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Formatted question description: https://leetcode.ca/all/2373.html

# 2373. Largest Local Values in a Matrix

• Difficulty: Easy.
• Related Topics: Array, Matrix.
• Similar Questions: .

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public int[][] largestLocal(int[][] grid) {
int n = grid.length;
int[][] res = new int[n - 2][n - 2];
for (int i = 0; i < n - 2; i++) {
for (int j = 0; j < n - 2; j++) {
for (int p = i; p < i + 3; p++) {
for (int q = j; q < j + 3; q++) {
res[i][j] = Math.max(res[i][j], grid[p][q]);
}
}
}
}
return res;
}
}

############

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;
}
}

• 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

############

# 2373. Largest Local Values in a Matrix
# https://leetcode.com/problems/largest-local-values-in-a-matrix/

class Solution:
def largestLocal(self, grid: List[List[int]]) -> List[List[int]]:
n = len(grid)
res = [[0] * (n - 2) for _ in range(n - 2)]

for i in range(n - 2):
for j in range(n - 2):
mmax = 0
for di in range(i, i + 3):
for dj in range(j, j + 3):
mmax = max(mmax, grid[di][dj])

res[i][j] = mmax

return res


• class Solution {
public:
vector<vector<int>> largestLocal(vector<vector<int>>& grid) {
int n = grid.size();
vector<vector<int>> ans(n - 2, vector<int>(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] = max(ans[i][j], grid[x][y]);
}
}
}
}
return ans;
}
};

• 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• 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;
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).