Number of Submatrices That Sum to Target in C++

C++Server Side ProgrammingProgramming

Suppose we have a matrix, and a target value, we have to find the number of non-empty submatrices that sum is same as target. Here a submatrix [(x1, y1), (x2, y2)] is the set of all cells matrix[x][y] with x in range x1 and x2 and y in range y1 and y2. Two submatrices [(x1, y1), (x2, y2)] and [(x1', y1'), (x2', y2')] are different if they have some coordinate that is different: like, if x1 is not same as x1'.

So, if the input is like

010
111
010

and target = 0, then the output will be 4, this is because four 1x1 submatrices that only contain 0.

To solve this, we will follow these steps −

  • ans := 0

  • col := number of columns

  • row := number of rows

  • for initialize i := 0, when i < row, update (increase i by 1), do −

    • for initialize j := 1, when j < col, update (increase j by 1), do −

      • matrix[i, j] := matrix[i, j] + matrix[i, j - 1]

  • Define one map m

  • for initialize i := 0, when i < col, update (increase i by 1), do −

    • for initialize j := i, when j < col, update (increase j by 1), do −

      • clear the map m

      • m[0] := 1

      • sum := 0

    • for initialize k := 0, when k < row, update (increase k by 1), do −

      • current := matrix[k, j]

      • if i - 1 >= 0, then −

        • current := current - matrix[k, i - 1]

      • sum := sum + current

      • ans := ans + m[target - sum]

      • increase m[-sum] by 1

  • 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 numSubmatrixSumTarget(vector<vector<int>>& matrix, int
   target) {
      int ans = 0;
      int col = matrix[0].size();
      int row = matrix.size();
      for(int i = 0; i < row; i++){
         for(int j = 1; j < col; j++){
            matrix[i][j] += matrix[i][j - 1];
         }
      }
      unordered_map <int, int> m;
      for(int i = 0; i < col; i++){
         for(int j = i; j < col; j++){
            m.clear();
            m[0] = 1;
            int sum = 0;
            for(int k = 0; k < row; k++){
               int current = matrix[k][j];
               if(i - 1 >= 0)current -= matrix[k][i - 1];
               sum += current;
               ans += m[target - sum];
               m[-sum]++;
            }
         }
      }
      return ans;
   }
};
main(){
   Solution ob;
   vector<vector<int>> v = {{0,1,0},{1,1,1},{0,1,0}};
   cout << (ob.numSubmatrixSumTarget(v, 0));
}

Input

{{0,1,0},{1,1,1},{0,1,0}}, 0

Output

4
raja
Published on 04-Jun-2020 11:08:36
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