942. DI String Match

Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length.

Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

• If S[i] == "I", then A[i] < A[i+1]

• If S[i] == "D", then A[i] > A[i+1]

Example 1:

Input: "IDID"
Output: [0,4,1,3,2]

Example 2:

Input: "III"
Output: [0,1,2,3]

Example 3:

Input: "DDI"
Output: [3,2,0,1]

Note:

1. 1 <= S.length <= 10000

2. S only contains characters "I" or "D".

vector<int> diStringMatch(string S) { // time: O(n); space: O(n)
    int n = S.length(), left = 0, right = n;
    vector<int> res(n + 1);
    for (int i = 0; i < n; ++i) {
        res[i] = S[i] == 'I' ? left++ : right--;
    }
    res[n] = left; // either left or right is okay
    return res;
}