Max Sum of Rectangle No Larger Than K in C++

Suppose we have a 2D matrix, and an integer k. We have to find the max sum of a rectangle in the matrix, such that its sum is not greater than k. So, if the input is like −

And k = 3, then the output will be 3, as the sum of marked rectangle is 3.

To solve this, we will follow these steps −

• Define a function maxSumSubmatrix(), this will take one 2D array matrix and k,
• n := row number, m := column number
• ans := -inf
• for initialize l := 0, when l < m, update (increase l by 1), do −
• Define an array rowSum of size n
• for initialize r := l, when r < m, update (increase r by 1), do −
  • for initialize i := 0, when i < n, update (increase i by 1), do −
    • rowSum[i] := rowSum[i] + matrix[i, r]
  • Define one set s
  • insert 0 into s
  • currSum := 0
  • for initialize i := 0, when i < n, update (increase i by 1), do −
    • currSum := currSum + rowSum[i]
    • it := first element of set that is not greater than currSum – k
    • if it is not equal to last element of s, then −
      • ans := maximum of ans and (currSum - it)
    • insert currSum into s
• return ans

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
      int n = matrix.size();
      int m = matrix[0].size();
      int ans = INT_MIN;
      for(int l = 0; l < m; l++){
         vector <int> rowSum(n);
         for(int r = l; r < m; r++){
            for(int i = 0; i < n; i++)rowSum[i] += matrix[i][r];
            set < int > s;
            s.insert(0);
            int currSum = 0;
            for(int i = 0; i < n; i++){
               currSum += rowSum[i];
               set <int> :: iterator it = s.lower_bound(currSum - k);
               if(it != s.end()){
                  ans = max(ans, (currSum - *it));
               }
               s.insert(currSum);
            }
         }
      }
      return ans;
   }
};
main(){
   Solution ob;
   vector<vector<int>> v = {{1,0,1},{0,-3,2}};
   cout << (ob.maxSumSubmatrix(v, 3));
}

Input

[{1,0,1},{0,-3,2}]
3

Output

3