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Given a n x n grid where each cell contains a character representing its color, determine if it is possible to make a square with equal sides using the same color. To create this square, one must choose any four identical corners in the grid. You are required to check if such a square can be formed. The grid is represented as a list of strings where each string is a row of the grid.

Example:

Input:
grid = [
  "abba",
  "aaaa",
  "bbbb",
  "bbaa"
]

Output:
True

Clarifying Questions

  1. Constraints on the size of the grid?
    • The problem statement mentions n x n, but actual constraints (e.g., maximum value for n) need to be clear.
  2. Character Set for Colors?
    • Should we consider that any character could be in the grid, or are we limited to specific colors?
  3. Single Color Square Only?
    • Are we considering squares with the same color only, or different character squares are allowed, but of the same character?

For now, I’ll assume any character can be present and we’ll focus on finding any character’s matching square.

Strategy

  1. Iterate over each possible n x n square’s top-left corner.
  2. For each starting point, check for different possible side lengths.
  3. Specifically, for each character in the corners of a candidate square, confirm if they match.

Code

def can_form_square(grid):
    n = len(grid)
    
    for i in range(n):
        for j in range(n):
            for k in range(1, n):
                if i + k < n and j + k < n:
                    if (grid[i][j] == grid[i][j + k] and 
                        grid[i][j] == grid[i + k][j] and 
                        grid[i][j] == grid[i + k][j + k]):
                        return True
    return False

# Example usage:
grid = [
  "abba",
  "aaaa",
  "bbbb",
  "bbaa"
]
print(can_form_square(grid))  # Output: True

Time Complexity

The solution involves nested loops:

  1. The first level of nested loops iterate over each cell in the grid.
  2. The third loop iterates over possible side lengths for squares.

Time Complexity: O(n^3), where n is the dimension of the grid: two loops for grid cells, and one for possible side lengths for each cell.

Thus this approach may work efficiently for small grids. For larger grids, optimization strategies could be considered.

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