# 791. Custom Sort String

`S`and`T`are strings composed of lowercase letters. In`S`, no letter occurs more than once.

`S`was sorted in some custom order previously. We want to permute the characters of`T`so that they match the order that`S`was sorted. More specifically, if`x`occurs before`y`in`S`, then`x`should occur before`y`in the returned string.

Return any permutation of`T`(as a string) that satisfies this property.

``````Example :
Input:

S = "cba"
T = "abcd"

Output:

Explanation:

"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.
``````

Note:

• `S`has length at most`26`, and no character is repeated in`S`.
• `T`has length at most`200`.
• `S`and`T`consist 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
else:
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

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 {
public:
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 {
public:
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 {
public:
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();
}
``````