LeetCode #2373 — EASY

Largest Local Values in a Matrix

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given an n x n integer matrix grid.

Generate an integer matrix maxLocal of size (n - 2) x (n - 2) such that:

maxLocal[i][j] is equal to the largest value of the 3 x 3 matrix in grid centered around row i + 1 and column j + 1.

In other words, we want to find the largest value in every contiguous 3 x 3 matrix in grid.

Return the generated matrix.

Example 1:

Input: grid = [[9,9,8,1],[5,6,2,6],[8,2,6,4],[6,2,2,2]]
Output: [[9,9],[8,6]]
Explanation: The diagram above shows the original matrix and the generated matrix.
Notice that each value in the generated matrix corresponds to the largest value of a contiguous 3 x 3 matrix in grid.

Example 2:

Input: grid = [[1,1,1,1,1],[1,1,1,1,1],[1,1,2,1,1],[1,1,1,1,1],[1,1,1,1,1]]
Output: [[2,2,2],[2,2,2],[2,2,2]]
Explanation: Notice that the 2 is contained within every contiguous 3 x 3 matrix in grid.

Constraints:

n == grid.length == grid[i].length
3 <= n <= 100
1 <= grid[i][j] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given an n x n integer matrix grid. Generate an integer matrix maxLocal of size (n - 2) x (n - 2) such that: maxLocal[i][j] is equal to the largest value of the 3 x 3 matrix in grid centered around row i + 1 and column j + 1. In other words, we want to find the largest value in every contiguous 3 x 3 matrix in grid. Return the generated matrix.

Baseline thinking

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

Pattern signal: Array

Example 1

[[9,9,8,1],[5,6,2,6],[8,2,6,4],[6,2,2,2]]

Example 2

[[1,1,1,1,1],[1,1,1,1,1],[1,1,2,1,1],[1,1,1,1,1],[1,1,1,1,1]]

Step 02

Core Insight

What unlocks the optimal approach

Use nested loops to run through all possible 3 x 3 windows in the matrix.
For each 3 x 3 window, iterate through the values to get the maximum value within the window.

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 #2373: Largest Local Values in a Matrix
class Solution {
    public int[][] largestLocal(int[][] grid) {
        int n = grid.length;
        int[][] ans = new int[n - 2][n - 2];
        for (int i = 0; i < n - 2; ++i) {
            for (int j = 0; j < n - 2; ++j) {
                for (int x = i; x <= i + 2; ++x) {
                    for (int y = j; y <= j + 2; ++y) {
                        ans[i][j] = Math.max(ans[i][j], grid[x][y]);
                    }
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2373: Largest Local Values in a Matrix
func largestLocal(grid [][]int) [][]int {
	n := len(grid)
	ans := make([][]int, n-2)
	for i := range ans {
		ans[i] = make([]int, n-2)
		for j := 0; j < n-2; j++ {
			for x := i; x <= i+2; x++ {
				for y := j; y <= j+2; y++ {
					ans[i][j] = max(ans[i][j], grid[x][y])
				}
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #2373: Largest Local Values in a Matrix
class Solution:
    def largestLocal(self, grid: List[List[int]]) -> List[List[int]]:
        n = len(grid)
        ans = [[0] * (n - 2) for _ in range(n - 2)]
        for i in range(n - 2):
            for j in range(n - 2):
                ans[i][j] = max(
                    grid[x][y] for x in range(i, i + 3) for y in range(j, j + 3)
                )
        return ans

// Accepted solution for LeetCode #2373: Largest Local Values in a Matrix
// 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 #2373: Largest Local Values in a Matrix
// class Solution {
//     public int[][] largestLocal(int[][] grid) {
//         int n = grid.length;
//         int[][] ans = new int[n - 2][n - 2];
//         for (int i = 0; i < n - 2; ++i) {
//             for (int j = 0; j < n - 2; ++j) {
//                 for (int x = i; x <= i + 2; ++x) {
//                     for (int y = j; y <= j + 2; ++y) {
//                         ans[i][j] = Math.max(ans[i][j], grid[x][y]);
//                     }
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2373: Largest Local Values in a Matrix
function largestLocal(grid: number[][]): number[][] {
    const n = grid.length;
    const res = Array.from({ length: n - 2 }, () => new Array(n - 2).fill(0));
    for (let i = 0; i < n - 2; i++) {
        for (let j = 0; j < n - 2; j++) {
            let max = 0;
            for (let k = i; k < i + 3; k++) {
                for (let z = j; z < j + 3; z++) {
                    max = Math.max(max, grid[k][z]);
                }
            }
            res[i][j] = max;
        }
    }
    return res;
}

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.