LeetCode 942. DI String Match

Description

https://leetcode.com/problems/di-string-match/

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'.

Explanation

Create the number list of S length + 1 integers. Iterate the string, if the character is ‘I’, pop the smallest number if the character is ‘D’, pop the biggest number.

Python Solution

class Solution:
    def diStringMatch(self, S: str) -> List[int]:
        n = len(S)
        
        numbers = [i for i in range(n + 1)]
        
        results = []
        
        for c in S:
            if c == 'I':                
                results.append(numbers.pop(0))
            elif c == 'D':
                results.append(numbers.pop())
        
        results.append(numbers.pop())
        
        return results
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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