# Lexicographically Smallest Equivalent String

# 1061. Lexicographically Smallest Equivalent String

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". ```

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

/** * @param {string} s1 * @param {string} s2 * @param {string} baseStr * @return {string} */ const smallestEquivalentString = function(s1, s2, baseStr) { const findPosition = (u, arr) => { return arr[u] === u ? u : findPosition(arr[u], arr); }; const alphabet = [...Array(26).keys()]; for (let i = 0; i < s1.length; i++) { const u = findPosition(s1.charCodeAt(i) - 97, alphabet); const v = findPosition(s2.charCodeAt(i) - 97, alphabet); if (u < v) { alphabet[v] = u; } else if (u > v) { alphabet[u] = v; } } let out = ""; for (let i = 0; i < baseStr.length; i++) { out += String.fromCharCode(findPosition(baseStr.charCodeAt(i) - 97, alphabet) + 97); } return out; };