What data structures are used to solve Construct String with Minimum Cost?

This problem uses Array, String, Dynamic Programming, Trie. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Construct String with Minimum Cost?

Insight 1: Dynamic Programming is suitable for this problem because we can build the solution incrementally, considering prefixes of the target string and the minimum cost to construct them. Insight 2: A Trie data structure can efficiently check if a substring of the target string matches any of the given words.

Which companies ask LeetCode 3213. Construct String with Minimum Cost in interviews?

This problem is frequently asked at Mitsogo.

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Construct String with Minimum Cost - LeetCode 3213 Solution

Q: What is the time complexity of Construct String with Minimum Cost?

The optimal solution has a time complexity of O(n * m), where n is the length of the target string and m is the average length of the words. The nested loop iterates through all possible substrings of the target, and in the worst case, the trie lookup for each substring takes O(m) time..

Q: What is the space complexity of Construct String with Minimum Cost?

The space complexity is O(T + n), where T is the total number of nodes in the Trie and n is the length of the target string. The Trie can store all the words, and the DP array requires O(n) space..

Construct String with Minimum Cost - Complete Solution Guide

Construct String with Minimum Cost is LeetCode problem 3213, 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

You are given a string target , an array of strings words , and an integer array costs , both arrays of the same length. Imagine an empty string s . You can perform the following operation any number of times (including zero ): Choose an index i in the range [0, words.length - 1] . Append words[i] to s . The cost of operation is costs[i] . Return the minimum cost to make s equal to target . If it's not possible, return -1 . Example 1: Input: target = "abcdef", words = ["abdef","abc","d","def","e

Detailed Explanation

The problem asks us to construct a target string 'target' by concatenating strings from a given array 'words'. Each word in 'words' has an associated cost in the 'costs' array. The goal is to find the minimum total cost to construct the 'target' string. If it's impossible to construct the 'target' string using the given words, we should return -1.

Solution Approach

The solution combines dynamic programming and a Trie data structure. We use dynamic programming to find the minimum cost to construct each prefix of the target string. The dp array stores the minimum cost to construct the target string up to a certain index. A Trie is used to efficiently search the 'words' array to find possible matches within the target string.

Step-by-Step Algorithm

Step 1: Create a Trie data structure and insert all the words from the 'words' array into the Trie. Before inserting, store the minimum cost for each unique word using a hash map (or an array and lookup for C implementation). This avoids processing the same word multiple times with varying costs.
Step 2: Initialize a DP array 'dp' of size n+1 (where n is the length of the target string) with infinity. dp[0] = 0, representing the cost to create an empty string.
Step 3: Iterate through the target string from index i = 0 to n-1. If dp[i] is infinity, it means that the prefix target[0...i-1] cannot be constructed, so skip to the next iteration.
Step 4: For each index i, traverse the Trie starting from the root. For each subsequent character target[j] from i to n-1, check if the character exists as a child in the current Trie node.
Step 5: If the character exists, move to the child node. If the child node represents the end of a word (i.e., has a cost associated with it), update dp[j+1] to be the minimum of its current value and dp[i] + cost, where cost is the cost of the word.
Step 6: If the character does not exist in the Trie, stop traversing the Trie for the current value of i.
Step 7: After iterating through the target string, dp[n] will contain the minimum cost to construct the entire target string. If dp[n] is still infinity, it means the target string cannot be constructed, so return -1. Otherwise, return dp[n].

Key Insights

Insight 1: Dynamic Programming is suitable for this problem because we can build the solution incrementally, considering prefixes of the target string and the minimum cost to construct them.
Insight 2: A Trie data structure can efficiently check if a substring of the target string matches any of the given words.
Insight 3: We should only consider the minimum cost for each distinct word, to avoid redundant computations. If a word appears multiple times with different costs, take the smallest cost.

Complexity Analysis

Time Complexity: O(n * m), where n is the length of the target string and m is the average length of the words. The nested loop iterates through all possible substrings of the target, and in the worst case, the trie lookup for each substring takes O(m) time.

Space Complexity: O(T + n), where T is the total number of nodes in the Trie and n is the length of the target string. The Trie can store all the words, and the DP array requires O(n) space.

Topics

This problem involves: Array, String, Dynamic Programming, Trie.

Companies

Asked at: Mitsogo.

Construct String with Minimum Cost | LeetCode 3213

Hard

Array String Dynamic Programming Trie

Time: O(n * m), where n is the length of the target string and m is the average length of the words. The nested loop iterates through all possible substrings of the target, and in the worst case, the trie lookup for each substring takes O(m) time.Space: O(T + n), where T is the total number of nodes in the Trie and n is the length of the target string. The Trie can store all the words, and the DP array requires O(n) space.

Solution Code

Available in:

import java.util.*;

class Solution {
    public int minimumCost(String target, String[] words, int[] costs) {
        int n = target.length();

        Map<Character, Object> trie = new HashMap<>();

        // Use a HashMap to store the minimum cost for each unique word
        Map<String, Integer> wordCosts = new HashMap<>();
        for (int i = 0; i < words.length; i++) {
            String word = words[i];
            int cost = costs[i];
            if (!wordCosts.containsKey(word) || cost < wordCosts.get(word)) {
                wordCosts.put(word, cost);
            }
        }

        // Build the Trie
        for (Map.Entry<String, Integer> entry : wordCosts.entrySet()) {
            String word = entry.getKey();
            int cost = entry.getValue();
            Map<Character, Object> node = trie;
            for (char charr : word.toCharArray()) {
                if (!node.containsKey(charr)) {
                    node.put(charr, new HashMap<Character, Object>());
                }
                node = (Map<Character, Object>) node.get(charr);
            }
            // Store the cost at the terminal node
            node.put("cost", cost);
        }

        int[] dp = new int[n + 1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;

        for (int i = 0; i < n; i++) {
            if (dp[i] == Integer.MAX_VALUE) {
                continue;
            }

            Map<Character, Object> node = trie;
            // Traverse the trie starting from target[i]
            for (int j = i; j < n; j++) {
                char charr = target.charAt(j);
                if (!node.containsKey(charr)) {
                    break;
                }
                node = (Map<Character, Object>) node.get(charr);
                if (node.containsKey("cost")) {
                    int cost = (int) node.get("cost");
                    // Update the dp table for the new prefix
                    dp[j + 1] = Math.min(dp[j + 1], dp[i] + cost);
                }
            }
        }

        int result = dp[n];
        return result != Integer.MAX_VALUE ? result : -1;
    }
}

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Problem Info

Description

Problem 3213: Construct String with Minimum Cost - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming Trie

Companies

Mitsogo

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3213

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

7 steps

1
Step 1: Create a Trie data structure and insert all the words from the 'words' array into the Trie. Before inserting, store the minimum cost for each unique word using a hash map (or an array and lookup for C implementation). This avoids processing the same word multiple times with varying costs.
2
Step 2: Initialize a DP array 'dp' of size n+1 (where n is the length of the target string) with infinity. dp[0] = 0, representing the cost to create an empty string.
3
Step 3: Iterate through the target string from index i = 0 to n-1. If dp[i] is infinity, it means that the prefix target[0...i-1] cannot be constructed, so skip to the next iteration.
4
Step 4: For each index i, traverse the Trie starting from the root. For each subsequent character target[j] from i to n-1, check if the character exists as a child in the current Trie node.
5
Step 5: If the character exists, move to the child node. If the child node represents the end of a word (i.e., has a cost associated with it), update dp[j+1] to be the minimum of its current value and dp[i] + cost, where cost is the cost of the word.
6
Step 6: If the character does not exist in the Trie, stop traversing the Trie for the current value of i.
7
Step 7: After iterating through the target string, dp[n] will contain the minimum cost to construct the entire target string. If dp[n] is still infinity, it means the target string cannot be constructed, so return -1. Otherwise, return dp[n].

💡

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(n * m), where n is the length of the target string and m is the average length of the words. The nested loop iterates through all possible substrings of the target, and in the worst case, the trie lookup for each substring takes O(m) time.

💾

Space Complexity

O(T + n), where T is the total number of nodes in the Trie and n is the length of the target string. The Trie can store all the words, and the DP array requires O(n) space.

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops in the dynamic programming and Trie traversal. The outer loop iterates 'n' times, and the inner loop iterates at most 'n' times. Within the inner loop, Trie lookups take O(m) on average. The space complexity is dominated by the Trie, which can store all the words, and the dp array.

📖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

Brute Force

Try all possible combinations of words to form the target string. This involves trying all possible subsets of 'words' and checking if their concatenation equals 'target'.

Time: O(2^k * n), where k is the number of words and n is the length of the target. The 2^k comes from the potential number of subsets, and the n comes from the string comparison.Space: O(n), primarily for storing the concatenated string being built.

Trade-offs: This approach is simple to understand but highly inefficient for larger input sizes. The DP approach is significantly more efficient.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable for this problem because we can build the solution incrementally, considering prefixes of the target string and the minimum cost to construct them.
2Insight 2: A Trie data structure can efficiently check if a substring of the target string matches any of the given words.
3Insight 3: We should only consider the minimum cost for each distinct word, to avoid redundant computations. If a word appears multiple times with different costs, take the smallest cost.

📚

Key Terms Explained

For Students

Array

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

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: When the target string cannot be constructed from the given words, the solution should return -1. The DP initialization with infinity and checking dp[n] handles this case.
Edge case 2: Empty target string: The algorithm handles this correctly because dp[0] is initialized to 0, which is the correct minimum cost for an empty string.
Edge case 3: If the words list is empty or contains only empty strings, and the target is not empty, the target can't be constructed; this will result in dp[target.length()] remaining infinity, and the function will return -1 as intended.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the minimum cost for a particular word. The solution should prioritize the minimum cost if a word appears multiple times in the input with different costs.

✗Common mistake 2: Not handling the case where the target string cannot be constructed. The DP array initialization and final check for infinity handles this case.

✗Common mistake 3: Incorrect Trie implementation, leading to incorrect matching of substrings.

✅

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: Text segmentation and cost optimization. For example, breaking down a sentence into known phrases with associated costs (e.g., using an API with different pricing for different phrases) to minimize the overall cost.

🏦Real-world scenario 2: DNA sequence assembly, where fragments of DNA (words) need to be assembled into a complete sequence (target) with minimum error rate (cost).

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: November 1, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

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 "Construct String with Minimum Cost"?

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 string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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