LeetCode 691: Stickers to Spell Word Solution

Master LeetCode problem 691 (Stickers to Spell Word), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

691. Stickers to Spell Word

Hard

Array Hash Table String Dynamic Programming Backtracking Bit Manipulation Memoization Bitmask

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming with bitmasking. We can represent the state of the stickers using a bitmask, where each bit corresponds to whether a sticker is used or not. We can iterate through all possible states of stickers and try to form the target string using the stickers. If a state can form the target string, we update the minimum number of stickers required to reach that state.

Create a bitmask for each sticker to represent whether a sticker is used or not.
Use dynamic programming to iterate through all possible states of stickers.
For each state, try to form the target string using the stickers.
Update the minimum number of stickers required to form the target string.

Time complexity: O(2^n * m * t), where n is the number of stickers, m is the maximum length of a sticker, and t is the length of the target string. Space complexity: O(2^n)

Solution Code

class Solution {
    public int minStickers(String[] stickers, String target) {
        int n = stickers.length;
        int[] dp = new int[1 << target.length()];
        Arrays.fill(dp, -1);
        dp[0] = 0;

        for (int state = 0; state < (1 << target.length()); state++) {
            if (dp[state] == -1) continue;

            for (String sticker : stickers) {
                int newState = state;
                for (char c : sticker.toCharArray()) {
                    for (int i = 0; i < target.length(); i++) {
                        if (((newState >> i) & 1) == 0 && target.charAt(i) == c) {
                            newState |= (1 << i);
                            break;
                        }
                    }
                }

                if (dp[newState] == -1 || dp[newState] > dp[state] + 1) {
                    dp[newState] = dp[state] + 1;
                }
            }
        }

        return dp[(1 << target.length()) - 1];
    }
}

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