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 :



title-img




                        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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