algoadvance

You are given an integer array matchsticks where matchsticks[i] is the length of the i-th matchstick. You want to use all the matchsticks to form a square. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Return true if you can make this square and false otherwise.

Clarifying Questions:

  1. Can matchsticks have zero length?
    • No, the lengths of the matchsticks are positive integers.
  2. Can the length of matchsticks be very large?
    • The lengths can be as large as typical integer values within the system constraints.
  3. Is the order of matchsticks in the input array significant?
    • No, the order does not matter.
  4. Can we assume that the input list will not be empty?
    • Yes, the input list will have at least one matchstick.
  5. Is it guaranteed that the number of matchsticks will be within a manageable range for brute force or backtracking solutions?
    • Yes, generally the number of matchsticks will be in a range that can be handled with these techniques.

Strategy:

To determine if the matchsticks can form a square, we can follow these steps:

  1. Check Basic Criteria:
    • Calculate the total length of all matchsticks. If this total is not divisible by 4, it is impossible to form a square.
    • Let’s call the side length of the square side_length = total_length / 4.
  2. Sorting for Optimization:
    • Sort matchsticks in descending order, which helps to place longer matchsticks first and prune the search space more efficiently.
  3. Backtracking Approach:
    • Use backtracking to try and place each matchstick into one of the 4 sides of the square. Track the lengths of the current sides being formed.
    • If at any point a side exceeds the required side length, backtrack.

  4. Implementation:

    Here’s the implementation in Python:

from typing import List

def makesquare(matchsticks: List[int]) -> bool:
    if not matchsticks:
        return False

    total_length = sum(matchsticks)
    if total_length % 4 != 0:
        return False

    side_length = total_length // 4
    matchsticks.sort(reverse=True)  # Sorting matchsticks to optimize the backtracking part

    sides = [0] * 4

    def backtrack(index):
        if index == len(matchsticks):
            # Check if all sides are equal to required side length
            return all(side == side_length for side in sides)
        
        for i in range(4):
            if sides[i] + matchsticks[index] <= side_length:
                sides[i] += matchsticks[index]
                if backtrack(index + 1):
                    return True
                sides[i] -= matchsticks[index]
            
            # If this stick didn't fit in any of the previous sides and the current side is zero, no need to try further.
            if sides[i] == 0:
                break

        return False

    return backtrack(0)

Time Complexity:

Thus, while the worst-case time complexity is high, practical performance is often acceptable due to optimizations from sorting and pruning unnecessary paths.

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