1061. Lexicographically Smallest Equivalent String

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: Union Find

• Time complexity: O(m+n)
• Space complexity: O(26+m -> m)

 

JavaScript

/**
 * @param {string} s1
 * @param {string} s2
 * @param {string} baseStr
 * @return {string}
 */
const smallestEquivalentString = function (s1, s2, baseStr) {
  const n = s1.length;
  const BASE_CODE = 'a'.charCodeAt(0);
  const uf = new UnionFind();
  const result = [];

  for (let index = 0; index < n; index++) {
    const codeA = s1[index].charCodeAt(0) - BASE_CODE;
    const codeB = s2[index].charCodeAt(0) - BASE_CODE;

    uf.union(codeA, codeB);
  }

  for (const letter of baseStr) {
    const code = letter.charCodeAt(0) - BASE_CODE;
    const smallestLetter = String.fromCharCode(uf.find(code) + BASE_CODE);

    result.push(smallestLetter);
  }

  return result.join('');
};

class UnionFind {
  constructor() {
    this.groups = Array.from({ length: 26 }, (_, index) => index);
  }

  find(node) {
    if (this.groups[node] === node) return node;

    this.groups[node] = this.find(this.groups[node]);

    return this.groups[node];
  }

  union(a, b) {
    const groupA = this.find(a);
    const groupB = this.find(b);

    if (groupA === groupB) return;
    if (groupA > groupB) {
      this.groups[groupA] = groupB;
    } else {
      this.groups[groupB] = groupA;
    }
  }
}