int i = 0; i <= n; ++i)
{
soFar.push_back(i);
combinations(n - i, columns - 1, soFar);
soFar.pop_back();
}
}
void combinations(int n, int columns)
{
std::vector<int> soFar;
combinations(n, columns, soFar);
}
Basically, you keep dividing the number into two subparts, till you reach your depth limit (the number of columns in your case).
To keep printing the previous numbers on the way back up, I store them in the soFar vector, pushing and popping them accordingly.
Here's the output for combinations(2, 4):
0 0 0 2
0 0 1 1
0 0 2 0
0 1 0 1
0 1 1 0
0 2 0 0
1 0 0 1
1 0 1 0
1 1 0 0
2 0 0 0
This is a straight combinatorics question. If you have m columns, then you have m-1 dividers between columns. With the number n, you want all the ways to order m-1 dividers and n elements. For example, with n=5 and m=3, one possible arrangement is xx,x,xx -- and you are looking at 7 choose 2.
So the general solution is m+n-1 choose m-1, or equivalently, m+n-1 choose n.
The formula for x choose y is x! / [y! * (x-y)!]
Consider splitting the problem into two subproblems:
1) Find all the combinations of numbers that add to your number:
i.e: 2-column case for "3": (2,1) and (3,0)
2) Permute all the combinations you've found:
i.e: (2,1) -> (2,1), (1,2) and (3,0) -> (3,0), (0,3)
For part 1), you get the problem of big numbers and many columns, say 5 with 4 columns (i know, they're unfathomably huge numbers):
5 = 4 + 1
5 = 3 + 2
5 = 3 + 1 + 1
5 = 2 + 1 + 1 + 1
5 = 1 + 1 + 1 + 1 + 1
If you look carefully, you have a possibility for recursion. As in, for 5 = 3 + 2: find the combinations for 3 and the combinations for 2 and so on... until you get to 1
As soon as you say recursion, tree structures start to sound interesting. This is how I'd approach the problem.
I am fairly new to C++, and am struggling through a problem that seems to have a solid solution but I just can't seem to find it. I have a contiguous array of ints starting at zero:
int i[6] = { 0, 1, 2, 3, 4, 5 }; // this is actually from an iterator
I would like to partition the array into groups of three. The design is to have two methods, j and k, such that given an i they will return the other two elements from the same group of three. For example:
i j(i) k(i)
0 1 2
1 0 2
2 0 1
3 4 5
4 3 5
5 3 4
The solution seems to involve summing the i with its value mod three and either plus or minus one, but I can't quite seem to work out the logic.
This should work:
int d = i % 3;
int j = i - d + ( d == 0 );
int k = i - d + 2 - ( d == 2 );
or following statement for k could be more readable:
int k = i - d + ( d == 2 ? 1 : 2 );
This should do it:
int j(int i)
{
int div = i / 3;
if (i%3 != 0)
return 3*div;
else
return 3*div+1;
}
int k(int i)
{
int div = i / 3;
if (i%3 != 2)
return 3*div+2;
else
return 3*div+1;
}
Test.
If you want shorter functions:
int j(int i)
{
return i/3*3 + (i%3 ? 0 : 1);
}
int k(int i)
{
return i/3*3 + (i%3-2 ? 2 : 1);
}
Well, first, notice that
j(i) == j(3+i) == j(6+i) == j(9+i) == ...
k(i) == k(3+i) == k(6+i) == k(9+i) == ...
In other words, you only need to find a formula for
j(i), i = 0, 1, 2
k(i), i = 0, 1, 2
and then for the rest of the cases simply plug in i mod 3.
From there, you'll have trouble finding a simple formula because your "rotation" isn't standard. Instead of
i j(i) k(i)
0 1 2
1 2 0
2 0 1
for which the formula would have been
j(i) = (i + 1) % 3
k(i) = (i + 2) % 3
you have
i j(i) k(i)
0 1 2
1 0 1
2 0 2
for which the only formula I can think of at the moment is
j(i) = (i == 0 ? 1 : 0)
k(i) = (i == 1 ? 1 : 2)
If the values of your array (let's call it arr, not i in order to avoid confusion with the index i) do not coincide with their respective index, you have to perform a reverse lookup to figure out their index first. I propose using an std::map<int,size_t> or an std::unordered_map<int,size_t>.
That structure reflects the inverse of arr and you can extra the index for a particular value with its subscript operator or the at member function. From then, you can operate purely on the indices, and use modulo (%) to access the previous and the next element as suggested in the other answers.