Optimal Partition of String — LeetCode 2405 Python Solution
MediumGreedyHash TableString
- Problem
- #2405
- Pattern
- Greedy
- Reading time
- 2 min
- Source
- leetcode.com
The problem
Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.
Example
- Input
- s = "abacaba"
- Output
- 4
- Explanation
- Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba").
Python solution
Python
class Solution:
def partitionString(self, s: str) -> int:
ans, mask = 1, 0
for x in map(lambda c: ord(c) - ord("a"), s):
if mask >> x & 1:
ans += 1
mask = 0
mask |= 1 << x
return ansComplexity
| Measure | Complexity |
|---|---|
| Time | O(n), where n is the length of the string s |
| Space | O(1) auxiliary |
Pattern: Greedy
Take the locally best option every time — when you can prove that never costs you later. LeetCode 2405. Optimal Partition of String is filed here because LeetCode tags it Greedy, which is the vocabulary this hub collects.
The greedy guide has the Python template for the pattern and the 346 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 2405. Optimal Partition of String?
- LeetCode 2405. Optimal Partition of String is rated Medium on LeetCode.
- What is the time complexity of LeetCode 2405. Optimal Partition of String?
- The Python solution on this page runs in O(n), where n is the length of the string s.
- What is the space complexity of LeetCode 2405. Optimal Partition of String?
- The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 2405. Optimal Partition of String cover?
- LeetCode 2405. Optimal Partition of String is tagged Greedy, Hash Table and String on LeetCode.