942. Di String Match

Easy

A permutation perm of n + 1 integers of all the integers in the range [0, n] can be represented as a string s of length n where:

s[i] == 'I' if perm[i] < perm[i + 1], and
s[i] == 'D' if perm[i] > perm[i + 1].

Given a string s, reconstruct the permutation perm and return it. If there are multiple valid permutations perm, return any of them.

Example 1:

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

Example 2:

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

Example 3:

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

Constraints:

1 <= s.length <= 10⁵
s[i] is either 'I' or 'D'.

 class Solution:
    def diStringMatch(self, s: str) -> List[int]:

        l, h = 0, len(s)
        res = []

        for i, c in enumerate(s):

            if c == "I":
                res.append(l)
                l+=1
            else:
                res.append(h)
                h-=1

        return res+[l]

942. Di String Match

Random Note