Question

2373. Largest Local Values in a Matrix

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Description

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

Solutions

Solution 1

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

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 {
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
}

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;
}