Minimum Absolute Difference in Sliding Submatrix

Minimum Absolute Difference in Sliding Submatrix - Problem

You are given an m × n integer matrix grid and an integer k.

For every contiguous k × k submatrix of grid, compute the minimum absolute difference between any two distinct values within that submatrix.

Return a 2D array ans of size (m - k + 1) × (n - k + 1), where ans[i][j] is the minimum absolute difference in the submatrix whose top-left corner is (i, j) in grid.

Note: If all elements in the submatrix have the same value, the answer will be 0.

Input & Output

Example 1 — Basic 3×3 Grid with k=2

$ Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 2

› Output: [[1,1],[1,1]]

💡 Note: Four 2×2 submatrices: top-left [[1,2],[4,5]] has min diff = min(|1-2|,|1-4|,|1-5|,|2-4|,|2-5|,|4-5|) = 1, top-right [[2,3],[5,6]] has min diff = 1, bottom-left [[4,5],[7,8]] has min diff = 1, bottom-right [[5,6],[8,9]] has min diff = 1

Example 2 — Same Values

$ Input: grid = [[1,1,1],[1,1,1],[1,1,1]], k = 2

› Output: [[0,0],[0,0]]

💡 Note: All elements are identical, so minimum absolute difference is 0 for all 2×2 submatrices

Example 3 — Single Submatrix

$ Input: grid = [[1,5],[3,7]], k = 2

› Output: [[2]]

💡 Note: Only one 2×2 submatrix: [[1,5],[3,7]]. Check pairs: |1-5|=4, |1-3|=2, |1-7|=6, |5-3|=2, |5-7|=2, |3-7|=4. Minimum is 2

Constraints

m == grid.length
n == grid[i].length
1 ≤ k ≤ min(m, n)
1 ≤ grid[i][j] ≤ 10⁹

Visualization

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Understanding the Visualization

Input Grid

3×3 matrix with k=2 sliding window

Extract Submatrices

Find all possible 2×2 submatrices

Compute Results

Calculate minimum difference for each submatrix

Key Takeaway

🎯 Key Insight: Minimum absolute difference in any set of numbers occurs between adjacent elements when sorted

Asked in

G Google 25 a Amazon 18 M Microsoft 12 A Apple 8

The key insight is that after sorting values in each submatrix, the minimum absolute difference must occur between adjacent elements. The optimal approach extracts all k² values from each submatrix, sorts them, then checks only adjacent pairs. Time: O(m×n×k²×log k), Space: O(k²)

Common Approaches

Approach	Time	Space	Notes
✓ Sort Then Compare Adjacent	O(m×n×k²×log(k²))	O(k²)	Sort values in each submatrix, then only check adjacent pairs for minimum difference
Brute Force	O(m×n×k⁴)	O(k²)	Check all pairs in each k×k submatrix to find minimum difference

Sort Then Compare Adjacent — Algorithm Steps

Extract all k² values from each submatrix
Sort the extracted values
Check only adjacent pairs in sorted array

Visualization

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Step-by-Step Walkthrough

Extract & Sort

Get k² values and sort them

Adjacent Check

Only compare neighboring values

Find Minimum

Track smallest adjacent difference

Code -

solution.c — C

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

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

int** solution(int** grid, int m, int n, int k, int* returnSize, int** returnColumnSizes) {
    int resultRows = m - k + 1;
    int resultCols = n - k + 1;
    
    int** result = (int**)malloc(resultRows * sizeof(int*));
    *returnColumnSizes = (int*)malloc(resultRows * sizeof(int));
    
    for (int i = 0; i < resultRows; i++) {
        result[i] = (int*)malloc(resultCols * sizeof(int));
        (*returnColumnSizes)[i] = resultCols;
        
        for (int j = 0; j < resultCols; j++) {
            // Extract all values from k×k submatrix
            int values[k * k];
            int valueCount = 0;
            for (int r = i; r < i + k; r++) {
                for (int c = j; c < j + k; c++) {
                    values[valueCount++] = grid[r][c];
                }
            }
            
            // Sort values
            qsort(values, valueCount, sizeof(int), compare);
            
            // Find minimum difference between adjacent elements
            int minDiff = INT_MAX;
            for (int x = 0; x < valueCount - 1; x++) {
                int diff = values[x + 1] - values[x];
                if (diff < minDiff) {
                    minDiff = diff;
                }
            }
            
            result[i][j] = (minDiff == INT_MAX) ? 0 : minDiff;
        }
    }
    
    *returnSize = resultRows;
    return result;
}

int main() {
    // Simplified parsing for demo
    int grid_data[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    int* grid[3] = {grid_data[0], grid_data[1], grid_data[2]};
    int k = 2;
    
    int returnSize;
    int* returnColumnSizes;
    int** result = solution(grid, 3, 3, k, &returnSize, &returnColumnSizes);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("[");
        for (int j = 0; j < returnColumnSizes[i]; j++) {
            printf("%d", result[i][j]);
            if (j < returnColumnSizes[i] - 1) printf(",");
        }
        printf("]");
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m×n×k²×log(k²))

For each position, sort k² elements taking O(k²log k) time

⚡ Linearithmic

Space Complexity

O(k²)

Temporary array to store and sort k² values from each submatrix

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sort Then Compare Adjacent — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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