LeetCode #1061 — MEDIUM

Lexicographically Smallest Equivalent String

Move from brute-force thinking to an efficient approach using union-find strategy.

Solve on LeetCode

The Problem

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 = "cde", "acd" and "aab" are equivalent strings of baseStr = "eed", and "aab" is the lexicographically smallest equivalent string of baseStr.

Return the lexicographically smallest equivalent string of baseStr by using the equivalency information from s1 and s2.

Example 1:

Input: s1 = "parker", s2 = "morris", baseStr = "parser"
Output: "makkek"
Explanation: Based on the equivalency information in s1 and s2, we can group their characters as [m,p], [a,o], [k,r,s], [e,i].
The characters in each group are equivalent and sorted in lexicographical order.
So the answer is "makkek".

Example 2:

Input: s1 = "hello", s2 = "world", baseStr = "hold"
Output: "hdld"
Explanation: Based on the equivalency information in s1 and s2, we can group their characters as [h,w], [d,e,o], [l,r].
So only the second letter 'o' in baseStr is changed to 'd', the answer is "hdld".

Example 3:

Input: s1 = "leetcode", s2 = "programs", baseStr = "sourcecode"
Output: "aauaaaaada"
Explanation: We group the equivalent characters in s1 and s2 as [a,o,e,r,s,c], [l,p], [g,t] and [d,m], thus all letters in baseStr except 'u' and 'd' are transformed to 'a', the answer is "aauaaaaada".

Constraints:

1 <= s1.length, s2.length, baseStr <= 1000
s1.length == s2.length
s1, s2, and baseStr consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: 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 = "cde", "acd" and "aab" are equivalent strings of baseStr = "eed", and "aab" is the lexicographically smallest equivalent string of baseStr. Return the lexicographically smallest equivalent string of baseStr by using the equivalency information from s1 and s2.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Union-Find

Example 1

"parker"
"morris"
"parser"

Example 2

"hello"
"world"
"hold"

Example 3

"leetcode"
"programs"
"sourcecode"

Core Insight

What unlocks the optimal approach

Model these equalities as edges of a graph.
Group each connected component of the graph and assign each node of this component to the node with the lowest lexicographically character.
Finally convert the string with the precalculated information.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1061: Lexicographically Smallest Equivalent String
class Solution {
    private final int[] p = new int[26];

    public String smallestEquivalentString(String s1, String s2, String baseStr) {
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < s1.length(); ++i) {
            int x = s1.charAt(i) - 'a';
            int y = s2.charAt(i) - 'a';
            int px = find(x), py = find(y);
            if (px < py) {
                p[py] = px;
            } else {
                p[px] = py;
            }
        }
        char[] s = baseStr.toCharArray();
        for (int i = 0; i < s.length; ++i) {
            s[i] = (char) ('a' + find(s[i] - 'a'));
        }
        return String.valueOf(s);
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

// Accepted solution for LeetCode #1061: Lexicographically Smallest Equivalent String
func smallestEquivalentString(s1 string, s2 string, baseStr string) string {
	p := make([]int, 26)
	for i := 0; i < 26; i++ {
		p[i] = i
	}

	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}

	for i := 0; i < len(s1); i++ {
		x := int(s1[i] - 'a')
		y := int(s2[i] - 'a')
		px := find(x)
		py := find(y)
		if px < py {
			p[py] = px
		} else {
			p[px] = py
		}
	}

	var s []byte
	for i := 0; i < len(baseStr); i++ {
		s = append(s, byte('a'+find(int(baseStr[i]-'a'))))
	}

	return string(s)
}

# Accepted solution for LeetCode #1061: Lexicographically Smallest Equivalent String
class Solution:
    def smallestEquivalentString(self, s1: str, s2: str, baseStr: str) -> str:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(26))
        for a, b in zip(s1, s2):
            x, y = ord(a) - ord("a"), ord(b) - ord("a")
            px, py = find(x), find(y)
            if px < py:
                p[py] = px
            else:
                p[px] = py
        return "".join(chr(find(ord(c) - ord("a")) + ord("a")) for c in baseStr)

// Accepted solution for LeetCode #1061: Lexicographically Smallest Equivalent String
impl Solution {
    pub fn smallest_equivalent_string(s1: String, s2: String, base_str: String) -> String {
        fn find(x: usize, p: &mut Vec<usize>) -> usize {
            if p[x] != x {
                p[x] = find(p[x], p);
            }
            p[x]
        }

        let mut p = (0..26).collect::<Vec<_>>();
        for (a, b) in s1.bytes().zip(s2.bytes()) {
            let x = (a - b'a') as usize;
            let y = (b - b'a') as usize;
            let px = find(x, &mut p);
            let py = find(y, &mut p);
            if px < py {
                p[py] = px;
            } else {
                p[px] = py;
            }
        }

        base_str
            .bytes()
            .map(|c| (b'a' + find((c - b'a') as usize, &mut p) as u8) as char)
            .collect()
    }
}

// Accepted solution for LeetCode #1061: Lexicographically Smallest Equivalent String
function smallestEquivalentString(s1: string, s2: string, baseStr: string): string {
    const p: number[] = Array.from({ length: 26 }, (_, i) => i);

    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };

    for (let i = 0; i < s1.length; i++) {
        const x = s1.charCodeAt(i) - 'a'.charCodeAt(0);
        const y = s2.charCodeAt(i) - 'a'.charCodeAt(0);
        const px = find(x);
        const py = find(y);
        if (px < py) {
            p[py] = px;
        } else {
            p[px] = py;
        }
    }

    const s: string[] = [];
    for (let i = 0; i < baseStr.length; i++) {
        const c = baseStr.charCodeAt(i) - 'a'.charCodeAt(0);
        s.push(String.fromCharCode('a'.charCodeAt(0) + find(c)));
    }
    return s.join('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + m)

Space

O(|\Sigma|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.