What is the time complexity of Stickers to Spell Word?

The optimal solution has a time complexity of O(2^t * n).

What is the space complexity of Stickers to Spell Word?

The space complexity is O(2^t + n*s).

What data structures are used to solve Stickers to Spell Word?

This problem uses Array, Hash Table, String, Dynamic Programming, Backtracking, Bit Manipulation, Memoization, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Stickers to Spell Word?

Insight 1: The problem can be modeled as a state-space search problem. The state represents the remaining characters needed to form the target string, and the goal is to reach a state where no characters are needed. Insight 2: Dynamic programming (memoization) is crucial for avoiding redundant computations. The number of possible states is exponential in the length of the target string, so a naive recursive approach would be too slow.

Which companies ask LeetCode 691. Stickers to Spell Word in interviews?

This problem is frequently asked at IXL.

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Stickers to Spell Word - LeetCode 691 Solution

Stickers to Spell Word - Complete Solution Guide

Stickers to Spell Word is LeetCode problem 691, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 random

Detailed Explanation

The problem asks us to find the minimum number of stickers needed to spell out a given target string. We are given a set of stickers, each containing a lowercase English word. We can use each sticker an unlimited number of times, cutting out individual letters and rearranging them to form the target string. If it's impossible to spell the target string using the given stickers, we should return -1.

Solution Approach

The provided code uses a top-down dynamic programming approach with memoization. The `solve` function recursively explores the state space, representing the remaining characters needed to form the target string. It returns the minimum number of stickers needed to reach the base case (an empty string, requiring 0 stickers) or -1 if it's impossible. The memoization table stores the results of previously computed states to avoid redundant calculations. A pre-check is also performed to determine if the target string can even be formed by the stickers given.

Step-by-Step Algorithm

Step 1: Preprocess the stickers: Count the occurrences of each character in each sticker.
Step 2: Initialize memoization table: Use a dictionary (Python) or hash map (Java, C++) to store the minimum number of stickers needed for each possible target string.
Step 3: Define the recursive solve function: This function takes the remaining target string as input.
Step 4: Base case: If the target string is empty, return 0 (no stickers needed).
Step 5: Memoization check: If the result for the current target string is already in the memoization table, return it.
Step 6: Find the first character in target string that needs coverage.
Step 7: Iterate through stickers: For each sticker, check if it contains the first needed character of the target string. If it does, simulate using that sticker by reducing the counts of needed characters. Recursively call the solve function with the new target state.
Step 8: Update minimum stickers: Keep track of the minimum number of stickers needed among all possible sticker choices.
Step 9: Store result in memoization table: Store the minimum number of stickers (or -1 if impossible) for the current target string in the memoization table.
Step 10: Check if the target string is even possible by verifying if all target characters exist in at least one of the stickers.
Step 11: Return the result: Return the minimum number of stickers needed to spell out the target string.

Key Insights

Insight 1: The problem can be modeled as a state-space search problem. The state represents the remaining characters needed to form the target string, and the goal is to reach a state where no characters are needed.
Insight 2: Dynamic programming (memoization) is crucial for avoiding redundant computations. The number of possible states is exponential in the length of the target string, so a naive recursive approach would be too slow.
Insight 3: Optimizing the sticker selection process is essential. Instead of trying all stickers for each state, we only need to consider stickers that contain the leftmost character of the remaining target string. This significantly reduces the branching factor of the search tree.
Insight 4: Pre-processing the stickers to count the occurrences of each character can speed up the calculation of the next state. Converting the target string to a character count array provides efficient access to the characters still needed.

Complexity Analysis

Time Complexity: O(2^t * n)

Space Complexity: O(2^t + n*s)

Topics

This problem involves: Array, Hash Table, String, Dynamic Programming, Backtracking, Bit Manipulation, Memoization, Bitmask.

Companies

Asked at: IXL.

Stickers to Spell Word | LeetCode 691

Hard

Array Hash Table String Dynamic Programming Backtracking Bit Manipulation Memoization Bitmask

Time: O(2^t * n)Space: O(2^t + n*s)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int minStickers(String[] stickers, String target) {
        int n = stickers.length;
        int[][] sticker_counts = new int[n][26];
        for (int i = 0; i < n; i++) {
            String sticker = stickers[i];
            for (char char_ : sticker.toCharArray()) {
                sticker_counts[i][char_ - 'a']++;
            }
        }

        Map<String, Integer> memo = new HashMap<>();

        
        
        
        
        
        
        

        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        

        

        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        

        int[] initial_target_counts = new int[26];
        for (char char_ : target.toCharArray()) {
            initial_target_counts[char_ - 'a']++;
        }

        int[] all_sticker_chars_coverage = new int[26];
        for (int[] sc : sticker_counts) {
            for (int i = 0; i < 26; i++) {
                if (sc[i] > 0) {
                    all_sticker_chars_coverage[i] = 1;
                }
            }
        }

        for (int i = 0; i < 26; i++) {
            if (initial_target_counts[i] > 0 && all_sticker_chars_coverage[i] == 0) {
                return -1;
            }
        }

        int result = solve(arrToString(initial_target_counts), sticker_counts, n, memo);
        return result;
    }

    private int solve(String target_tuple, int[][] sticker_counts, int n, Map<String, Integer> memo) {
        int[] targetArr = stringToArr(target_tuple);
        if (Arrays.stream(targetArr).sum() == 0) {
            return 0;
        }

        if (memo.containsKey(target_tuple)) {
            return memo.get(target_tuple);
        }

        int min_stickers = Integer.MAX_VALUE;

        int first_char_idx = -1;
        for (int i = 0; i < 26; i++) {
            if (targetArr[i] > 0) {
                first_char_idx = i;
                break;
            }
        }

        for (int i = 0; i < n; i++) {
            if (sticker_counts[i][first_char_idx] == 0) {
                continue;
            }

            int[] next_target = Arrays.copyOf(targetArr, targetArr.length);
            for (int j = 0; j < 26; j++) {
                next_target[j] = Math.max(0, next_target[j] - sticker_counts[i][j]);
            }

            int res = solve(arrToString(next_target), sticker_counts, n, memo);
            if (res != -1) {
                min_stickers = Math.min(min_stickers, 1 + res);
            }
        }

        int result = (min_stickers == Integer.MAX_VALUE) ? -1 : min_stickers;
        memo.put(target_tuple, result);
        return result;
    }

    private String arrToString(int[] arr) {
        return Arrays.toString(arr);
    }

    private int[] stringToArr(String str) {
        str = str.substring(1, str.length() - 1);
        String[] values = str.split(", ");
        int[] arr = new int[values.length];
        for (int i = 0; i < values.length; i++) {
            arr[i] = Integer.parseInt(values[i]);
        }
        return arr;
    }
}

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Problem Info

Description

Problem 691: Stickers to Spell Word - Detailed explanation and solution approaches.

Topics

Array Hash Table String Dynamic Programming Backtracking Bit Manipulation Memoization Bitmask

Companies

IXL

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:691

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Preprocess the stickers: Count the occurrences of each character in each sticker.
2
Step 2: Initialize memoization table: Use a dictionary (Python) or hash map (Java, C++) to store the minimum number of stickers needed for each possible target string.
3
Step 3: Define the recursive solve function: This function takes the remaining target string as input.
4
Step 4: Base case: If the target string is empty, return 0 (no stickers needed).
5
Step 5: Memoization check: If the result for the current target string is already in the memoization table, return it.
6
Step 6: Find the first character in target string that needs coverage.
7
Step 7: Iterate through stickers: For each sticker, check if it contains the first needed character of the target string. If it does, simulate using that sticker by reducing the counts of needed characters. Recursively call the solve function with the new target state.
8
Step 8: Update minimum stickers: Keep track of the minimum number of stickers needed among all possible sticker choices.
9
Step 9: Store result in memoization table: Store the minimum number of stickers (or -1 if impossible) for the current target string in the memoization table.
10
Step 10: Check if the target string is even possible by verifying if all target characters exist in at least one of the stickers.
11
Step 11: Return the result: Return the minimum number of stickers needed to spell out the target string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(2^t * n)

💾

Space Complexity

O(2^t + n*s)

📝

Why?

The time complexity is derived from the recursive nature of the `solve` function. In the worst case, each character in the target can either be provided by a sticker, or not, leading to O(2^t) possible states to explore. For each state, we iterate through all `n` stickers, leading to a factor of `n`. The space complexity is from the memoization which in the worst case stores results for all possible subsets of target characters, and the sticker counts.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

Use BFS to explore the state space, starting from the initial target string and iteratively reducing it by applying stickers. This approach can be more efficient if the optimal solution is found relatively quickly.

Time: O(V + E), where V is the number of possible target states and E is the number of possible transitions between states. Which is approximately O(2^t * n) in worst case.Space: O(V), where V is number of possible states i.e. O(2^t) in worst case.

Trade-offs: BFS can be more efficient than DFS with memoization in some cases, especially when the optimal solution is found early in the search. However, BFS can require more memory if the state space is very large.

A* Search

Use A* search to find the minimum number of stickers. A heuristic function can estimate the remaining cost (number of stickers) to reach the empty target. This approach prioritizes more promising paths.

Time: Depends on the quality of the heuristic and the structure of the problem. In the worst case, it can be exponential. O(2^t * n) in worst case.Space: O(V), where V is the number of visited nodes. Can be O(2^t) in worst case.

Trade-offs: A* can be much faster than BFS or DFS with memoization if a good heuristic is available. However, designing an effective heuristic can be challenging.

✨

Key Insights

1Insight 1: The problem can be modeled as a state-space search problem. The state represents the remaining characters needed to form the target string, and the goal is to reach a state where no characters are needed.
2Insight 2: Dynamic programming (memoization) is crucial for avoiding redundant computations. The number of possible states is exponential in the length of the target string, so a naive recursive approach would be too slow.
3Insight 3: Optimizing the sticker selection process is essential. Instead of trying all stickers for each state, we only need to consider stickers that contain the leftmost character of the remaining target string. This significantly reduces the branching factor of the search tree.
4Insight 4: Pre-processing the stickers to count the occurrences of each character can speed up the calculation of the next state. Converting the target string to a character count array provides efficient access to the characters still needed.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Target string is empty. The solution should return 0 because no stickers are needed.
Edge case 2: It is impossible to form the target string using the given stickers. The solution should return -1. This edge case is handled by checking coverage of all characters of the target before doing the recursion.
Edge case 3: A sticker is empty. This is not explicitly handled but will not cause an error because the inner loop avoids stickers with a zero count for the needed character.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization. This will lead to exponential time complexity and a time limit exceeded error.

✗Common mistake 2: Not handling the case where it's impossible to form the target string. This requires checking if all the letters of the target string are in at least one of the stickers.

✗Common mistake 3: Incorrectly updating character counts when applying stickers. It's important to ensure that the counts are correctly reduced and that the resulting counts are not negative.

✗Common mistake 4: Using the wrong type of key in the memoization table (e.g., using a string instead of a hashable object in Python, like a tuple).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Inventory management. Determining the minimum number of different types of materials (stickers) needed to fulfill a specific order (target string).

🏦Real-world scenario 2: Text generation. Finding the smallest set of text snippets (stickers) that can be combined to generate a desired message (target string), like composing marketing copy.

🎮Real-world scenario 3: Resource allocation. Given a set of resources (stickers) with different capabilities, finding the minimum number of resources needed to achieve a specific goal (target string of capabilities needed).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Stickers to Spell Word"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?