What is the time complexity of Minimum Cost to Convert String II?

The optimal solution has a time complexity of O(N^3 + M * S^2).

What is the space complexity of Minimum Cost to Convert String II?

The space complexity is O(N^2 + T).

What data structures are used to solve Minimum Cost to Convert String II?

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

What are the key insights for solving Minimum Cost to Convert String II?

Insight 1: The problem's constraint on operations (disjoint or identical substrings) allows for the use of dynamic programming, as the minimum cost to convert a prefix of the string can be computed based on the minimum cost to convert smaller prefixes. Insight 2: Building a graph of conversions between substrings and finding all-pairs shortest paths using the Floyd-Warshall algorithm is crucial for determining the cost of converting between any two given strings.

Which companies ask LeetCode 2977. Minimum Cost to Convert String II in interviews?

This problem is frequently asked at Atlassian.

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Minimum Cost to Convert String II - LeetCode 2977 Solution

Minimum Cost to Convert String II - Complete Solution Guide

Minimum Cost to Convert String II is LeetCode problem 2977, 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 two 0-indexed strings source and target , both of length n and consisting of lowercase English characters. You are also given two 0-indexed string arrays original and changed , and an integer array cost , where cost[i] represents the cost of converting the string original[i] to the string changed[i] . You start with the string source . In one operation, you can pick a substring x from the string, and change it to y at a cost of z if there exists any index j such that cost[j] == z ,

Detailed Explanation

The problem asks us to find the minimum cost to convert a string `source` to a string `target`. We are given a set of allowed conversions, where converting a substring `original[i]` to `changed[i]` costs `cost[i]`. We can perform multiple conversions as long as the substrings picked are either disjoint or identical. The goal is to return the minimum possible total cost to transform `source` into `target`, or -1 if the transformation is impossible.

Solution Approach

The solution uses a dynamic programming approach combined with the Floyd-Warshall algorithm and a Trie data structure. First, it builds a graph representing the cost to convert one string to another using the given `original`, `changed`, and `cost` arrays. The Floyd-Warshall algorithm is then applied to find the shortest path (minimum cost) between any two strings in the graph. Next, a Trie is constructed to efficiently find all occurrences of convertible substrings in both the `source` and `target` strings. Finally, dynamic programming is used to calculate the minimum cost to convert the `source` string to the `target` string. The `dp[i]` value represents the minimum cost to convert the first `i` characters of `source` to the first `i` characters of `target`.

Step-by-Step Algorithm

Step 1: **Build the Conversion Graph:** Create a graph where nodes represent strings (unique strings from `original` and `changed`). Edges represent the cost to convert one string to another as given in the `cost` array.
Step 2: **Find All-Pairs Shortest Paths (Floyd-Warshall):** Use the Floyd-Warshall algorithm to compute the minimum cost to convert any string to any other string within the graph. This allows us to efficiently determine the cost of converting `original[i]` to `changed[i]` even if there isn't a direct edge.
Step 3: **Construct the Trie:** Build a Trie data structure using the strings from `original` and `changed`. The Trie enables fast substring searching within `source` and `target` strings. Each leaf node stores the index of the string.
Step 4: **Identify Matching Substrings:** Traverse `source` and `target`, using the Trie to identify all possible convertible substrings at each position. Store the indices in source_matches and target_matches where conversion can occur at different length.
Step 5: **Dynamic Programming:** Initialize a DP array `dp` of size `n+1`, where `dp[i]` represents the minimum cost to convert `source[0...i-1]` to `target[0...i-1]`. The base case is `dp[0] = 0`. Iterate from `i = 1` to `n`. For each `i`, consider two cases: (1) If `source[i-1] == target[i-1]`, then `dp[i] = min(dp[i], dp[i-1])` (no operation). (2) For each possible substring ending at `i`, check if a conversion is possible using source_matches and target_matches. If a conversion is possible and has a cost from the Floyd-Warshall result, update `dp[i]` with the minimum cost so far.
Step 6: **Return Result:** After the DP iteration, `dp[n]` contains the minimum cost to convert `source` to `target`. If `dp[n]` is infinity, it means the conversion is impossible, so return -1. Otherwise, return `dp[n]`.

Key Insights

Insight 1: The problem's constraint on operations (disjoint or identical substrings) allows for the use of dynamic programming, as the minimum cost to convert a prefix of the string can be computed based on the minimum cost to convert smaller prefixes.
Insight 2: Building a graph of conversions between substrings and finding all-pairs shortest paths using the Floyd-Warshall algorithm is crucial for determining the cost of converting between any two given strings.
Insight 3: A Trie data structure efficiently identifies all possible substrings of `source` and `target` that can be converted according to the provided `original` and `changed` arrays.

Complexity Analysis

Time Complexity: O(N^3 + M * S^2)

Space Complexity: O(N^2 + T)

Topics

This problem involves: Array, String, Dynamic Programming, Graph, Trie, Shortest Path.

Companies

Asked at: Atlassian.

Minimum Cost to Convert String II | LeetCode 2977

Hard

Array String Dynamic Programming Graph Trie Shortest Path

Time: O(N^3 + M * S^2)Space: O(N^2 + T)

Solution Code

Available in:

import java.util.*;

class Solution {
    public long minimumCost(String source, String target, String[] original, String[] changed, int[] cost) {
        double inf = Double.POSITIVE_INFINITY;

        // Step 1: Build the graph and find all-pairs shortest paths using Floyd-Warshall
        Set<String> all_strings = new HashSet<>(Arrays.asList(original));
        all_strings.addAll(Arrays.asList(changed));
        Map<String, Integer> str_to_idx = new HashMap<>();
        Map<Integer, String> idx_to_str = new HashMap<>();
        int i = 0;
        for (String s : all_strings) {
            str_to_idx.put(s, i);
            idx_to_str.put(i, s);
            i++;
        }
        int num_strings = all_strings.size();

        double[][] dist = new double[num_strings][num_strings];
        for (int row = 0; row < num_strings; row++) {
            for (int col = 0; col < num_strings; col++) {
                dist[row][col] = inf;
            }
        }
        for (int row = 0; row < num_strings; row++) {
            dist[row][row] = 0;
        }

        for (i = 0; i < original.length; i++) {
            int u = str_to_idx.get(original[i]);
            int v = str_to_idx.get(changed[i]);
            dist[u][v] = Math.min(dist[u][v], cost[i]);
        }

        for (int k = 0; k < num_strings; k++) {
            for (int row = 0; row < num_strings; row++) {
                for (int col = 0; col < num_strings; col++) {
                    if (dist[row][k] != inf && dist[k][col] != inf) {
                        dist[row][col] = Math.min(dist[row][col], dist[row][k] + dist[k][col]);
                    }
                }
            }
        }

        // Step 2: Find all occurrences of convertible substrings using a Trie
        Map<Character, Object> trie = new HashMap<>();
        for (Map.Entry<Integer, String> entry : idx_to_str.entrySet()) {
            int s_idx = entry.getKey();
            String s = entry.getValue();
            Map<Character, Object> node = trie;
            for (char char_ : s.toCharArray()) {
                if (!node.containsKey(char_)) {
                    node.put(char_, new HashMap<Character, Object>());
                }
                node = (Map<Character, Object>) node.get(char_);
            }
            ((Map<Character, Object>) node).put('#', s_idx);
        }

        Map<Integer, List<Integer>> source_matches = new HashMap<>();
        for (i = 0; i < source.length(); i++) {
            Map<Character, Object> node = trie;
            for (int j = i; j < source.length(); j++) {
                char char_ = source.charAt(j);
                if (!node.containsKey(char_)) {
                    break;
                }
                node = (Map<Character, Object>) node.get(char_);
                if (((Map<Character, Object>) node).containsKey('#')) {
                    int index = j + 1;
                    source_matches.computeIfAbsent(index, k -> new ArrayList<>()).add((Integer) ((Map<Character, Object>) node).get('#'));
                }
            }
        }

        Map<Integer, List<Integer>> target_matches = new HashMap<>();
        for (i = 0; i < target.length(); i++) {
            Map<Character, Object> node = trie;
            for (int j = i; j < target.length(); j++) {
                char char_ = target.charAt(j);
                if (!node.containsKey(char_)) {
                    break;
                }
                node = (Map<Character, Object>) node.get(char_);
                if (((Map<Character, Object>) node).containsKey('#')) {
                    int index = j + 1;
                    target_matches.computeIfAbsent(index, k -> new ArrayList<>()).add((Integer) ((Map<Character, Object>) node).get('#'));
                }
            }
        }

        // Step 3: Dynamic Programming
        int n = source.length();
        double[] dp = new double[n + 1];
        Arrays.fill(dp, inf);
        dp[0] = 0;

        for (i = 1; i <= n; i++) {
            // Case 1: Single character match (no-op)
            if (source.charAt(i - 1).equals(target.charAt(i - 1))) {
                if (dp[i - 1] != inf) {
                    dp[i] = Math.min(dp[i], dp[i - 1]);
                }
            }

            // Case 2: Substring conversion
            if (source_matches.containsKey(i) && target_matches.containsKey(i)) {
                List<Integer> s_idx_list = source_matches.get(i);
                List<Integer> t_idx_list = target_matches.get(i);

                for (int s_idx : s_idx_list) {
                    for (int t_idx : t_idx_list) {
                        String s_str = idx_to_str.get(s_idx);
                        String t_str = idx_to_str.get(t_idx);

                        if (s_str.length() == t_str.length()) {
                            int k = s_str.length();
                            int j = i - k;
                            if (dp[j] != inf) {
                                double conversion_cost = dist[s_idx][t_idx];
                                if (conversion_cost != inf) {
                                    dp[i] = Math.min(dp[i], dp[j] + conversion_cost);
                                }
                            }
                        }
                    }
                }
            }
        }

        double result = dp[n];
        return (long) (result != inf ? result : -1);
    }
}

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Problem Info

Description

Problem 2977: Minimum Cost to Convert String II - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming Graph Trie Shortest Path

Companies

Atlassian

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2977

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

6 steps

1
Step 1: **Build the Conversion Graph:** Create a graph where nodes represent strings (unique strings from `original` and `changed`). Edges represent the cost to convert one string to another as given in the `cost` array.
2
Step 2: **Find All-Pairs Shortest Paths (Floyd-Warshall):** Use the Floyd-Warshall algorithm to compute the minimum cost to convert any string to any other string within the graph. This allows us to efficiently determine the cost of converting `original[i]` to `changed[i]` even if there isn't a direct edge.
3
Step 3: **Construct the Trie:** Build a Trie data structure using the strings from `original` and `changed`. The Trie enables fast substring searching within `source` and `target` strings. Each leaf node stores the index of the string.
4
Step 4: **Identify Matching Substrings:** Traverse `source` and `target`, using the Trie to identify all possible convertible substrings at each position. Store the indices in source_matches and target_matches where conversion can occur at different length.
5
Step 5: **Dynamic Programming:** Initialize a DP array `dp` of size `n+1`, where `dp[i]` represents the minimum cost to convert `source[0...i-1]` to `target[0...i-1]`. The base case is `dp[0] = 0`. Iterate from `i = 1` to `n`. For each `i`, consider two cases: (1) If `source[i-1] == target[i-1]`, then `dp[i] = min(dp[i], dp[i-1])` (no operation). (2) For each possible substring ending at `i`, check if a conversion is possible using source_matches and target_matches. If a conversion is possible and has a cost from the Floyd-Warshall result, update `dp[i]` with the minimum cost so far.
6
Step 6: **Return Result:** After the DP iteration, `dp[n]` contains the minimum cost to convert `source` to `target`. If `dp[n]` is infinity, it means the conversion is impossible, 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^3 + M * S^2)

💾

Space Complexity

O(N^2 + T)

📝

Why?

Floyd-Warshall's triple nested loop accounts for the O(N^3) term. Trie construction's iteration through all strings contributes to the O(N*S). Matching substrings within source/target dominates because of nested loops to iterate different lengths, giving O(M * S^2). Dynamic programming requires O(M) space for storing prefix costs.

📖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

Dijkstra's Algorithm

Instead of Floyd-Warshall, Dijkstra's algorithm could be used to find the shortest paths from each node to all other nodes. However, since we need all-pairs shortest paths, using Dijkstra's from each node results in similar complexity to Floyd-Warshall.

Time: O(N^3 log N) if using a heap-based priority queue, or O(N^3) with Fibonacci heaps.Space: O(N^2)

Trade-offs: Dijkstra's is more efficient than Floyd-Warshall if the graph is sparse. However, in this case, there is no guarantee of sparseness and using a heap for each Dijkstra execution adds overhead. Floyd-Warshall is generally preferred for dense graphs due to its simpler implementation and constant factors.

✨

Key Insights

1Insight 1: The problem's constraint on operations (disjoint or identical substrings) allows for the use of dynamic programming, as the minimum cost to convert a prefix of the string can be computed based on the minimum cost to convert smaller prefixes.
2Insight 2: Building a graph of conversions between substrings and finding all-pairs shortest paths using the Floyd-Warshall algorithm is crucial for determining the cost of converting between any two given strings.
3Insight 3: A Trie data structure efficiently identifies all possible substrings of `source` and `target` that can be converted according to the provided `original` and `changed` arrays.

📚

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.

Graph

Nodes connected by edges. Like a social network or city map!

String

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

Edge Cases

Edge case 1: `source` and `target` are identical. The algorithm correctly returns 0 as no conversions are needed, because the DP correctly initializes dp[0] = 0, and when source[i-1] == target[i-1], dp[i] = dp[i-1].
Edge case 2: It's impossible to convert `source` to `target`. The algorithm returns -1 because the DP array `dp[n]` remains infinity, indicating that no sequence of conversions can transform `source` to `target`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling disjoint substring conversion constraints correctly. If the conditions aren't explicitly checked, incorrect substring operations might be chained leading to incorrect costs.

✗Common mistake 2: Forgetting to use Floyd-Warshall to precompute all-pairs shortest path leads to only direct edges being considered, and the graph might be disconnected if all original -> changed costs are not directly available.

✅

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: Compiler optimization. In compiler design, this approach can be used to determine the minimum cost to transform one code snippet into another, based on a set of predefined code transformations and associated costs.

🏦Real-world scenario 2: Natural Language Processing (NLP). The problem of converting one string to another is analogous to machine translation. The costs represent the complexities associated with mapping words and phrases from one language to another.

💡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: September 6, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

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 "Minimum Cost to Convert String II"?

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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