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(As the direct problem statement from LeetCode isn’t available, we can infer a reasonable problem based on typical grid-related conditions.)

Check if Grid Satisfies Conditions

Given an m x n grid, determine if the grid satisfies certain conditions provided below. You have to write a function to check if the grid follows:

  1. Each row or column in the grid contains only even numbers.
  2. Each even number in the grid (if it exists) should be unique within that row or column.

Function Signature:

def checkGridConditions(grid: List[List[int]]) -> bool:

Example:

Constraints:

  1. Grid dimensions (m and n) are >= 1 and <= 100.
  2. Grid elements are integers in the range [0, 100].

Clarifying Questions:

  1. Non-Integer Values: Are non-integer values allowed?
    • Assumption: No, inputs will only be integers as given by constraints.
  2. Negative Numbers: Are negative numbers part of the grid?
    • Assumption: No, only positive integers [0, 100].
  3. Handling Zeros: What should be done with the zero values in the grid?
    • Assumption: Zero values are considered invalid if they appear in the grid, as the problem might intend to check only for positive even numbers.

Strategy:

  1. Traverse Row/Column: We need to do a row-by-row and column-by-column inspection.
  2. Check Even and Unique:
    • Ensure each number is even in the row/column.
    • Ensure no duplicates of even numbers within the same row/column.
  3. Use Data Structures: Utilize sets for checking uniqueness within rows and columns.
  4. Return Results:
    • If all checks pass, return True.
    • If any condition fails, return False.

Code:

from typing import List

def checkGridConditions(grid: List[List[int]]) -> bool:
    m, n = len(grid), len(grid[0])
    
    # Check Rows
    for i in range(m):
        seen = set()
        for j in range(n):
            if grid[i][j] % 2 != 0 or grid[i][j] in seen:
                return False
            seen.add(grid[i][j])
    
    # Check Columns
    for j in range(n):
        seen = set()
        for i in range(m):
            if grid[i][j] % 2 != 0 or grid[i][j] in seen:
                return False
            seen.add(grid[i][j])
    
    return True

# Time Complexity:
# - O(m * n): It involves iterating through each cell of the m x n grid once for rows check and once for columns check.

# Example Usage:
if __name__ == "__main__":
    grid1 = [[2, 4], [4, 2]]
    print(checkGridConditions(grid1))  # Output: True
    grid2 = [[1, 2], [3, 4]]
    print(checkGridConditions(grid2))  # Output: False

This approach ensures every number is checked twice (once row-wise and once column-wise), but it avoids unnecessary nested loops and thus maintains efficiency. Each element is checked once per row and per column, rendering an overall O(m * n) time complexity.

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