791. Custom Sort String791
1. Description
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.
2. Example
Example 1:
Input:
S = “cba”
T = “abcd”
Output: “cbad”
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.
3. 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.
4. Solutions
My Accepted Solution
n = todo.size()
Time complexity: O(n)
Space complexity: O(ALPHABET_COUNT)
class Solution {
public:
// string customSortString(string S, string T)
string customSortString(const string& i_template, string todo) {
array<int, ALPHABET_COUNT> lettersCount{};
for (auto letter : todo) {
++lettersCount[toIndex(letter)];
}
// now, letters in todo is useless, we insert sorted letters
int insertIndex = 0;
for (auto letter : i_template) {
while (lettersCount[toIndex(letter)] > 0) {
todo[insertIndex] = letter;
--lettersCount[toIndex(letter)];
++insertIndex;
}
}
// insert left letters
for (int i = 0; i < ALPHABET_COUNT; ++i) {
while (lettersCount[i] > 0) {
todo[insertIndex] = toLetter(i);
--lettersCount[i];
++insertIndex;
}
}
return todo;
}
private:
// I hate magic numbers
static const int ALPHABET_COUNT = 26;
inline int toIndex(char letter) {
return letter - 'a';
}
inline char toLetter(int index) {
return index + 'a';
}
};