Count Submatrices That Sum Target - Google Top Interview Questions


Problem Statement :


You are given a two-dimensional integer matrix and an integer target.

 Return the number of submatrices in matrix whose sum equals target.

Constraints

0 ≤ n, m ≤ 100 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [

    [0, -1],

    [0, 0]

]

target = 0

Output

5

Explanation

We have three 1 x 1 submatrices, one 2 x 1 submatrix and one 1 x 2 submatrix.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<vector<int>>& matrix, int target) {
    int n = matrix.size();
    int m = n == 0 ? 0 : matrix[0].size();

    if (m > n) {
        vector<vector<int>> transposeMatrix(m, vector<int>(n));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                transposeMatrix[j][i] = matrix[i][j];
            }
        }

        return solve(transposeMatrix, target);
    }

    int res = 0;
    for (int l = 0; l < m; l++) {
        vector<int> a(n);
        for (int r = l; r < m; r++) {
            for (int i = 0; i < n; i++) {
                a[i] += matrix[i][r];
            }

            unordered_map<int, int> prefCnt = {{0, 1}};
            int pref = 0;
            for (int i = 0; i < n; i++) {
                pref += a[i];
                auto it = prefCnt.find(pref - target);
                if (it != end(prefCnt)) res += it->second;
                prefCnt[pref]++;
            }
        }
    }
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[][] matrix, int target) {
        int ret = 0;
        RangeSumMatrix rsm = new RangeSumMatrix(matrix);
        for (int i = 0; i < matrix.length; i++) {
            for (int j = i; j < matrix.length; j++) {
                for (int a = 0; a < matrix[0].length; a++) {
                    for (int b = a; b < matrix[0].length; b++) {
                        if (rsm.total(i, a, j, b) == target) {
                            ret++;
                        }
                    }
                }
            }
        }
        return ret;
    }
}

class RangeSumMatrix {
    private int[][] pf;
    public RangeSumMatrix(int[][] matrix) {
        pf = new int[matrix.length + 1][matrix[0].length + 1];
        for (int i = 1; i <= matrix.length; ++i) {
            for (int j = 1; j <= matrix[0].length; ++j) {
                pf[i][j] = pf[i - 1][j] + pf[i][j - 1] - pf[i - 1][j - 1] + matrix[i - 1][j - 1];
            }
        }
    }

    public int total(int row0, int col0, int row1, int col1) {
        return pf[row1 + 1][col1 + 1] - pf[row0][col1 + 1] - pf[row1 + 1][col0] + pf[row0][col0];
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, matrix, target):
        matrix = [list(accumulate(row, initial=0)) for row in matrix]
        N, M, res = len(matrix), len(matrix[0]), 0

        for col1 in range(1, M):
            for col2 in range(col1, M):
                prev_sums = defaultdict(int, {0: 1})
                run_sum = 0
                for row in range(N):
                    run_sum += matrix[row][col2] - matrix[row][col1 - 1]
                    res += prev_sums[run_sum - target]
                    prev_sums[run_sum] += 1
        return res


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