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Problem solution · Python

Difference of Number of Distinct Values on Diagonals

Difference of Number of Distinct Values on Diagonals: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
37 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Difference of Number of Distinct Values on Diagonals, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 37 lines of Python from the credited upstream file 2711.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeDifference of Number of Distinct Values on Diagonals · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def differenceOfDistinctValues(self, grid: list[list[int]]) -> list[list[int]]:    m = len(grid)    n = len(grid[0])    ans = [[0] * n for _ in range(m)]     def fillInDiagonal(i: int, j: int) -> None:      topLeft = set()      bottomRight = set()       # Fill in the diagonal from the top-left to the bottom-right.      while i < len(grid) and j < len(grid[0]):        ans[i][j] = len(topLeft)        # Post-addition, so this information can be utilized in subsequent cells.        topLeft.add(grid[i][j])        i += 1        j += 1       i -= 1      j -= 1       # Fill in the diagonal from the bottom-right to the top-left.      while i >= 0 and j >= 0:        ans[i][j] = abs(ans[i][j] - len(bottomRight))        # Post-addition, so this information can be utilized in subsequent cells.        bottomRight.add(grid[i][j])        i -= 1        j -= 1     for i in range(m):      fillInDiagonal(i, 0)     for j in range(1, n):      fillInDiagonal(0, j)     return ans

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