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'
ifperm[i] < perm[i + 1]
, ands[i] == 'D'
ifperm[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).