LeetCode #3567 — MEDIUM

Minimum Absolute Difference in Sliding Submatrix

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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)

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1,8],[3,-2]]
2

Example 2

[[3,-1]]
1

Example 3

[[1,-2,3],[2,3,5]]
2

Step 02

Core Insight

What unlocks the optimal approach

Use bruteforce over the submatrices

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3567: Minimum Absolute Difference in Sliding Submatrix
class Solution {
    public int[][] minAbsDiff(int[][] grid, int k) {
        int m = grid.length, n = grid[0].length;
        int[][] ans = new int[m - k + 1][n - k + 1];
        for (int i = 0; i <= m - k; i++) {
            for (int j = 0; j <= n - k; j++) {
                List<Integer> nums = new ArrayList<>();
                for (int x = i; x < i + k; x++) {
                    for (int y = j; y < j + k; y++) {
                        nums.add(grid[x][y]);
                    }
                }
                Collections.sort(nums);
                int d = Integer.MAX_VALUE;
                for (int t = 1; t < nums.size(); t++) {
                    int a = nums.get(t - 1);
                    int b = nums.get(t);
                    if (a != b) {
                        d = Math.min(d, Math.abs(a - b));
                    }
                }
                ans[i][j] = (d == Integer.MAX_VALUE) ? 0 : d;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3567: Minimum Absolute Difference in Sliding Submatrix
func minAbsDiff(grid [][]int, k int) [][]int {
	m, n := len(grid), len(grid[0])
	ans := make([][]int, m-k+1)
	for i := range ans {
		ans[i] = make([]int, n-k+1)
	}
	for i := 0; i <= m-k; i++ {
		for j := 0; j <= n-k; j++ {
			var nums []int
			for x := i; x < i+k; x++ {
				for y := j; y < j+k; y++ {
					nums = append(nums, grid[x][y])
				}
			}
			sort.Ints(nums)
			d := math.MaxInt
			for t := 1; t < len(nums); t++ {
				if nums[t] != nums[t-1] {
					diff := abs(nums[t] - nums[t-1])
					if diff < d {
						d = diff
					}
				}
			}
			if d != math.MaxInt {
				ans[i][j] = d
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #3567: Minimum Absolute Difference in Sliding Submatrix
class Solution:
    def minAbsDiff(self, grid: List[List[int]], k: int) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        ans = [[0] * (n - k + 1) for _ in range(m - k + 1)]
        for i in range(m - k + 1):
            for j in range(n - k + 1):
                nums = []
                for x in range(i, i + k):
                    for y in range(j, j + k):
                        nums.append(grid[x][y])
                nums.sort()
                d = min((abs(a - b) for a, b in pairwise(nums) if a != b), default=0)
                ans[i][j] = d
        return ans

// Accepted solution for LeetCode #3567: Minimum Absolute Difference in Sliding Submatrix
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3567: Minimum Absolute Difference in Sliding Submatrix
// class Solution {
//     public int[][] minAbsDiff(int[][] grid, int k) {
//         int m = grid.length, n = grid[0].length;
//         int[][] ans = new int[m - k + 1][n - k + 1];
//         for (int i = 0; i <= m - k; i++) {
//             for (int j = 0; j <= n - k; j++) {
//                 List<Integer> nums = new ArrayList<>();
//                 for (int x = i; x < i + k; x++) {
//                     for (int y = j; y < j + k; y++) {
//                         nums.add(grid[x][y]);
//                     }
//                 }
//                 Collections.sort(nums);
//                 int d = Integer.MAX_VALUE;
//                 for (int t = 1; t < nums.size(); t++) {
//                     int a = nums.get(t - 1);
//                     int b = nums.get(t);
//                     if (a != b) {
//                         d = Math.min(d, Math.abs(a - b));
//                     }
//                 }
//                 ans[i][j] = (d == Integer.MAX_VALUE) ? 0 : d;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3567: Minimum Absolute Difference in Sliding Submatrix
function minAbsDiff(grid: number[][], k: number): number[][] {
    const m = grid.length;
    const n = grid[0].length;
    const ans: number[][] = Array.from({ length: m - k + 1 }, () => Array(n - k + 1).fill(0));
    for (let i = 0; i <= m - k; i++) {
        for (let j = 0; j <= n - k; j++) {
            const nums: number[] = [];
            for (let x = i; x < i + k; x++) {
                for (let y = j; y < j + k; y++) {
                    nums.push(grid[x][y]);
                }
            }
            nums.sort((a, b) => a - b);
            let d = Number.MAX_SAFE_INTEGER;
            for (let t = 1; t < nums.length; t++) {
                if (nums[t] !== nums[t - 1]) {
                    d = Math.min(d, Math.abs(nums[t] - nums[t - 1]));
                }
            }
            ans[i][j] = d === Number.MAX_SAFE_INTEGER ? 0 : d;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.