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You are given an integer array forts of length n representing the positions of enemy forts, your forts, and empty positions between them. The value of forts[i] can be one of three possible scenarios:

• 1 represents your fort at position i
• -1 represents an enemy fort at position i
• 0 represents an empty position at position i

You want to calculate the maximum number of enemy forts that can be captured in a single move. A move involves capturing all enemy forts between two of your forts (i.e., between two 1s) without encountering any other obstruction (enemy forts are not obstructed in this case). Therefore, the forts you capture must be uninterrupted. The task is to return the maximum number of contiguous -1 values that can be captured in a single move.

Clarifying Questions

• Does a move have to start and end with your forts (i.e., 1 on both ends)?
  • Yes, that is correct.
• Are there any constraints on the length of the array or input values?
  • Constraints: n == forts.length where 1 <= n <= 1000 and the values in forts can only be -1, 0, or 1.

Strategy

1. Traverse through the array to identify possible segments between every two 1s.
2. For each segment, compute the number of -1s it contains.
3. Keep track of the maximum number of enemy forts (-1s) that can be captured between any two 1s.
4. Return the maximum count found.

This algorithm involves a single pass through the array (O(n)) with some additional operations which makes it efficient given the constraints.

Code

def captureForts(forts):
    n = len(forts)
    max_enemies = 0
    i = 0
    
    while i < n:
        if forts[i] == 1:
            j = i + 1
            enemy_count = 0
            while j < n and forts[j] != 1:
                if forts[j] == -1:
                    enemy_count += 1
                j += 1
            if j < n and forts[j] == 1:  # Valid segment between two '1's
                max_enemies = max(max_enemies, enemy_count)
            i = j  # Move to the next potential starting '1'
        else:
            i += 1
    return max_enemies

Time Complexity

• Time Complexity: O(n), where n is the length of the forts array. This linear complexity arises from traversing the array once and checking each segment.
• Space Complexity: O(1), as we are using a constant amount of extra space, just variables for counting and tracking the maximum.

By following these steps, the function efficiently determines the maximum number of contiguous enemy forts that can be captured in a single move.

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