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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).

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