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.

解題

Runtime: 0 ms, faster than 100%

Memory Usage: 2.1 MB, less than 100%

func smallestEquivalentString(s1 string, s2 string, baseStr string) string {
    parents := make([]int, 26) // 儲存parents
    for i:=0; i<26; i++ {
        parents[i] = i
    }

    for i:=0; i<len(s1); i++ {
        a, b := int(s1[i] - 'a'), int(s2[i] - 'a')
        pa, pb := find(a, &parents), find(b, &parents)
        
        if pa < pb {
            parents[pb] = pa // 紀錄比較小的當parent
        } else if pa > pb {
            parents[pa] = pb
        }
    }

    ans := make([]byte, 0)
    for i:=0; i<len(baseStr); i++ {
        ans = append(ans, byte(find(int(baseStr[i] - 'a'), &parents) + 'a'))
    }

    return string(ans)
}

func find(x int, parents *[]int) int {
    if (*parents)[x] == x { return x }
    (*parents)[x] = find((*parents)[x], parents)
    return (*parents)[x]
}