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