What is the time complexity of Lexicographically Smallest Equivalent String?

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

What data structures are used to solve Lexicographically Smallest Equivalent String?

This problem uses String, Union Find. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Lexicographically Smallest Equivalent String?

Insight 1: The core problem involves identifying equivalence classes of characters. The Union-Find data structure is perfectly suited to efficiently manage these equivalence relations. Insight 2: The lexicographically smallest character within each equivalence class needs to be determined quickly for each character in `baseStr`. Union-Find with path compression ensures efficient finding of the 'root' or representative of each equivalence class, which will always be the smallest character.

Which companies ask LeetCode 1061. Lexicographically Smallest Equivalent String in interviews?

This problem is frequently asked at Cloudera.

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Lexicographically Smallest Equivalent String - LeetCode 1061 Solution

Lexicographically Smallest Equivalent String - Complete Solution Guide

Lexicographically Smallest Equivalent String is LeetCode problem 1061, 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 strings of the same length s1 and s2 and a string baseStr . We say s1[i] and s2[i] are equivalent characters. For example, if s1 = "abc" and s2 = "cde" , then we have 'a' == 'c' , 'b' == 'd' , and 'c' == 'e' . Equivalent characters follow the usual rules of any equivalence relation: Reflexivity: 'a' == 'a' . Symmetry: 'a' == 'b' implies 'b' == 'a' . Transitivity: 'a' == 'b' and 'b' == 'c' implies 'a' == 'c' . For example, given the equivalency information from s1 = "abc" and s2

Detailed Explanation

The problem requires finding the lexicographically smallest equivalent string of a given `baseStr` based on equivalency information provided by two strings `s1` and `s2` of the same length. Two characters `s1[i]` and `s2[i]` are considered equivalent. This equivalence relation follows reflexivity, symmetry, and transitivity. The goal is to replace each character in `baseStr` with the lexicographically smallest character in its equivalence class derived from the `s1` and `s2` pairs.

Solution Approach

The provided code uses the Union-Find algorithm to solve the problem. First, the algorithm iterates through `s1` and `s2`, establishing equivalency relationships between characters using the `union` operation. This operation merges the equivalence classes of two characters. Crucially, the `union` operation always links the character with the *larger* ASCII value to the character with the *smaller* ASCII value. This ensures that the root node of each equivalent set represents the smallest character in that set. Second, the algorithm iterates through `baseStr`. For each character, it finds the root of its equivalence class using the `find` operation (with path compression for efficiency). The `find` operation returns the representative (smallest) character of the equivalence class. Finally, the algorithm replaces the character in `baseStr` with the representative character.

Step-by-Step Algorithm

Step 1: Initialize a `parent` array of size 26, where `parent[i] = i`. This represents that initially, each character ('a' + i) is only equivalent to itself.
Step 2: Iterate through `s1` and `s2` simultaneously. For each index `i`, apply the `union` operation to the characters `s1[i]` and `s2[i]`. The `union` operation merges the equivalence classes of the characters represented by their indices.
Step 3: The `union` operation first determines the roots of the equivalence classes that contain the charaters using the `find` operation. The `find` operation recursively follows the `parent` pointers until it reaches the root of the tree (the representative of the equivalence class). Path compression is implemented to optimize future lookups.
Step 4: In the `union` operation, we compare the roots. The root with the larger ASCII value is made a child of the root with the smaller ASCII value. This ensures that the root will always represent the smallest character in the equivalent set.
Step 5: Iterate through `baseStr`. For each character `c`, find the root of its equivalence class by using `find(c - 'a')`.
Step 6: Append the character represented by the root node to the result string. `(char)(root + 'a')`.
Step 7: Return the resulting string.

Key Insights

Insight 1: The core problem involves identifying equivalence classes of characters. The Union-Find data structure is perfectly suited to efficiently manage these equivalence relations.
Insight 2: The lexicographically smallest character within each equivalence class needs to be determined quickly for each character in `baseStr`. Union-Find with path compression ensures efficient finding of the 'root' or representative of each equivalence class, which will always be the smallest character.
Insight 3: Initializing the parent array with each element being its own parent is crucial. This represents that initially, each character is only equivalent to itself.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Union Find.

Companies

Asked at: Cloudera.

Lexicographically Smallest Equivalent String | LeetCode 1061

Medium

String Union Find

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

Solution Code

Available in:

class Solution {
    public String smallestEquivalentString(String s1, String s2, String baseStr) {
        int[] parent = new int[26];
        for (int i = 0; i < 26; i++) {
            parent[i] = i;
        }

        function(parent);

        for (int i = 0; i < s1.length(); i++) {
            union(parent, s1.charAt(i) - 'a', s2.charAt(i) - 'a');
        }

        StringBuilder result = new StringBuilder();
        for (char c : baseStr.toCharArray()) {
            int root = find(parent, c - 'a');
            result.append((char) (root + 'a'));
        }

        return result.toString();
    }

    private int find(int[] parent, int i) {
        if (parent[i] == i) {
            return i;
        }
        parent[i] = find(parent, parent[i]);
        return parent[i];
    }

    private void union(int[] parent, int i, int j) {
        int root_i = find(parent, i);
        int root_j = find(parent, j);
        if (root_i < root_j) {
            parent[root_j] = root_i;
        } else {
            parent[root_i] = root_j;
        }
    }

    private void function(int[] parent){

    }
}

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Problem Info

Description

Problem 1061: Lexicographically Smallest Equivalent String - Detailed explanation and solution approaches.

Topics

String Union Find

Companies

Cloudera

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1061

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: Initialize a `parent` array of size 26, where `parent[i] = i`. This represents that initially, each character ('a' + i) is only equivalent to itself.
2
Step 2: Iterate through `s1` and `s2` simultaneously. For each index `i`, apply the `union` operation to the characters `s1[i]` and `s2[i]`. The `union` operation merges the equivalence classes of the characters represented by their indices.
3
Step 3: The `union` operation first determines the roots of the equivalence classes that contain the charaters using the `find` operation. The `find` operation recursively follows the `parent` pointers until it reaches the root of the tree (the representative of the equivalence class). Path compression is implemented to optimize future lookups.
4
Step 4: In the `union` operation, we compare the roots. The root with the larger ASCII value is made a child of the root with the smaller ASCII value. This ensures that the root will always represent the smallest character in the equivalent set.
5
Step 5: Iterate through `baseStr`. For each character `c`, find the root of its equivalence class by using `find(c - 'a')`.
6
Step 6: Append the character represented by the root node to the result string. `(char)(root + 'a')`.
7
Step 7: Return the resulting 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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the iterations through `s1`, `s2`, and `baseStr`. The Union-Find operations are effectively constant time due to path compression. The space complexity is constant because the size of the `parent` array does not depend on the input size.

📖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

Depth-First Search (DFS)

An alternative approach would be to use Depth-First Search (DFS) to build the equivalence classes. Create a graph where nodes are characters, and edges represent equivalency. For each character in baseStr, traverse the graph to find the smallest character in its connected component.

Time: O(n + m * 26), where n is the length of s1 and s2, and m is the length of baseStr. For each character in baseStr, we potentially visit all 26 characters.Space: O(26) for the adjacency list representation of the graph.

Trade-offs: DFS can be less efficient than Union-Find, especially with path compression, due to the potential for multiple traversals of the same connected component. Union-Find is generally preferred for its amortized constant-time performance.

✨

Key Insights

1Insight 1: The core problem involves identifying equivalence classes of characters. The Union-Find data structure is perfectly suited to efficiently manage these equivalence relations.
2Insight 2: The lexicographically smallest character within each equivalence class needs to be determined quickly for each character in `baseStr`. Union-Find with path compression ensures efficient finding of the 'root' or representative of each equivalence class, which will always be the smallest character.
3Insight 3: Initializing the parent array with each element being its own parent is crucial. This represents that initially, each character is only equivalent to itself.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty strings `s1`, `s2`, and `baseStr`. The algorithm still functions correctly because the loops will simply not execute, and an empty string will be returned if `baseStr` is empty.
Edge case 2: `s1` and `s2` are the same. The algorithm still correctly identifies that all characters at corresponding positions are equivalent to themselves, and the `baseStr` remains unchanged if no other equivalencies exist.
Edge case 3: `s1` and `s2` have no equivalencies (i.e., all character pairs are distinct). The Union-Find will not merge any sets, and each character in `baseStr` will remain unchanged because its equivalence class contains only itself.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not implementing path compression in the `find` operation, which can lead to significantly degraded performance, especially for larger input strings.

✗Common mistake 2: Incorrectly implementing the `union` operation, such as not consistently choosing the smallest representative as the root of the merged equivalence class, or performing unions without finding roots first.

✗Common mistake 3: Failing to allocate memory for the result string in C, leading to memory errors.

✗Common mistake 4: Not understanding that the 'parent' array relates to index of the character. Misinterpreting it as containing the character itself will not merge correctly.

✅

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: Data normalization. This algorithm could be used to normalize different representations of the same data, ensuring consistency across a dataset. For example, unifying different spellings of the same name.

🏦Real-world scenario 2: Network connectivity analysis. Union-Find can be used to determine connected components in a network, which can be useful for identifying network outages or bottlenecks. By considering two systems connected in s1 and s2 strings, you can find what systems are equivalent in baseStr by mapping them to the lowest label system.

🎮Real-world scenario 3: Image processing. Determining connected regions of pixels with similar colors for image segmentation. Each pixel can be a node and equivalency may relate to color thresholds.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 9, 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

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Lexicographically Smallest Equivalent String"?

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

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