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Given heights of n towers and a value k. We need to either increase or decrease height of every tower by k (only once) where k > 0. The task is to minimize the difference between the heights of the longest and the shortest tower after modifications, and output this difference.
I get the intuition behind the solution but I can not comment on the correctness of the solution below.
// C++ program to find the minimum possible
// difference between maximum and minimum
// elements when we have to add/subtract
// every number by k
#include <bits/stdc++.h>
using namespace std;
// Modifies the array by subtracting/adding
// k to every element such that the difference
// between maximum and minimum is minimized
int getMinDiff(int arr[], int n, int k)
{
if (n == 1)
return 0;
// Sort all elements
sort(arr, arr+n);
// Initialize result
int ans = arr[n-1] - arr[0];
// Handle corner elements
int small = arr[0] + k;
int big = arr[n-1] - k;
if (small > big)
swap(small, big);
// Traverse middle elements
for (int i = 1; i < n-1; i ++)
{
int subtract = arr[i] - k;
int add = arr[i] + k;
// If both subtraction and addition
// do not change diff
if (subtract >= small || add <= big)
continue;
// Either subtraction causes a smaller
// number or addition causes a greater
// number. Update small or big using
// greedy approach (If big - subtract
// causes smaller diff, update small
// Else update big)
if (big - subtract <= add - small)
small = subtract;
else
big = add;
}
return min(ans, big - small);
}
// Driver function to test the above function
int main()
{
int arr[] = {4, 6};
int n = sizeof(arr)/sizeof(arr[0]);
int k = 10;
cout << "\nMaximum difference is "
<< getMinDiff(arr, n, k);
return 0;
}
Can anyone help me provide the correct solution to this problem?
The codes above work, however I don't find much explanation so I'll try to add some in order to help develop intuition.
For any given tower, you have two choices, you can either increase its height or decrease it.
Now if you decide to increase its height from say Hi to Hi + K, then you can also increase the height of all shorter towers as that won't affect the maximum. Similarly, if you decide to decrease the height of a tower from Hi to Hi − K, then you can also decrease the heights of all taller towers.
We will make use of this, we have n buildings, and we'll try to make each of the building the highest and see making which building the highest gives us the least range of heights(which is our answer). Let me explain:
So what we want to do is - 1) We first sort the array(you will soon see why).
2) Then for every building from i = 0 to n-2[1] , we try to make it the highest (by adding K to the building, adding K to the buildings on its left and subtracting K from the buildings on its right).
So say we're at building Hi, we've added K to it and the buildings before it and subtracted K from the buildings after it. So the minimum height of the buildings will now be min(H0 + K, Hi+1 - K), i.e. min(1st building + K, next building on right - K).
(Note: This is because we sorted the array. Convince yourself by taking a few examples.)
Likewise, the maximum height of the buildings will be max(Hi + K, Hn-1 - K), i.e. max(current building + K, last building on right - K).
3) max - min gives you the range.
[1]Note that when i = n-1. In this case, there is no building after the current building, so we're adding K to every building, so the range will merely be
height[n-1] - height[0] since K is added to everything, so it cancels out.
Here's a Java implementation based on the idea above:
class Solution {
int getMinDiff(int[] arr, int n, int k) {
Arrays.sort(arr);
int ans = arr[n-1] - arr[0];
int smallest = arr[0] + k, largest = arr[n-1]-k;
for(int i = 0; i < n-1; i++){
int min = Math.min(smallest, arr[i+1]-k);
int max = Math.max(largest, arr[i]+k);
if (min < 0) continue;
ans = Math.min(ans, max-min);
}
return ans;
}
}
int getMinDiff(int a[], int n, int k) {
sort(a,a+n);
int i,mx,mn,ans;
ans = a[n-1]-a[0]; // this can be one possible solution
for(i=0;i<n;i++)
{
if(a[i]>=k) // since height of tower can't be -ve so taking only +ve heights
{
mn = min(a[0]+k, a[i]-k);
mx = max(a[n-1]-k, a[i-1]+k);
ans = min(ans, mx-mn);
}
}
return ans;
}
This is C++ code, it passed all the test cases.
This python code might be of some help to you. Code is self explanatory.
def getMinDiff(arr, n, k):
arr = sorted(arr)
ans = arr[-1]-arr[0] #this case occurs when either we subtract k or add k to all elements of the array
for i in range(n):
mn=min(arr[0]+k, arr[i]-k) #after sorting, arr[0] is minimum. so adding k pushes it towards maximum. We subtract k from arr[i] to get any other worse (smaller) minimum. worse means increasing the diff b/w mn and mx
mx=max(arr[n-1]-k, arr[i]+k) # after sorting, arr[n-1] is maximum. so subtracting k pushes it towards minimum. We add k to arr[i] to get any other worse (bigger) maximum. worse means increasing the diff b/w mn and mx
ans = min(ans, mx-mn)
return ans
Here's a solution:-
But before jumping on to the solution, here's some info that is required to understand it. In the best case scenario, the minimum difference would be zero. This could happen only in two cases - (1) the array contain duplicates or (2) for an element, lets say 'x', there exists another element in the array which has the value 'x + 2*k'.
The idea is pretty simple.
First we would sort the array.
Next, we will try to find either the optimum value (for which the answer would come out to be zero) or at least the closest number to the optimum value using Binary Search
Here's a Javascript implementation of the algorithm:-
function minDiffTower(arr, k) {
arr = arr.sort((a,b) => a-b);
let minDiff = Infinity;
let prev = null;
for (let i=0; i<arr.length; i++) {
let el = arr[i];
// Handling case when the array have duplicates
if (el == prev) {
minDiff = 0;
break;
}
prev = el;
let targetNum = el + 2*k; // Lets say we have an element 10. The difference would be zero when there exists an element with value 10+2*k (this is the 'optimum value' as discussed in the explaination
let closestMatchDiff = Infinity; // It's not necessary that there would exist 'targetNum' in the array, so we try to find the closest to this number using Binary Search
let lb = i+1;
let ub = arr.length-1;
while (lb<=ub) {
let mid = lb + ((ub-lb)>>1);
let currMidDiff = arr[mid] > targetNum ? arr[mid] - targetNum : targetNum - arr[mid];
closestMatchDiff = Math.min(closestMatchDiff, currMidDiff);
if (arr[mid] == targetNum) break; // in this case the answer would be simply zero, no need to proceed further
else if (arr[mid] < targetNum) lb = mid+1;
else ub = mid-1;
}
minDiff = Math.min(minDiff, closestMatchDiff);
}
return minDiff;
}
Here is the C++ code, I have continued from where you left. The code is self-explanatory.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int minDiff(int arr[], int n, int k)
{
// If the array has only one element.
if (n == 1)
{
return 0;
}
//sort all elements
sort(arr, arr + n);
//initialise result
int ans = arr[n - 1] - arr[0];
//Handle corner elements
int small = arr[0] + k;
int big = arr[n - 1] - k;
if (small > big)
{
// Swap the elements to keep the array sorted.
int temp = small;
small = big;
big = temp;
}
//traverse middle elements
for (int i = 0; i < n - 1; i++)
{
int subtract = arr[i] - k;
int add = arr[i] + k;
// If both subtraction and addition do not change the diff.
// Subtraction does not give new minimum.
// Addition does not give new maximum.
if (subtract >= small or add <= big)
{
continue;
}
// Either subtraction causes a smaller number or addition causes a greater number.
//Update small or big using greedy approach.
// if big-subtract causes smaller diff, update small Else update big
if (big - subtract <= add - small)
{
small = subtract;
}
else
{
big = add;
}
}
return min(ans, big - small);
}
int main(void)
{
int arr[] = {1, 5, 15, 10};
int n = sizeof(arr) / sizeof(arr[0]);
int k = 3;
cout << "\nMaximum difference is: " << minDiff(arr, n, k) << endl;
return 0;
}
class Solution {
public:
int getMinDiff(int arr[], int n, int k) {
sort(arr, arr+n);
int diff = arr[n-1]-arr[0];
int mine, maxe;
for(int i = 0; i < n; i++)
arr[i]+=k;
mine = arr[0];
maxe = arr[n-1]-2*k;
for(int i = n-1; i > 0; i--){
if(arr[i]-2*k < 0)
break;
mine = min(mine, arr[i]-2*k);
maxe = max(arr[i-1], arr[n-1]-2*k);
diff = min(diff, maxe-mine);
}
return diff;
}
};
class Solution:
def getMinDiff(self, arr, n, k):
# code here
arr.sort()
res = arr[-1]-arr[0]
for i in range(1, n):
if arr[i]>=k:
# at a time we can increase or decrease one number only.
# Hence assuming we decrease ith elem, we will increase i-1 th elem.
# using this we basically find which is new_min and new_max possible
# and if the difference is smaller than res, we return the same.
new_min = min(arr[0]+k, arr[i]-k)
new_max = max(arr[-1]-k, arr[i-1]+k)
res = min(res, new_max-new_min)
return res
Background:
The problem is from leetcode:
In an N by N square grid, each cell is either empty (0) or blocked
(1).
A clear path from top-left to bottom-right has length k if and
only if it is composed of cells C_1, C_2, ..., C_k such that:
Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
C_1 is at location (0, 0) (ie. has value grid[0][0])
C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).
Return the length of the shortest such clear path from top-left to
bottom-right. If such a path does not exist, return -1.
Question:
I was quite certain that my algorithm was correct but for this test case:
[[0,1,0,0,0],[0,1,0,0,0],[0,0,0,0,1],[0,1,1,1,0],[0,1,0,0,0]]
I get 9, and the correct answer is 7. Is there something I am doing wrong in the code below?
Code:
class Solution {
public:
std::vector<std::vector<int>> dirs = {{0,1},{1,0},{-1,0},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}};
int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
if(grid.empty())
return 0;
if(grid[0][0] == 1 || grid[grid.size()-1][grid.size()-1] == 1)
return -1;
int m = grid.size(), n = grid[0].size();
std::pair<int, int> start = {0,0};
std::pair<int, int> end = {m-1, n-1};
std::vector<std::vector<bool>> visited(m, std::vector<bool>(n, false));
std::priority_queue<std::pair<int,int>> q;
q.push(start);
visited[start.first][start.second] = true;
int count = 1;
while(!q.empty())
{
auto cur = q.top();
q.pop();
if(cur.first == end.first && cur.second == end.second)
return count;
for(auto dir : dirs)
{
int x = cur.first, y = cur.second;
if(isValid(grid, x + dir[0], y + dir[1]))
x += dir[0], y += dir[1];
if(!visited[x][y])
{
visited[x][y] = true;
q.push({x,y});
}
}
count++;
}
return -1;
}
bool isValid(std::vector<std::vector<int>>& grid, int i, int j)
{
if(i < 0 || i >= grid.size() || j < 0 || j >= grid[i].size() || grid[i][j] != 0)
return false;
return true;
}
};
This is not a problem for which you would use Dijkstra's algorithm. That algorithm is targetting weighted graphs, while the problem you are dealing with is unweighted. Moreover, the way you use a priority queue is wrong. A C++ priority queue will by default pop the element that is largest, but since you provide it coordinates, that means it will pop the element with the largest coordinates. This is obviously not what you need. In fact, you do not have anything to order nodes by, since this problem is about an unweighted graph.
Secondly, count is counting the total number of nodes you visit. That cannot be right, since you surely also visit nodes that are not on the shortest path that you eventually find.
This kind of problem is solved with a standard depth-first search. You can do it with two vectors (no need for stack, queue or deque, ...): the second vector gets populated with the unvisited neighbors of all the nodes in the first. Once that cycle is completed, you replace the first vector with the second, create a new second vector, and repeat... until you find the target node. The number of times you do this (outer) repetition corresponds to the length of the path.
Here is your shortestPathBinaryMatrix function with the necessary adaptations to make it work:
int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
if(grid.empty())
return 0;
if(grid[0][0] == 1 || grid[grid.size()-1][grid.size()-1] == 1)
return -1;
int m = grid.size(), n = grid[0].size();
pair<int, int> start = {0,0};
pair<int, int> end = {m-1, n-1};
vector<vector<bool>> visited(m, vector<bool>(n, false));
// no priority queue needed: the graph is not weighted
vector<std::pair<int,int>> q;
q.push_back(start);
visited[start.first][start.second] = true;
int count = 1;
while(!q.empty())
{
// just iterate the vector and populate a new one
vector<std::pair<int,int>> q2;
for(auto const& cur: q) {
if(cur.first == end.first && cur.second == end.second)
return count;
for(auto dir : dirs)
{
int x = cur.first, y = cur.second;
if(isValid(grid, x + dir[0], y + dir[1]))
x += dir[0], y += dir[1];
if(!visited[x][y])
{
visited[x][y] = true;
q2.push_back({x,y});
}
}
}
count++;
q = q2; // prepare for next iteration
}
return -1;
}
Not allowed to modify the array ( The array is read only ).
Using constant extra space is allowed.
ex:
A : [2 1 4 3 2]
k : 3
answer : 2
I did it below way. The answer is correct but need to be more memory efficient.
void insert_sorted(vector<int> &B, int a,int k)
{
for(int i=0;i<k;i++)
{
if(B[i]>=a)
{
for(int j=k-1;j>i;j--)
B[j]=B[j-1];
B[i]=a;
return;
}
}
}
int Solution::kthsmallest(const vector<int> &A, int k) {
vector <int> B;
for(int i=0;i<k;i++)
{
B.push_back(INT_MAX);
}
int l=A.size();
for(int i=0;i<l;i++)
{
if(B[k-1]>=A[i])
insert_sorted(B,A[i],k);
}
return B[k-1];
}
One possible solution is binary search.
Let A be the input array; we want to find a number b such that exactly k items in A are smaller than b.
Obviously, b must be inside the range [0, max(A)].
And we do binary search starting with this range.
Suppose we are searching within range [lo, hi].
Let c = (lo + hi)/2 which is the middle pivot.
There are three cases:
number of items in A less than c are less than k.
In this case the number we search for should be larger than c, so it should be in range (c, hi]
number of items in A less than c are larger than k.
Similarly, the number we search for is in range [lo, c)
number of items in A less than c equals to k.
In this case, the answer is the minimum element in A that is greater than or equals to c. This can be find by doing a linear search in A again
The complexity is O(n log m), where m is the max element in A.
/* assume k is 0 based, i.e. 0 <= k < n */
int kth_element(const vector<int> &A, int k){
int lo = 0, hi = *max_element(A.begin(), A.end());
while (lo <= hi){
int mid = (lo + hi) / 2;
int rank_lo = count_if(A.begin(), A.end(), [=](int i){ return i < mid;});
int rank_hi = count_if(A.begin(), A.end(), [=](int i){ return i <= mid;});
if (rank_lo <= k && k < rank_hi)
return mid;
if (k >= rank_hi)
lo = mid + 1;
else
hi = mid - 1;
}
}
Although it's not the answer to this particular problem (as it requires a modifiable collection), there is a function called std::nth_element, which rearranges the elements so that the kth element is at position k, and all elements at positions less than k are smaller than or equal to the kth element, where k is a input parameter.
The question does not ask for any time constraints. An O(nk) solution is fairly simple, by iterating the array k times (at most), and discarding one element (and its duplicates) each time.
int FindKthSmallesr(const std::vector<int>& v, int k) {
// assuming INT_MIN cannot be a value. Could be relaxed by an extra iteration.
int last_min = INT_MIN;
while (k > 0) {
int current_min = INT_MAX;
for (int x : v) {
if (x <= last_min) continue;
current_min = std::min(current_min, x);
}
last_min = current_min;
for (int x : v) {
if (x == current_min) k--;
}
}
return last_min;
}
Code on ideone: http://ideone.com/RjRIkM
If only constant extra space is allowed, we can use a simple O(n*k) algorithm.
int kth_smallest(const vector<int>& v, int k) {
int curmin = -1;
int order = -1;
while (order < k) { // while kth element wasn't reached
curmin = *min_element(v.begin(), v.end(), [curmin](int a, int b) {
if (a <= curmin) return false;
if (b <= curmin) return true;
return a < b;
}); // find minimal number among not counted yet
order += count(v.begin(), v.end(), curmin); // count all 'minimal' numbers
}
return curmin;
}
online version to play with: http://ideone.com/KNMYxA
I am trying to find the minimal distance in the Manhattan metric (x,y). I am searching for information about this. But I haven't found anything.
#include<bits/stdc++.h>
using namespace std;
#define st first
#define nd second
pair<int, int> pointsA[1000001];
pair<int, int> pointsB[1000001];
int main() {
int n, t;
unsigned long long dist;
scanf("%d", &t);
while(t-->0) {
dist = 4000000000LL;
scanf("%d", &n);
for(int i = 0; i < n; i++) {
scanf("%d%d", &pointsA[i].st, &pointsA[i].nd);
}
for(int i = 0; i < n; i++) {
scanf("%d%d", &pointsB[i].st, &pointsB[i].nd);
}
for(int i = 0; i < n ;i++) {
for(int j = 0; j < n ; j++) {
if(abs(pointsA[i].st - pointsB[j].st) + abs(pointsA[i].nd - pointsB[j].nd) < dist) {
dist = abs(pointsA[i].st - pointsB[j].st) + abs(pointsA[i].nd - pointsB[j].nd);
}
}
printf("%lld\n", dist);
}
}
}
My code works in O(n^2) but is too slow. I do not know whether it will be useful but y in pointsA always be > 0 and y in pointsB always be < 0. My code compare actually distance to next and chose smallest.
for example:
input:
2
3
-2 2
1 3
3 1
0 -1
-1 -2
1 -2
1
1 1
-1 -1
Output:
5
4
My solution (note for simplicity I do not care about overflow in manhattan_dist and for that reason it does not work with unsigned long long):
#include <cstdlib>
#include <cstdio>
#include <cassert>
#include <vector>
#include <limits>
#include <algorithm>
typedef std::pair<int, int> Point;
typedef std::vector<std::pair<int, int> > PointsList;
static inline bool cmp_by_x(const Point &a, const Point &b)
{
if (a.first < b.first) {
return true;
} else if (a.first > b.first) {
return false;
} else {
return a.second < b.second;
}
}
static inline bool cmp_by_y(const Point &a, const Point &b)
{
if (a.second < b.second) {
return true;
} else if (a.second > b.second) {
return false;
} else {
return a.first < b.first;
}
}
static inline unsigned manhattan_dist(const Point &a, const Point &b)
{
return std::abs(a.first - b.first) +
std::abs(a.second - b.second);
}
int main()
{
unsigned int n_iter = 0;
if (scanf("%u", &n_iter) != 1) {
std::abort();
}
for (unsigned i = 0; i < n_iter; ++i) {
unsigned int N = 0;
if (scanf("%u", &N) != 1) {
std::abort();
}
if (N == 0) {
continue;
}
PointsList pointsA(N);
for (PointsList::iterator it = pointsA.begin(), endi = pointsA.end(); it != endi; ++it) {
if (scanf("%d%d", &it->first, &it->second) != 2) {
std::abort();
}
assert(it->second > 0);
}
PointsList pointsB(N);
for (PointsList::iterator it = pointsB.begin(), endi = pointsB.end(); it != endi; ++it) {
if (scanf("%d%d", &it->first, &it->second) != 2) {
std::abort();
}
assert(it->second < 0);
}
std::sort(pointsA.begin(), pointsA.end(), cmp_by_y);
std::sort(pointsB.begin(), pointsB.end(), cmp_by_y);
const PointsList::const_iterator min_a_by_y = pointsA.begin();
const PointsList::const_iterator max_b_by_y = (pointsB.rbegin() + 1).base();
assert(*max_b_by_y == pointsB.back());
unsigned dist = manhattan_dist(*min_a_by_y, *max_b_by_y);
const unsigned diff_x = std::abs(min_a_by_y->first - max_b_by_y->first);
const unsigned best_diff_y = dist - diff_x;
const int max_y_for_a = max_b_by_y->second + dist;
const int min_y_for_b = min_a_by_y->second - dist;
PointsList::iterator it;
for (it = pointsA.begin() + 1; it != pointsA.end() && it->second <= max_y_for_a; ++it) {
}
if (it != pointsA.end()) {
pointsA.erase(it, pointsA.end());
}
PointsList::reverse_iterator rit;
for (rit = pointsB.rbegin() + 1; rit != pointsB.rend() && rit->second >= min_y_for_b; ++rit) {
}
if (rit != pointsB.rend()) {
pointsB.erase(pointsB.begin(), (rit + 1).base());
}
std::sort(pointsA.begin(), pointsA.end(), cmp_by_x);
std::sort(pointsB.begin(), pointsB.end(), cmp_by_x);
for (size_t j = 0; diff_x > 0 && j < pointsA.size(); ++j) {
const Point &cur_a_point = pointsA[j];
assert(max_y_for_a >= cur_a_point.second);
const int diff_x = dist - best_diff_y;
const int min_x = cur_a_point.first - diff_x + 1;
const int max_x = cur_a_point.first + diff_x - 1;
const Point search_term = std::make_pair(max_x, std::numeric_limits<int>::min());
PointsList::const_iterator may_be_near_it = std::lower_bound(pointsB.begin(), pointsB.end(), search_term, cmp_by_x);
for (PointsList::const_reverse_iterator rit(may_be_near_it); rit != pointsB.rend() && rit->first >= min_x; ++rit) {
const unsigned cur_dist = manhattan_dist(cur_a_point, *rit);
if (cur_dist < dist) {
dist = cur_dist;
}
}
}
printf("%u\n", dist);
}
}
Benchmark on my machine (Linux + i7 2.70 GHz + gcc -Ofast -march=native):
$ make bench
time ./test1 < data.txt > test1_res
real 0m7.846s
user 0m7.820s
sys 0m0.000s
time ./test2 < data.txt > test2_res
real 0m0.605s
user 0m0.590s
sys 0m0.010s
test1 is your variant, and test2 is mine.
You'll need to learn how to write functions and how to use containers. With your current coding style, it's infeasible to get a better solution.
The problem is that the better solution is a recursive method. Sort the points by X coordinate. Now recursively split the set in half and determine the closest distance within each half as well as the closest distance between a pair of points from either half.
The last part is efficient because both halves are sorted by X. Comparing the last values from the left half with the first value of the right half gives a good upper bound on the distance.
So there's a really simple optimization you can make that can cut a ton of time off.
Since you state that all points in set A have y > 0, and all points in set B have y < 0, you can immediately discard all points in A whose y > mindist and all points in B whose y < -mindist so far. These points can never be closer than the current closest pair:
for(int i = 0; i < n ;i++) {
if (pointsA[i].nd > dist)
continue; // <- this is the big one
for(int j = 0; j < n ; j++) {
if (pointsB[j].nd < -dist)
continue; // <- helps although not as much
if(abs(pointsA[i].st - pointsB[j].st) + abs(pointsA[i].nd - pointsB[j].nd) < dist) {
dist = abs(pointsA[i].st - pointsB[j].st) + abs(pointsA[i].nd - pointsB[j].nd);
}
}
printf("%lld\n", dist);
}
For a test of 40000 points per set, on my machine with gcc and -O2 this reduces the time from 8.2 seconds down to roughly 0.01 seconds (and yields correct results)! (measured with QueryPerformanceCounter on Windows).
Not too shabby.
Fwiw, computing your distance twice isn't actually that big of a deal. First of all that "second" calculation doesn't actually happen all that often, it only happens when a closer distance is found.
And secondly, for reasons I can't explain, storing it in a variable and only calculating it once actually consistently seems to add about 20% to the total run time, raising it from an average of 8.2 sec to about 10.5 seconds for the above set.
I'd say discarding points based on your assumptions about the Y values is by far the biggest bang for your buck you can get without significantly changing your algorithm.
You may be able to take advantage of that further by pre-sorting A and B in order of increasing Y values for A, and decreasing Y values for B, before finding the distances, to maximize the chance of skipping over subsequent point sets.
Keep a list of candidates in group A and group B, initially containing the whole input. Take min y of A and max y B for the closest pair in y, calculate the Manhattan distance, and eliminate any with y greater than the upper bound from the candidate lists. This might slash the input or it might have essentially no effect, but it's O(N) and a cheap first step.
Now sort the remaining candidates in x and y. This gives you a separated list in y ,and a mixed list in x, and is O(N log N), where N has been cut down, hopefully but not necessarily, by step one.
For each point, now calculate its closest neighbour in y (trivial) and closest in x (a bit harder), then calculate its minimum possible Manhattan distance, assuming the closest in x is also the closest in y. Eliminate any points further than your bound from the candidate list. Now sort again, by minimum possible. That's another N log N operation.
Now start with your best candidate and find its genuine minimum distance, by trying the closest point in either direction in x or y, and terminating when either delta x or delta y goes above the best so far, or delta x or delta y goes above your maximum bound. If you have better candidate pair then the current candidate pair, purge the candidate list of everything with a worse minimum possible. If the best candidate point doesn't form half of a candidate pair, you just purge that one point.
When you've purged a certain number of candidates, recalculate the lists. I'm not sure what the best value to use would be, certainly if you get to the worst candidate you must do that and then start again at the best. Maybe use 50%.
Eventually you are left with only one candidate pair. I'm not quite sure what the analysis is - worst case I suppose you only eliminate a few candidates on each test. But for most inputs you should get the candidate list down to a small value pretty fast.
I was solving problems on HackerRank when I got stuck at this one.
Problem Statement
You are given a 2D matrix, a, of dimension MxN and a positive integer R. You have to rotate the matrix R times and print the resultant matrix. Rotation should be in anti-clockwise direction.
Rotation of a 4x5 matrix is represented by the following figure. Note that in one rotation, you have to shift elements by one step only (refer sample tests for more clarity).
It is guaranteed that the minimum of M and N will be even.
Input
First line contains three space separated integers, M, N and R, where M is the number of rows, N is number of columns in matrix, and R is the number of times the matrix has to be rotated.
Then M lines follow, where each line contains N space separated positive integers. These M lines represent the matrix.
Output
Print the rotated matrix.
Constraints
2 <= M, N <= 300
1 <= R <= 10^9
min(M, N) % 2 == 0
1 <= aij <= 108, where i ∈ [1..M] & j ∈ [1..N]'
What I tried to do was store the circles in a 1D array. Something like this.
while(true)
{
k = 0;
for(int j = left; j <= right; ++j) {temp[k] = a[top][j]; ++k;}
top++;
if(top > down || left > right) break;
for(int i = top; i <= down; ++i) {temp[k] = a[i][right]; ++k;}
right--;
if(top > down || left > right) break;
for(int j = right; j >= left; --j) {temp[k] = a[down][j] ; ++k;}
down--;
if(top > down || left > right) break;
for(int i = down; i >= top; --i) {temp[k] = a[i][left]; ++k;}
left++;
if(top > down || left > right) break;
}
Then I could easily rotate the 1D matrix by calculating its length modulo R. But then how do I put it back in matrix form? Using a loop again would possibly cause a timeout.
Please don't provide code, but only give suggestions. I want to do it myself.
Solution Created :
#include <iostream>
using namespace std;
int main() {
int m,n,r;
cin>>m>>n>>r;
int a[300][300];
for(int i = 0 ; i < m ; ++i){
for(int j = 0; j < n ; ++j)
cin>>a[i][j];
}
int left = 0;
int right = n-1;
int top = 0;
int down = m-1;
int tleft = 0;
int tright = n-1;
int ttop = 0;
int tdown = m-1;
int b[300][300];
int k,size;
int temp[1200];
while(true){
k=0;
for(int i = left; i <= right ; ++i)
{
temp[k] = a[top][i];
// cout<<temp[k]<<" ";
++k;
}
++top;
if(top > down || left > right)
break;
for(int i = top; i <= down ; ++i)
{
temp[k]=a[i][right];
// cout<<temp[k]<<" ";
++k;
}
--right;
if(top > down || left > right)
break;
for(int i = right; i >= left ; --i)
{
temp[k] = a[down][i];
// cout<<temp[k]<<" ";
++k;
}
--down;
if(top > down || left > right)
break;
for(int i = down; i >= top ; --i)
{
temp[k] = a[i][left];
// cout<<temp[k]<<" ";
++k;
}
++left;
if(top > down || left > right)
break;
//________________________________\\
size = k;
k=0;
// cout<<size<<endl;
for(int i = tleft; i <= tright ; ++i)
{
b[ttop][i] = temp[(k + (r%size))%size];
// cout<<(k + (r%size))%size<<" ";
// int index = (k + (r%size))%size;
// cout<<index;
++k;
}
++ttop;
for(int i = ttop; i <= tdown ; ++i)
{
b[i][tright]=temp[(k + (r%size))%size];
++k;
}
--tright;
for(int i = tright; i >= tleft ; --i)
{
b[tdown][i] = temp[(k + (r%size))%size];
++k;
}
--tdown;
for(int i = tdown; i >= ttop ; --i)
{
b[i][tleft] = temp[(k + (r%size))%size];
++k;
}
++tleft;
}
size=k;
k=0;
if(top != ttop){
for(int i = tleft; i <= tright ; ++i)
{
b[ttop][i] = temp[(k + (r%size))%size];
++k;
}
++ttop;
}
if(right!=tright){
for(int i = ttop; i <= tdown ; ++i)
{
b[i][tright]=temp[(k + (r%size))%size];
++k;
}
--tright;
}
if(down!=tdown){
for(int i = tright; i >= tleft ; --i)
{
b[tdown][i] = temp[(k + (r%size))%size];
++k;
}
--tdown;
}
if(left!=tleft){
for(int i = tdown; i >= ttop ; --i)
{
b[i][tleft] = temp[(k + (r%size))%size];
++k;
}
++tleft;
}
for(int i = 0 ; i < m ;++i){
for(int j = 0 ; j < n ;++j)
cout<<b[i][j]<<" ";
cout<<endl;
}
return 0;
}
You need to break down this problem (remind me of an interview question from gg and fb) :
Solve first rotating a sequence one a single position
Then solve rotating a sequence N times
Model each "circle" or ring as an array. You may or may not actually need to store in a separate data
Iterate over each ring and apply the rotating algorithm
Lets consider the case of an array of length L which needs to be rotated R time. Observe that if R is a multiple of L, the array will be unchanged.
Observe too that rotating x times to the right is the same as rotating L - x to the left (and vice versa).
Thus you can first design an algorithm able to rotate once either left or right one exactly one position
Reduce the problem of rotating R times to the left to rotating R modulo L to the left
If you want to go further reduce the problem of rotating R modulo L to the left to rotating left R modulo L or rotating right L - R modulo L. Which means if you have 100 elements and you have to do 99 rotations left, you better do 1 rotation right and be done with it.
So the complexity will be O ( Number of circles x Circle Length x Single Rotation Cost)
With an array in-place it means O( min(N,m) * (N * M)^2 )
If you use a doubly linked list as temporary storage, a single rotation sequence is done by removing the front and putting it at the tail (or vice versa to rotate right). So what you can do is copy all data first to a linked list. Run the single rotation algorithm R modulo L times, copy back the linked list on the ring position, and move on the next right till all rings are processed.
Copy ring data to list is O(L), L <= N*M
Single Rotation Cost is O(1)
All rotations R modulo L is O(L)
Repeat on all min(N,m) rings
With a spare double linked list it means complexity of O( min(N,m) * (N * M))
I would start with a simplifying assumption: M is less than or equal to N. Thus, you are guaranteed to have an even number of rows. (What if M > N? Then transpose the matrix, carry out the algorithm, and transpose the matrix again.)
Because you have an even number of rows, you can easily find the corners of each cycle within the matrix. The outermost cycle has these corners:
a1,1 → aM,1 → aM,N → a1,N
To find the next cycle, move each corner inward, which means incrementing or decrementing the index at each corner as appropriate.
Knowing the sequence of corners allows you to iterate over each cycle and store the values in a one-dimensional vector. In each such vector a, start from index R % a.size() and increment the index a.size() - 1 times to iterate over the rotated elements of the cycle. Copy each element a[i % a.size()] back to the cycle.
Note that we don't actually rotate the vector. We accomplish the rotation by starting from an offset index when we copy elements back to the matrix. Thus, the overall running time of the algorithm is O(MN), which is optimal because it costs O(MN) just to read the input matrix.
I would treat this as a problem that divides the matrix into submatrices. You could probably write a function that shifts the matrices (and submatrices) outer rows and columns by one each time you call it. Take care to handle the four corners of the matrix appropriately.
Check this out for suggestions how to shift the columns.
Edit (more detailed):
Read each matrix circle in as a vector, use std::rotate on it R % length.vector times, write back. Maximally 150 operations.
Each element moves uniquely according to one of four formulas, adding five movements of known sizes (I'll leave the size calculation out since you wanted to figure it out):
formula (one of these four):
left + down + right + up + left
down + right + up + left + down
right + up + left + down + right
up + left + down + right + up
Since the smallest side of the matrix is even, we know there is not an element remaining in place. After R rotations, the element has circled around floor (R / formula) times but still needs to undergo extra = R % formula shifts. Once you know extra, simply calculate the appropriate placement for the element.