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Largest Local Values in a Matrix

Easy
53.9%
Updated 6/1/2025

Largest Local Values in a Matrix

1. What is this problem about?

The Largest Local Values in a Matrix is a fundamental matrix processing problem. You are given an n×nn \times n integer matrix. Your task is to generate a new (n2)×(n2)(n-2) \times (n-2) matrix where each element at index (i,j)(i, j) is the maximum value found in the 3×33 \times 3 submatrix centered around the corresponding cell in the original matrix. This is essentially a "max-pooling" operation with a 3×33 \times 3 kernel and a stride of 1, often used in image processing and convolutional neural networks.

2. Why is this asked in interviews?

This Array, Matrix interview pattern is common at companies like Meta and Google for entry-level positions or as a warm-up. It tests basic 2D array indexing, nested loop implementation, and attention to detail regarding grid boundaries. It's a test of whether a candidate can translate a spatial requirement into clean, bug-free code.

3. Algorithmic pattern used

The pattern used is a Sliding Window on a 2D Grid. Since the kernel size is fixed at 3×33 \times 3, we iterate through the original matrix from i=0i=0 to n3n-3 and j=0j=0 to n3n-3. For each (i,j)(i, j), we examine all 9 cells in the 3×33 \times 3 block starting at that position (from row=irow=i to i+2i+2 and col=jcol=j to j+2j+2) and find the maximum value. We then store this maximum in our result matrix.

4. Example explanation

Given a 4×44 \times 4 matrix:

9 9 8 1
5 6 2 6
8 2 6 4
6 2 2 2

The result will be a 2×22 \times 2 matrix.

  • For result(0,0): Max of the top-left 3×33 \times 3 block [[9,9,8],[5,6,2],[8,2,6]] is 9.
  • For result(0,1): Max of the top-right 3×33 \times 3 block [[9,8,1],[6,2,6],[2,6,4]] is 9.
  • For result(1,0): Max of the bottom-left 3×33 \times 3 block [[5,6,2],[8,2,6],[6,2,2]] is 8.
  • For result(1,1): Max of the bottom-right 3×33 \times 3 block [[6,2,6],[2,6,4],[2,2,2]] is 6. Result:
9 9
8 6

5. Common mistakes candidates make

  • Index out of bounds: Not correctly calculating the limits for the outer loops (n2n-2) or the inner 3×33 \times 3 loops.
  • Incorrect initialization: Initializing the "max" variable to 0 when the matrix could potentially contain negative numbers (though usually the problem specifies positive integers).
  • Off-by-one errors: Miscounting the size of the output matrix.
  • Redundant calculations: While not strictly a mistake for 3×33 \times 3, for larger kernels, one might forget that row-wise or column-wise maximums can be precomputed to save time.

6. Interview preparation tip

When working with matrices, always visualize the coordinates on paper. Write down the start and end indices for a few sample submatrices to ensure your loop boundaries are correct. Mastery of nested loops is the foundation for more complex grid-based DP or Graph problems.

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