LeetCode 1061. Lexicographically Smallest Equivalent String Solution in Java, C++, Python & More | Explanation + Code

Description

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.

Solutions

Solution 1: Union Find

We can use Union Find (Disjoint Set Union, DSU) to handle the equivalence relations between characters. Each character can be regarded as a node, and the equivalence relations can be seen as edges connecting these nodes. With Union Find, we can group all equivalent characters together and quickly find the representative element for each character during queries. When performing union operations, we always set the representative element to be the lexicographically smallest character. This ensures that the final string is the lexicographically smallest equivalent string.

The time complexity is O((n + m) × log |Σ|) and the space complexity is O(|Σ|), where n is the length of strings s1 and s2, m is the length of baseStr, and |Σ| is the size of the character set, which is 26 in this problem.

PythonJavaC++GoTypeScriptRustC#

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)(code-box)

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]; } }(code-box)

class Solution { public: string smallestEquivalentString(string s1, string s2, string baseStr) { vector<int> p(26); iota(p.begin(), p.end(), 0); auto find = [&](this auto&& find, int x) -> int { if (p[x] != x) { p[x] = find(p[x]); } return p[x]; }; for (int i = 0; i < s1.length(); ++i) { int x = s1[i] - 'a'; int y = s2[i] - 'a'; int px = find(x), py = find(y); if (px < py) { p[py] = px; } else { p[px] = py; } } string s; for (char c : baseStr) { s.push_back('a' + find(c - 'a')); } return s; } };(code-box)

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) }(code-box)

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(''); }(code-box)

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() } }(code-box)

public class Solution { public string SmallestEquivalentString(string s1, string s2, string baseStr) { int[] p = new int[26]; for (int i = 0; i < 26; i++) { p[i] = i; } int Find(int x) { if (p[x] != x) { p[x] = Find(p[x]); } return p[x]; } for (int i = 0; i < s1.Length; i++) { int x = s1[i] - 'a'; int y = s2[i] - 'a'; int px = Find(x); int py = Find(y); if (px < py) { p[py] = px; } else { p[px] = py; } } var res = new System.Text.StringBuilder(); foreach (char c in baseStr) { int idx = Find(c - 'a'); res.Append((char)(idx + 'a')); } return res.ToString(); } }(code-box)