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Stickers to Spell Word

Difficulty: Hard

Problem Description

Given n different types of stickers, each with a lowercase English word, you need to spell out the string target by cutting individual letters from your collection of stickers. You can use each sticker an infinite number of times. The goal is to return the minimum number of stickers needed to spell out target. If the task is impossible, return -1.

Key Insights

  • Each sticker can be used multiple times.
  • The problem can be approached using dynamic programming or backtracking with memoization.
  • Counting the frequency of letters in both stickers and the target is essential.
  • A bitmask can be used to represent the letters needed for the target.
  • The solution involves finding the minimal combination of stickers that can cover the needed letters.

Space and Time Complexity

Time Complexity: O(n * 2^m), where n is the number of stickers and m is the length of the target. Space Complexity: O(2^m), for storing the memoization results based on the bitmask of the target.

Solution

The approach involves using a recursive function with memoization to explore combinations of stickers. Each state is represented by a bitmask indicating which letters of the target have been covered. The recursive function tries to cover the remaining letters of the target using each sticker. If a sticker can contribute letters, it updates the state and recursively calls itself to check if the remaining letters can be covered.

Code Solutions

from collections import Counter

def minStickers(stickers, target):
    # Count the frequency of characters in each sticker
    sticker_counts = [Counter(sticker) for sticker in stickers]
    
    # Memoization dictionary
    memo = {}
    
    def dp(remaining_target):
        if remaining_target == "":
            return 0
        if remaining_target in memo:
            return memo[remaining_target]
        
        # Count the frequency of characters in the remaining target
        target_count = Counter(remaining_target)
        min_stickers = float('inf')
        
        # Try each sticker
        for sticker in sticker_counts:
            # If sticker can contribute characters
            if sticker[remaining_target[0]] > 0:  # Only consider stickers that can help
                new_target = []
                for char, count in target_count.items():
                    if count > sticker[char]:
                        new_target.append(char * (count - sticker[char]))
                new_target_str = ''.join(new_target)
                min_stickers = min(min_stickers, 1 + dp(new_target_str))
        
        memo[remaining_target] = min_stickers if min_stickers != float('inf') else -1
        return memo[remaining_target]
    
    result = dp(target)
    return result if result != float('inf') else -1
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