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

The optimal solution has a time complexity of O(n + m).

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

This problem uses Array, String, Graph, Shortest Path. Understanding these concepts is essential for solving this Medium problem efficiently.

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

Insight 1: The problem can be modeled as a shortest path problem on a graph where each character is a node, and the cost to change from one character to another is the weight of the edge connecting them. Insight 2: Since the number of characters is limited (lowercase English letters), we can efficiently use the Floyd-Warshall algorithm to compute the shortest paths between all pairs of characters.

Which companies ask LeetCode 2976. Minimum Cost to Convert String I in interviews?

This problem is frequently asked at Atlassian.

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Minimum Cost to Convert String I - LeetCode 2976 Solution

Minimum Cost to Convert String I - Complete Solution Guide

Minimum Cost to Convert String I is LeetCode problem 2976, a Medium 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 letters. You are also given two 0-indexed character arrays original and changed , and an integer array cost , where cost[i] represents the cost of changing the character original[i] to the character changed[i] . You start with the string source . In one operation, you can pick a character x from the string and change it to the character y at a cost of z if there exists any index j such th

Detailed Explanation

The problem asks us to find the minimum cost to convert a string `source` to a string `target` of the same length. We are given a set of possible character changes with associated costs in the `original`, `changed`, and `cost` arrays, respectively. We can apply these changes an unlimited number of times. The goal is to find the cheapest sequence of changes to transform `source` into `target`. If it's impossible to convert `source` to `target`, we return -1. The key constraint is that changes can only occur if the corresponding `original[i]`, `changed[i]`, and `cost[i]` exist in the input arrays.

Solution Approach

The solution utilizes the Floyd-Warshall algorithm to find the shortest paths between all pairs of characters. It first constructs a cost matrix (distance matrix) representing the cost of changing one character to another. Then, it applies the Floyd-Warshall algorithm to this matrix to compute the minimum cost between any two characters. Finally, it iterates through the `source` and `target` strings and sums up the minimum costs to convert each character in `source` to the corresponding character in `target`. If any character conversion is impossible (i.e., the minimum cost is infinite), it returns -1. Otherwise, it returns the total cost.

Step-by-Step Algorithm

Step 1: Initialize a 2D array `dist` of size 26x26 (for lowercase English letters) to represent the minimum cost between any two characters. Initialize all values to infinity, and set the cost of changing a character to itself to 0.
Step 2: Populate the `dist` array with the initial costs from the `original`, `changed`, and `cost` arrays. `dist[u][v]` will store the cost of changing character `u` to character `v`. If there are multiple ways to change from `u` to `v`, take the minimum cost.
Step 3: Apply the Floyd-Warshall algorithm to compute the shortest paths between all pairs of characters. This involves iterating through all possible intermediate nodes `k`, and updating the `dist[i][j]` value if going from `i` to `j` through `k` is cheaper than the current cost.
Step 4: Iterate through the `source` and `target` strings. For each character pair `source[i]` and `target[i]`, find the minimum cost to convert `source[i]` to `target[i]` using the `dist` array.
Step 5: If the minimum cost for any character conversion is infinity, it means the conversion is impossible. Return -1.
Step 6: Otherwise, sum up the minimum costs for all character conversions. Return the total cost.

Key Insights

Insight 1: The problem can be modeled as a shortest path problem on a graph where each character is a node, and the cost to change from one character to another is the weight of the edge connecting them.
Insight 2: Since the number of characters is limited (lowercase English letters), we can efficiently use the Floyd-Warshall algorithm to compute the shortest paths between all pairs of characters.
Insight 3: If the shortest path between two characters is infinite (i.e., there is no path), then the conversion is impossible. We need to check for this condition when calculating the total cost.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(1)

Topics

This problem involves: Array, String, Graph, Shortest Path.

Companies

Asked at: Atlassian.

Minimum Cost to Convert String I | LeetCode 2976

Medium

Array String Graph Shortest Path

Time: O(n + m)Space: O(1)

Solution Code

Available in:

class Solution {
    public long minimumCost(String source, String target, String[] original, String[] changed, int[] cost) {
        int num_chars = 26;
        double inf = Double.MAX_VALUE;
        
        double[][] dist = new double[num_chars][num_chars];
        
        for (int i = 0; i < num_chars; i++) {
            for (int j = 0; j < num_chars; j++) {
                dist[i][j] = inf;
            }
            dist[i][i] = 0;
        }
        
        for (int i = 0; i < original.length; i++) {
            int u = original[i].charAt(0) - 'a';
            int v = changed[i].charAt(0) - 'a';
            int c = cost[i];
            dist[u][v] = Math.min(dist[u][v], c);
        }
        
        for (int k = 0; k < num_chars; k++) {
            for (int i = 0; i < num_chars; i++) {
                for (int j = 0; j < num_chars; j++) {
                    if (dist[i][k] != inf && dist[k][j] != inf) {
                        dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
                    }
                }
            }
        }
        
        double total_cost = 0;
        for (int i = 0; i < source.length(); i++) {
            int u = source.charAt(i) - 'a';
            int v = target.charAt(i) - 'a';
            
            double conversion_cost = dist[u][v];
            
            if (conversion_cost == inf) {
                return -1;
            }
            
            total_cost += conversion_cost;
        }
        
        return (long) total_cost;
    }
}

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Problem Info

Description

Problem 2976: Minimum Cost to Convert String I - Detailed explanation and solution approaches.

Topics

Array String Graph Shortest Path

Companies

Atlassian

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2976

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: Initialize a 2D array `dist` of size 26x26 (for lowercase English letters) to represent the minimum cost between any two characters. Initialize all values to infinity, and set the cost of changing a character to itself to 0.
2
Step 2: Populate the `dist` array with the initial costs from the `original`, `changed`, and `cost` arrays. `dist[u][v]` will store the cost of changing character `u` to character `v`. If there are multiple ways to change from `u` to `v`, take the minimum cost.
3
Step 3: Apply the Floyd-Warshall algorithm to compute the shortest paths between all pairs of characters. This involves iterating through all possible intermediate nodes `k`, and updating the `dist[i][j]` value if going from `i` to `j` through `k` is cheaper than the current cost.
4
Step 4: Iterate through the `source` and `target` strings. For each character pair `source[i]` and `target[i]`, find the minimum cost to convert `source[i]` to `target[i]` using the `dist` array.
5
Step 5: If the minimum cost for any character conversion is infinity, it means the conversion is impossible. Return -1.
6
Step 6: Otherwise, sum up the minimum costs for all character conversions. Return the total cost.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the Floyd-Warshall algorithm, which is O(k^3) since we are iterating through each possible intermediate node between every pair of nodes. The space complexity is O(k^2) because the `dist` array stores the minimum cost between each pair of characters. 'm' and 'n' can be larger than k^3 but they are bounded by constraints. The k^3 will be 26*26*26 which is not significant compared to the input length in some cases.

📖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, we could use Dijkstra's algorithm for each character in the source string to find the shortest path to the corresponding target character. We would run Dijkstra's algorithm 26 times in the worst case.

Time: O(n*k*(m + k*log(k))), where k is the number of possible characters (26), n is the length of the string, and m is the number of edges/original changes.Space: O(k^2) to store adjacency list and distances

Trade-offs: Dijkstra's algorithm can be more efficient than Floyd-Warshall if the graph is sparse. However, in this problem, the number of characters is small, and the graph is likely to be relatively dense, so Floyd-Warshall is often preferred due to its simplicity and relatively good performance for small graphs.

✨

Key Insights

1Insight 1: The problem can be modeled as a shortest path problem on a graph where each character is a node, and the cost to change from one character to another is the weight of the edge connecting them.
2Insight 2: Since the number of characters is limited (lowercase English letters), we can efficiently use the Floyd-Warshall algorithm to compute the shortest paths between all pairs of characters.
3Insight 3: If the shortest path between two characters is infinite (i.e., there is no path), then the conversion is impossible. We need to check for this condition when calculating the total cost.

📚

Key Terms Explained

For Students

Array

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

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: When the source and target strings are already the same. The cost should be 0 because no conversions are needed. The Floyd-Warshall algorithm ensures that dist[i][i] = 0, so this case is handled correctly.
Edge case 2: When it's impossible to convert a character in the source string to the corresponding character in the target string. The algorithm checks for infinite cost and returns -1.
Edge case 3: When there is no cost to convert between characters and it is interpreted as infinite. The algorithm uses a minimum cost initialization and updates using Math.min to resolve this issue. The Floyd-Warshall will populate the minimum cost or maintain the infinite cost if no path is available.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where it's impossible to convert the source string to the target string, leading to incorrect results.

✗Common mistake 2: Using integer overflow issues when calculating the cumulative cost, especially if there are many character conversions with large costs. Using long long or double helps prevent this.

✗Common mistake 3: Forgetting to initialize the distance matrix correctly (setting diagonal elements to 0 and other elements to infinity).

✅

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: Spell checkers or text editors that suggest corrections by finding the minimum 'edit distance' between misspelled words and correctly spelled words.

🏦Real-world scenario 2: DNA sequencing and analysis, where finding the minimum cost to transform one DNA sequence to another is crucial in understanding evolutionary relationships or identifying mutations.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 5, 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 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Cost to Convert String I"?

For this medium-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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