solutions: 2373 - Largest Local Values in a Matrix (Easy)

solutions/2300-2399/2373-largest-local-values-in-a-matrix-easy.md

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+---
+description: 'Author: @wingkwong, @jit | https://leetcode.com/problems/largest-local-values-in-a-matrix/'
+tags: [Array, Matrix]
+---
+
+# 2373 - Largest Local Values in a Matrix (Easy) 
+
+## Problem Link
+
+https://leetcode.com/problems/largest-local-values-in-a-matrix/
+
+## 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`
+
+## Approach 1: Simulation
+
+<Tabs>
+<TabItem value="py" label="Python">
+<SolutionAuthor name="@wingkwong"/>
+
+```py
+class Solution:
+    def largestLocal(self, grid: List[List[int]]) -> List[List[int]]:
+        n = len(grid)
+        # the result matrix is always (n - 2) * (n - 2)
+        res = [[0] * (n - 2) for _ in range(n - 2)]
+        # iterate all possible 3 x 3 grids
+        for i in range(n - 2):
+            for j in range(n - 2):
+                # for (i, j) being at top-left, 
+                # iterate to check the max in this 3 x 3 grid
+                for ii in range(i, i + 3):
+                    for jj in range(j, j + 3):
+                        res[i][j] = max(res[i][j], grid[ii][jj])
+        return res
+```
+
+</TabItem>
+
+<TabItem value="scala" label="scala">
+<SolutionAuthor name="@jit"/>
+
+```scala
+object Solution:
+  def largestLocal(grid: Array[Array[Int]]): Array[Array[Int]] =
+    Array.tabulate(grid.size - 2, grid.size - 2): (i, j) =>
+      grid.slice(i, i + 3).map(_.slice(j, j + 3).max).max
+```
+
+</TabItem>
+</Tabs>