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20250113.py
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76 lines (51 loc) · 2.1 KB
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'''
https://leetcode.com/problems/minimum-length-of-string-after-operations/description/
Minimum Length of String After Operations
You are given a string s.
You can perform the following process on s any number of times:
Choose an index i in the string such that there is at least one character to the left of index i that is equal to s[i], and at least one character to the right that is also equal to s[i].
Delete the closest character to the left of index i that is equal to s[i].
Delete the closest character to the right of index i that is equal to s[i].
Return the minimum length of the final string s that you can achieve.
Example 1:
Input: s = "abaacbcbb"
Output: 5
Explanation:
We do the following operations:
Choose index 2, then remove the characters at indices 0 and 3. The resulting string is s = "bacbcbb".
Choose index 3, then remove the characters at indices 0 and 5. The resulting string is s = "acbcb".
Example 2:
Input: s = "aa"
Output: 2
Explanation:
We cannot perform any operations, so we return the length of the original string.
Constraints:
1 <= s.length <= 2 * 105
s consists only of lowercase English letters.
'''
from collections import defaultdict
class Solution:
def minimumLength(self, s: str) -> int:
occurrences = defaultdict(int)
characters = list(s)
for char in range(len(characters)):
occurrences[s[char]] += 1
for key in occurrences:
if occurrences[key] >= 3:
while occurrences[key] >= 3:
occurrences[key] = occurrences[key] - 2
return sum(occurrences.values())
solution = Solution()
print(solution.minimumLength("abaacbcbb")) # -> 5
print(solution.minimumLength("aa")) # -> 2
'''
Pseudocode:
- Precompile all letters and their occurrences
- Loop through the map of letters and their occurrences
- Locate the letters that have occurrences >= 3
- subtract 2 from the value until it the value is < 3
- Keep doing this until all characters in the mapping have < 3 occurrences
- Return the length of the string
- Sum of all values in the occurrences dict
- Note: O(n) time complexity, but could be refactored for better preformance
'''