• Easy
• Related Topic: Math

Problem

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 <= S.length <= 10000

S only contains characters “I” or “D”.

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• O(N)

class Solution:    def diStringMatch(self, S):        """        :type S: str        :rtype: List[int]        """        res = []        l = len(S)        h = 0        t = l        for g in S:            if g == "I":                res.append(h)                h += 1            if g == "D":                res.append(t)                t -= 1        res.append(h)        return res