Count Square Submatrices of 1s in C++

Suppose we have a 2d binary matrix, we have to find the total number of submatrices with all 1 s.

So, if the input is like

then the output will be 10, as there five 1 x 1 matrix, two 2 x 1 matrix. two 1 x 2 matrices. And one 2 x 2 matrix.

To solve this, we will follow these steps −

• Define a function getAns(), this will take an array a,

• ret := 0

• n := size of a

• Define an array v of size n

• Define one stack st

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

  • while (st is not empty and a[top element of st] >= a[i]), do −

    • pop from st

  • if st is not empty, then −

    • prev := top element of st

    • v[i] := v[i] + v[prev]

    • v[i] := v[i] + a[i] * (i - prev)

  • Otherwise

    • v[i] := v[i] + a[i] * (i + 1)

  • insert i into st

• for each i in v −

  • ret := ret + i

• return ret

• From the main method do the following −

• ret := 0

• n := size of v

• m := (if n is non-zero, then size of v[0], otherwise 0)

• Define an array temp of size m

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

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

    • temp[j] := (if v[i, j] is non-zero, then temp[j] + 1, otherwise 0)

  • ret := ret + getAns(temp)

• return ret

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int getAns(vector& a) {
      int ret = 0;
      int n = a.size();
      vector<int> v(n);
      stack<int> st;
      for (int i = 0; i < a.size(); i++) {
         while (!st.empty() && a[st.top()] >= a[i])
            st.pop();
         if(!st.empty()) {
            int prev = st.top();
            v[i] += v[prev];
            v[i] += a[i] * (i - prev);
         }
         else{
            v[i] += a[i] * (i + 1);
         }
         st.push(i);
      }
      for (int i : v) {
         ret += i;
      }
      return ret;
   }
   int solve(vector<vector<int>>& v) {
      int ret = 0;
      int n = v.size();
      int m = n ? v[0].size() : 0;
      vector<int> temp(m);
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < m; j++) {
            temp[j] = v[i][j] ? temp[j] + 1 : 0;
         }
         ret += getAns(temp);
      }
      return ret;
   }
};
int solve(vector<vector<int>>& matrix) {
   return (new Solution())->solve(matrix);
}
main(){
   vector<vector> matrix = {
      {1, 1, 0},
      {1, 1, 0},
      {0, 0, 1}
   };
   cout << solve(matrix);
}

Input

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

Output

10