**Count Rectangular Submatrices - Google Top Interview Questions**

### Problem Statement :

Given a two-dimensional list of integers matrix containing 1s and 0s, return the total number of submatrices with all 1 s. Constraints n, m ≤ 250 where n and m are the number of rows and columns in matrix. Example 1 Input matrix = [ [1, 1, 0], [1, 1, 0], [0, 0, 0] ] Output 9 Explanation There's four 1 x 1 matrices. Theres two 2 x 1 matrices. There's two 1 x 2 matrices. And there's one 2 x 2 matrix.

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
int n = matrix.size(), m = matrix[0].size();
vector<vector<int>> dp(n + 1, vector<int>(m));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (!matrix[i][j]) continue;
dp[i][j] = j == 0 ? 1 : dp[i][j - 1] + 1;
}
}
int ans = 0;
for (int j = 0; j < m; ++j) {
stack<pair<int, int>> st;
for (int i = 0; i <= n; ++i) {
int v = dp[i][j], len = 1;
while (!st.empty() && st.top().first > v) {
auto [h, w] = st.top();
st.pop();
int lo = v;
if (!st.empty() && st.top().first >= v) {
lo = st.top().first;
st.top().second += w;
} else
len += w;
ans += (h - lo) * w * (w + 1) / 2;
}
if (!st.empty() && st.top().first == v)
st.top().second++;
else
st.emplace(v, len);
cout << ans << " ";
}
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] a) {
if (a == null || a.length == 0 || a[0].length == 0) {
return 0;
}
int n = a.length;
int m = a[0].length;
int ans = 0;
int[] row = new int[m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (a[i][j] == 0) {
row[j] = 0;
} else {
row[j]++;
}
}
int[] dp = new int[m];
// [1, 2, 3, 7, 4*]
Stack<Integer> stack = new Stack<>();
for (int j = 0; j < m; j++) {
while (stack.size() > 0 && row[stack.peek()] >= row[j]) {
stack.pop();
}
dp[j] = (stack.size() > 0 ? (j - stack.peek()) * row[j] + dp[stack.peek()]
: row[j] * (j + 1));
ans += dp[j];
stack.push(j);
}
}
return ans;
}
}
```

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