How do you solve Stickers to Spell Word on LeetCode?

Stickers to Spell Word (LeetCode #691) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m * 2^m) time complexity and O(2^m) space complexity. Key insight: Using bitmasking allows us to efficiently track which letters of the target we have formed.

What is the brute force solution for Stickers to Spell Word?

The brute force approach for Stickers to Spell Word is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible combination of stickers to see if we can form the target word. This method is straightforward but inefficient, as it doesn't leverage any optimizations. The optimal Optimal Solution approach improves this to O(n * m * 2^m).

What are the interview tips for Stickers to Spell Word?

Always clarify the problem constraints and edge cases before diving into coding. Discuss your thought process and potential optimizations with the interviewer. Practice explaining your code and the reasoning behind your approach.

What are common mistakes when solving Stickers to Spell Word?

Not considering the case where some letters in the target cannot be formed from the stickers. Failing to optimize the search space by using techniques like bitmasking.

#691

Stickers to Spell Word

Hard

ArrayHash TableStringDynamic ProgrammingBacktrackingBit ManipulationMemoizationBitmaskBitmaskingDynamic ProgrammingBreadth-First Search (BFS)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m * 2^m)
Space	O(1)	O(2^m)

💡

Intuition

Time O(n * m * 2^m)Space O(2^m)

The optimal solution uses dynamic programming and bitmasking to efficiently track which letters of the target can be formed. By representing the target as a bitmask, we can quickly check combinations of stickers and avoid redundant calculations.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency count for each sticker and the target.
2Step 2: Use a bitmask to represent the letters in the target that have been formed.
3Step 3: Use a queue for BFS to explore combinations of stickers, updating the bitmask as we go.
4Step 4: Track the minimum number of stickers used to reach the full target bitmask.

solution.py30 lines

1from collections import defaultdict, deque
2
3def minStickers(stickers, target):
4    target_count = [0] * 26
5    for char in target:
6        target_count[ord(char) - ord('a')] += 1
7
8    sticker_count = []
9    for sticker in stickers:
10        count = [0] * 26
11        for char in sticker:
12            count[ord(char) - ord('a')] += 1
13        sticker_count.append(count)
14
15    target_mask = (1 << len(target)) - 1
16    queue = deque([(0, 0)])
17    visited = set([0])
18    while queue:
19        mask, steps = queue.popleft()
20        for sticker in sticker_count:
21            new_mask = mask
22            for i in range(len(target)):
23                if (mask & (1 << i)) == 0 and sticker[ord(target[i]) - ord('a')] > 0:
24                    new_mask |= (1 << i)
25            if new_mask == target_mask:
26                return steps + 1
27            if new_mask not in visited:
28                visited.add(new_mask)
29                queue.append((new_mask, steps + 1))
30    return -1

ℹ

Complexity note: This complexity arises from the number of states (2^m) we can have for the target letters, multiplied by the number of stickers (n) and the length of the target (m) for processing.

1Using bitmasking allows us to efficiently track which letters of the target we have formed.
2Dynamic programming helps avoid redundant calculations by storing previously computed results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.