Number of Submatrices That Sum to Target in C++

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
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
ans := 0
col := number of columns
col := number of columns
row := number of rows
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]
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]
for initialize j := 1, when j < col, update (increase j by 1), do −
matrix[i, j] := matrix[i, j] + matrix[i, j - 1]
matrix[i, j] := matrix[i, j] + matrix[i, j - 1]
Define one map m
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 mm[0] := 1sum := 0for 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 + currentans := ans + m[target - sum]increase m[-sum] by 1
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 mm[0] := 1sum := 0
for initialize j := i, when j < col, update (increase j by 1), do −
clear the map m
clear the map m
m[0] := 1
m[0] := 1
sum := 0
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 + currentans := ans + m[target - sum]increase m[-sum] by 1
for initialize k := 0, when k < row, update (increase k by 1), do −
current := matrix[k, j]
current := matrix[k, j]
if i - 1 >= 0, then −current := current - matrix[k, i - 1]
if i - 1 >= 0, then −
current := current - matrix[k, i - 1]
current := current - matrix[k, i - 1]
sum := sum + current
sum := sum + current
ans := ans + m[target - sum]
ans := ans + m[target - sum]
increase m[-sum] by 1
increase m[-sum] by 1
return ans
return ans
Let us see the following implementation to get better understanding −
 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));
}
{{0,1,0},{1,1,1},{0,1,0}}, 0
4