LeetCode in Kotlin

1061. Lexicographically Smallest Equivalent String

Medium

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.

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    lateinit var parent: IntArray

    fun smallestEquivalentString(s1: String, s2: String, baseStr: String): String? {
        parent = IntArray(26)
        val n = s1.length
        var result = ""
        for (i in 0..25) parent[i] = i
        for (i in 0 until n) {
            union(s1[i].code - 'a'.code, s2[i].code - 'a'.code)
        }
        val base = 'a'.code
        for (element in baseStr) {
            result += (base + find(element.code - 'a'.code)).toChar()
        }
        return result
    }

    private fun union(a: Int, b: Int) {
        val parentA = find(a)
        val parentB = find(b)
        if (parentA != parentB) {
            if (parentA < parentB) {
                parent[parentB] = parentA
            } else {
                parent[parentA] = parentB
            }
        }
    }

    private fun find(x: Int): Int {
        var x = x
        while (parent[x] != x) {
            x = parent[x]
        }
        return x
    }
}

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