# Program to count number of square submatix of 1s in the given matrix 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

 1 1 0 1 1 0 0 0 1

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

Updated on: 22-Dec-2020

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