DI String Match

Description

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".

Solutions

Two passes

class Solution {
public:
    vector<int> diStringMatch(string S) {
        int n = 0;
        vector<int> res(S.size() + 1);
        for (int i = S.size() - 1; i >= 0; --i)
            if (S[i] == 'D')
                res[i + 1] = n++;
        res[0] = n++;
        for (int i = 0; i < S.size(); ++i)
            if (S[i] == 'I')
                res[i + 1] = n++;
        return res;
    }
};

One pass:

class Solution {
public:
    vector<int> diStringMatch(string S) {
        int lo = 0, hi = S.size();
        vector<int> res(S.size() + 1);
        for (int i = S.size() - 1; i >= 0; --i)
            res[i + 1] = S[i] == 'I' ? hi-- : lo++;
        res[0] = lo;
        return res;
    }
};

Left to right:

class Solution {
public:
    vector<int> diStringMatch(string S) {
        int lo = 0, hi = S.size();
        vector<int> res(S.size() + 1);
        for (int i = 0; i < S.size(); ++i)
            res[i] = S[i] == 'I' ? lo++ : hi--;
        res.back() = lo;
        return res;
    }
};