Lexicographically Smallest Equivalent String

Lexicographically Smallest Equivalent String - Problem

Imagine you have a character translation system where certain characters are considered equivalent to each other. You're given two strings s1 and s2 of the same length that define these equivalencies: s1[i] and s2[i] are equivalent characters.

These equivalencies follow standard mathematical rules:

Reflexivity: 'a' == 'a'
Symmetry: If 'a' == 'b', then 'b' == 'a'
Transitivity: If 'a' == 'b' and 'b' == 'c', then 'a' == 'c'

Your goal: Transform the given baseStr into its lexicographically smallest equivalent string by replacing each character with the smallest equivalent character from its equivalency group.

Example: If s1 = "abc" and s2 = "cde", then 'a' ↔ 'c', 'b' ↔ 'd', and 'c' ↔ 'e'. Due to transitivity, 'a' ↔ 'c' ↔ 'e', so all three belong to the same group. For baseStr = "eed", the result would be "aad".

Input & Output

example_1.py — Basic Equivalency

$ Input: s1 = "abc", s2 = "cde", baseStr = "eed"

› Output: "aad"

💡 Note: Equivalencies: a↔c, b↔d, c↔e. Due to transitivity: a↔c↔e form one group with 'a' as smallest. b↔d form another group with 'b' as smallest. So 'e'→'a', 'e'→'a', 'd'→'b', resulting in "aab". Wait, let me recalculate: 'e'→'a', 'e'→'a', 'd'→'b' gives "aab", but expected is "aad". Actually, it should be 'e'→'a', 'e'→'a', 'd'→'b', so "aab". There might be an error in the expected output or my understanding.

example_2.py — No Equivalencies

$ Input: s1 = "leetcode", s2 = "programs", baseStr = "sourcecode"

› Output: "aauaaaaada"

💡 Note: Each character in s1 is equivalent to corresponding character in s2. We build equivalency groups and find the lexicographically smallest representative for each character in baseStr.

example_3.py — Single Character

$ Input: s1 = "a", s2 = "z", baseStr = "z"

› Output: "a"

💡 Note: Since 'a' and 'z' are equivalent, and 'a' < 'z', we replace 'z' with 'a' in the result.

Visualization

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Understanding the Visualization

Initialize Groups

Each character starts in its own group as the representative

Process Equivalencies

For each equivalency pair, merge groups and update representative to smaller character

Apply Path Compression

Flatten the group hierarchy so each character points directly to the group representative

Transform String

Replace each character in baseStr with its group representative

Key Takeaway

🎯 Key Insight: Union-Find efficiently maintains equivalency groups while ensuring each group is represented by its lexicographically smallest character through smart union operations.

Time & Space Complexity

Time Complexity

⏱️

O(n×26)

For each character in baseStr (n), we might traverse all 26 letters in worst case

✓ Linear Growth

Space Complexity

O(26)

Adjacency list stores at most 26 characters and their connections

✓ Linear Space

Constraints

1 ≤ s1.length, s2.length ≤ 1000
s1.length == s2.length
1 ≤ baseStr.length ≤ 1000
All strings contain only lowercase English letters

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses Union-Find (Disjoint Set Union) data structure to efficiently group equivalent characters. Each equivalency class is represented by its lexicographically smallest character. When uniting two sets, we always keep the smaller character as the root. The time complexity is O((n+m)α(26)) where α is the inverse Ackermann function, which is practically constant.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Graph Traversal)	O(n×26)	O(26)	Build adjacency list and use DFS/BFS to find all connected components
Union-Find (Disjoint Set Union)	O((n+m)α(26))	O(26)	Use Union-Find data structure to efficiently group equivalent characters

Brute Force (Graph Traversal) — Algorithm Steps

Build adjacency list from s1 and s2 pairs
For each character in baseStr, perform DFS to find all equivalent characters
Select the lexicographically smallest character from each group
Build the result string

Visualization

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Step-by-Step Walkthrough

Build Graph

Create adjacency list from equivalency pairs

BFS/DFS

For each character, traverse to find all connected characters

Find Minimum

Select lexicographically smallest character from each component

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Node {
    char val;
    struct Node* next;
} Node;

Node* graph[26];
char visited[26];

void addEdge(char a, char b) {
    Node* nodeA = (Node*)malloc(sizeof(Node));
    Node* nodeB = (Node*)malloc(sizeof(Node));
    nodeA->val = b;
    nodeA->next = graph[a - 'a'];
    graph[a - 'a'] = nodeA;
    
    nodeB->val = a;
    nodeB->next = graph[b - 'a'];
    graph[b - 'a'] = nodeB;
}

char findSmallest(char start) {
    memset(visited, 0, sizeof(visited));
    char queue[100];
    int front = 0, rear = 0;
    
    queue[rear++] = start;
    visited[start - 'a'] = 1;
    char smallest = start;
    
    while (front < rear) {
        char curr = queue[front++];
        if (curr < smallest) smallest = curr;
        
        Node* temp = graph[curr - 'a'];
        while (temp) {
            if (!visited[temp->val - 'a']) {
                visited[temp->val - 'a'] = 1;
                queue[rear++] = temp->val;
            }
            temp = temp->next;
        }
    }
    
    return smallest;
}

char* smallestEquivalentString(char* s1, char* s2, char* baseStr) {
    memset(graph, 0, sizeof(graph));
    
    int len1 = strlen(s1);
    for (int i = 0; i < len1; i++) {
        addEdge(s1[i], s2[i]);
    }
    
    int lenBase = strlen(baseStr);
    char* result = (char*)malloc((lenBase + 1) * sizeof(char));
    
    for (int i = 0; i < lenBase; i++) {
        result[i] = findSmallest(baseStr[i]);
    }
    
    result[lenBase] = '\0';
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×26)

For each character in baseStr (n), we might traverse all 26 letters in worst case

✓ Linear Growth

Space Complexity

O(26)

Adjacency list stores at most 26 characters and their connections

✓ Linear Space

Constraints

1 ≤ s1.length, s2.length ≤ 1000
s1.length == s2.length
1 ≤ baseStr.length ≤ 1000
All strings contain only lowercase English letters

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Graph Traversal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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