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;
};