LeetCode-in-Java

3567. Minimum Absolute Difference in Sliding Submatrix

Medium

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

For every contiguous k x 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) x (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.

A submatrix (x1, y1, x2, y2) is a matrix that is formed by choosing all cells matrix[x][y] where x1 <= x <= x2 and y1 <= y <= y2.

Example 1:

Input: grid = [[1,8],[3,-2]], k = 2

Output: [[2]]

Explanation:

• There is only one possible k x k submatrix: [[1, 8], [3, -2]].
• Distinct values in the submatrix are [1, 8, 3, -2].
• The minimum absolute difference in the submatrix is |1 - 3| = 2. Thus, the answer is [[2]].

Example 2:

Input: grid = [[3,-1]], k = 1

Output: [[0,0]]

Explanation:

• Both k x k submatrix has only one distinct element.
• Thus, the answer is [[0, 0]].

Example 3:

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

Output: [[1,2]]

Explanation:

• There are two possible k × k submatrix:
  • Starting at (0, 0): [[1, -2], [2, 3]].
    • Distinct values in the submatrix are [1, -2, 2, 3].
    • The minimum absolute difference in the submatrix is |1 - 2| = 1.
  • Starting at (0, 1): [[-2, 3], [3, 5]].
    • Distinct values in the submatrix are [-2, 3, 5].
    • The minimum absolute difference in the submatrix is |3 - 5| = 2.
• Thus, the answer is [[1, 2]].

Constraints:

• 1 <= m == grid.length <= 30
• 1 <= n == grid[i].length <= 30
• -105 <= grid[i][j] <= 105
• 1 <= k <= min(m, n)

Solution

import java.util.Arrays;

public class Solution {
    public int[][] minAbsDiff(int[][] grid, int k) {
        int rows = grid.length;
        int cols = grid[0].length;
        int[][] result = new int[rows - k + 1][cols - k + 1];
        for (int x = 0; x <= rows - k; x++) {
            for (int y = 0; y <= cols - k; y++) {
                int size = k * k;
                int[] elements = new int[size];
                int idx = 0;
                for (int i = x; i < x + k; i++) {
                    for (int j = y; j < y + k; j++) {
                        elements[idx++] = grid[i][j];
                    }
                }
                Arrays.sort(elements);
                int minDiff = Integer.MAX_VALUE;
                for (int i = 1; i < size; i++) {
                    if (elements[i] != elements[i - 1]) {
                        minDiff = Math.min(minDiff, elements[i] - elements[i - 1]);
                        if (minDiff == 1) {
                            break;
                        }
                    }
                }
                result[x][y] = minDiff == Integer.MAX_VALUE ? 0 : minDiff;
            }
        }
        return result;
    }
}

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