LeetCode #691 — HARD

Stickers to Spell Word

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

We are given n different types of stickers. Each sticker has a lowercase English word on it.

You would like to spell out the given string target by cutting individual letters from your collection of stickers and rearranging them. You can use each sticker more than once if you want, and you have infinite quantities of each sticker.

Return the minimum number of stickers that you need to spell out target. If the task is impossible, return -1.

Note: In all test cases, all words were chosen randomly from the 1000 most common US English words, and target was chosen as a concatenation of two random words.

Example 1:

Input: stickers = ["with","example","science"], target = "thehat"
Output: 3
Explanation:
We can use 2 "with" stickers, and 1 "example" sticker.
After cutting and rearrange the letters of those stickers, we can form the target "thehat".
Also, this is the minimum number of stickers necessary to form the target string.

Example 2:

Input: stickers = ["notice","possible"], target = "basicbasic"
Output: -1
Explanation:
We cannot form the target "basicbasic" from cutting letters from the given stickers.

Constraints:

n == stickers.length
1 <= n <= 50
1 <= stickers[i].length <= 10
1 <= target.length <= 15
stickers[i] and target consist of lowercase English letters.

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: We are given n different types of stickers. Each sticker has a lowercase English word on it. You would like to spell out the given string target by cutting individual letters from your collection of stickers and rearranging them. You can use each sticker more than once if you want, and you have infinite quantities of each sticker. Return the minimum number of stickers that you need to spell out target. If the task is impossible, return -1. Note: In all test cases, all words were chosen randomly from the 1000 most common US English words, and target was chosen as a concatenation of two random words.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

["with","example","science"]
"thehat"

Example 2

["notice","possible"]
"basicbasic"

Related Problems

Ransom Note (ransom-note)

Step 02

Core Insight

What unlocks the optimal approach

We want to perform an exhaustive search, but we need to speed it up based on the input data being random. For all stickers, we can ignore any letters that are not in the target word. When our candidate answer won't be smaller than an answer we have already found, we can stop searching this path. When a sticker dominates another, we shouldn't include the dominated sticker in our sticker collection. [Here, we say a sticker `A` dominates `B` if `A.count(letter) >= B.count(letter)` for all letters.]

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #691: Stickers to Spell Word
class Solution {
    public int minStickers(String[] stickers, String target) {
        int n = target.length();
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(0);
        boolean[] vis = new boolean[1 << n];
        vis[0] = true;
        for (int ans = 0; !q.isEmpty(); ++ans) {
            for (int m = q.size(); m > 0; --m) {
                int cur = q.poll();
                if (cur == (1 << n) - 1) {
                    return ans;
                }
                for (String s : stickers) {
                    int[] cnt = new int[26];
                    int nxt = cur;
                    for (char c : s.toCharArray()) {
                        ++cnt[c - 'a'];
                    }
                    for (int i = 0; i < n; ++i) {
                        int j = target.charAt(i) - 'a';
                        if ((cur >> i & 1) == 0 && cnt[j] > 0) {
                            --cnt[j];
                            nxt |= 1 << i;
                        }
                    }
                    if (!vis[nxt]) {
                        vis[nxt] = true;
                        q.offer(nxt);
                    }
                }
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #691: Stickers to Spell Word
func minStickers(stickers []string, target string) (ans int) {
	n := len(target)
	q := []int{0}
	vis := make([]bool, 1<<n)
	vis[0] = true
	for ; len(q) > 0; ans++ {
		for m := len(q); m > 0; m-- {
			cur := q[0]
			q = q[1:]
			if cur == 1<<n-1 {
				return
			}
			for _, s := range stickers {
				cnt := [26]int{}
				for _, c := range s {
					cnt[c-'a']++
				}
				nxt := cur
				for i, c := range target {
					if cur>>i&1 == 0 && cnt[c-'a'] > 0 {
						nxt |= 1 << i
						cnt[c-'a']--
					}
				}
				if !vis[nxt] {
					vis[nxt] = true
					q = append(q, nxt)
				}
			}
		}
	}
	return -1
}

# Accepted solution for LeetCode #691: Stickers to Spell Word
class Solution:
    def minStickers(self, stickers: List[str], target: str) -> int:
        n = len(target)
        q = deque([0])
        vis = [False] * (1 << n)
        vis[0] = True
        ans = 0
        while q:
            for _ in range(len(q)):
                cur = q.popleft()
                if cur == (1 << n) - 1:
                    return ans
                for s in stickers:
                    cnt = Counter(s)
                    nxt = cur
                    for i, c in enumerate(target):
                        if (cur >> i & 1) == 0 and cnt[c] > 0:
                            cnt[c] -= 1
                            nxt |= 1 << i
                    if not vis[nxt]:
                        vis[nxt] = True
                        q.append(nxt)
            ans += 1
        return -1

// Accepted solution for LeetCode #691: Stickers to Spell Word
use std::collections::{HashSet, VecDeque};

impl Solution {
    pub fn min_stickers(stickers: Vec<String>, target: String) -> i32 {
        let mut q = VecDeque::new();
        q.push_back(0);
        let mut ans = 0;
        let n = target.len();
        let mut vis = HashSet::new();
        vis.insert(0);
        while !q.is_empty() {
            for _ in 0..q.len() {
                let state = q.pop_front().unwrap();
                if state == (1 << n) - 1 {
                    return ans;
                }
                for s in &stickers {
                    let mut nxt = state;
                    let mut cnt = [0; 26];
                    for &c in s.as_bytes() {
                        cnt[(c - b'a') as usize] += 1;
                    }
                    for (i, &c) in target.as_bytes().iter().enumerate() {
                        let idx = (c - b'a') as usize;
                        if (nxt & (1 << i)) == 0 && cnt[idx] > 0 {
                            nxt |= 1 << i;
                            cnt[idx] -= 1;
                        }
                    }
                    if !vis.contains(&nxt) {
                        q.push_back(nxt);
                        vis.insert(nxt);
                    }
                }
            }
            ans += 1;
        }
        -1
    }
}

// Accepted solution for LeetCode #691: Stickers to Spell Word
function minStickers(stickers: string[], target: string): number {
    const n = target.length;
    const q: number[] = [0];
    const vis: boolean[] = Array(1 << n).fill(false);
    vis[0] = true;
    for (let ans = 0; q.length; ++ans) {
        const qq: number[] = [];
        for (const cur of q) {
            if (cur === (1 << n) - 1) {
                return ans;
            }
            for (const s of stickers) {
                const cnt: number[] = Array(26).fill(0);
                for (const c of s) {
                    cnt[c.charCodeAt(0) - 97]++;
                }
                let nxt = cur;
                for (let i = 0; i < n; ++i) {
                    const j = target.charCodeAt(i) - 97;
                    if (((cur >> i) & 1) === 0 && cnt[j]) {
                        nxt |= 1 << i;
                        cnt[j]--;
                    }
                }
                if (!vis[nxt]) {
                    vis[nxt] = true;
                    qq.push(nxt);
                }
            }
        }
        q.splice(0, q.length, ...qq);
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(2^n × m × (l + n)

Space

O(2^n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.