Count Submatrices with Top-Left Element and Sum Less Than k

Count Submatrices with Top-Left Element and Sum Less Than k - Problem

You are given a 0-indexed integer matrix grid and an integer k.

Return the number of submatrices that contain the top-left element of the grid, and have a sum less than or equal to k.

A submatrix is a rectangular region within the matrix. For this problem, we only consider submatrices that include the element at position (0,0).

Input & Output

Example 1 — Basic Case

$ Input: grid = [[7,6],[6,4]], k = 13

› Output: 4

💡 Note: All possible submatrices: (0,0)→sum=7≤13 ✓, (0,0)→(0,1)→sum=13≤13 ✓, (0,0)→(1,0)→sum=13≤13 ✓, (0,0)→(1,1)→sum=23>13 ✗. Total: 4 valid submatrices.

Example 2 — Single Element

$ Input: grid = [[7]], k = 6

› Output: 0

💡 Note: Only one submatrix: (0,0) with sum=7 > 6, so no valid submatrices.

Example 3 — All Valid

$ Input: grid = [[1,1],[1,1]], k = 10

› Output: 4

💡 Note: All submatrices have small sums: 1, 2, 2, 4 - all ≤ 10, so count = 4.

Constraints

m == grid.length
n == grid[i].length
1 ≤ m, n ≤ 100
0 ≤ grid[i][j] ≤ 1000
0 ≤ k ≤ 10⁶

Visualization

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Understanding the Visualization

Input

2D grid and threshold k

Process

Check all rectangles from (0,0) to (i,j)

Output

Count of valid submatrices

Key Takeaway

🎯 Key Insight: 2D prefix sums transform O(mn) rectangle sum queries into O(1) lookups

Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is to use 2D prefix sums for fast rectangle sum queries. Build a prefix sum array where prefix[i][j] = sum of rectangle from (0,0) to (i,j). Then check each possible submatrix in O(1) time. Best approach: 2D Prefix Sum. Time: O(mn), Space: O(mn).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(m²n²)	O(1)	Try every possible submatrix starting from (0,0), calculate sum directly
2D Prefix Sum	O(mn)	O(mn)	Pre-compute cumulative sums to get any rectangle sum in O(1) time

Brute Force — Algorithm Steps

For each row i from 0 to m-1
For each column j from 0 to n-1
Sum all elements from (0,0) to (i,j)
If sum ≤ k, increment count

Visualization

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Step-by-Step Walkthrough

Matrix Input

Given grid and k value

Try All Rectangles

For each bottom-right corner, sum rectangle

Count Valid

Count rectangles with sum ≤ k

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int** grid, int gridSize, int* gridColSize, int k) {
    int m = gridSize, n = gridColSize[0];
    int count = 0;
    
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            int total = 0;
            for (int row = 0; row <= i; row++) {
                for (int col = 0; col <= j; col++) {
                    total += grid[row][col];
                }
            }
            if (total <= k) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for small grids
    int** grid = malloc(100 * sizeof(int*));
    int gridColSize[100];
    int gridSize = 0;
    
    // Simplified parsing - assumes small matrix
    int rows = 0, cols = 0;
    for (int i = 0; line[i]; i++) {
        if (line[i] == '[' && i > 0) rows++;
    }
    
    // For demo, assume 2x2 grid
    gridSize = 2;
    gridColSize[0] = gridColSize[1] = 2;
    grid[0] = malloc(2 * sizeof(int));
    grid[1] = malloc(2 * sizeof(int));
    grid[0][0] = 7; grid[0][1] = 6;
    grid[1][0] = 6; grid[1][1] = 4;
    
    int k;
    scanf("%d", &k);
    
    int result = solution(grid, gridSize, gridColSize, k);
    printf("%d\n", result);
    
    free(grid[0]);
    free(grid[1]);
    free(grid);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m²n²)

For each of mn positions, we sum up to mn elements

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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