791. Custom Sort String

SandTare strings composed of lowercase letters. InS, no letter occurs more than once.

Swas sorted in some custom order previously. We want to permute the characters ofTso that they match the order thatSwas sorted. More specifically, ifxoccurs beforeyinS, thenxshould occur beforeyin the returned string.

Return any permutation ofT(as a string) that satisfies this property.

Example :

S = "cba"
T = "abcd"



"a", "b", "c" appear in S, so the order of "a", "b", "c" should be "c", "b", and "a". 
Since "d" does not appear in S, it can be at any position in T. "dcba", "cdba", "cbda" are also valid outputs.


  • Shas length at most26, and no character is repeated inS.
  • Thas length at most200.
  • SandTconsist of lowercase letters only.

Code : Python counting sort

class Solution(object):
    def customSortString(self, S, T):
        :type S: str
        :type T: str
        :rtype: str
        d = {}
        # 1. traverse T to build the counting 
        for c in T:
            if c not in d:
                d[c] = 1
                d[c] += 1

        # 2. traverse S to construct the relative order
        ans = ''
        for c in S:
            for _ in range(d.get(c,0)):
                ans+= c
            # reset the counting:
            d[c] = 0

        # adding in unfined letters
        for i in range(26):
            for j in range(d.get(chr(i + 97),0)):
                ans += chr(i + 97)
        return ans

Code: C++ with Lambda funciton:

class Solution {
    string customSortString(string S, string T) {
        sort(T.begin(), T.end(),
             [&](char a, char b){return S.find(a) < S.find(b);});
        return T;

Code: C++ with vector<int> as counting bucket

class Solution {
    string customSortString(string S, string T) {
        vector<int> cnt(26, 0);
        string res = "";
        // counting in T
        for(auto& c : T) cnt[c-'a']++;
        // retrive by order in S
        for(auto& c : S){
            res.append(cnt[c-'a'], c);
            cnt[c-'a'] = 0;

        // add the residual chars that appears in T but not in S
        for(int i = 0; i < 26; i++) if(cnt[i] > 0) res.append(cnt[i], 'a' + i);       
        return res;     

Code: C++ building map with transform & sort the T

class Solution {
    string customSortString(string S, string T) {
        unordered_map<char, int> d;
        // build the map
        int i = 0;
        transform(S.begin(), S.end(), inserter(d, d.end()),
                 [&](char &a){return make_pair(a, ++i);});

        sort(T.begin(), T.end(),
            [&](char a, char b){return d[a] < d[b];});

        return T;

Code: Java Counting Sort using int[]

 public String customSortString(String S, String T) {
        int[] count = new int[26];
        for (char c : T.toCharArray()) { ++count[c - 'a']; }  // count each char in T.
        StringBuilder sb = new StringBuilder();
        for (char c : S.toCharArray()) {                            
            while (count[c - 'a']-- > 0) { sb.append(c); }    // sort chars both in T and S by the order of S.
        for (char c = 'a'; c <= 'z'; ++c) {
            while (count[c - 'a']-- > 0) { sb.append(c); }   // group chars in T but not in S.
        return sb.toString();

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