Divide Chocolate — LeetCode 1231 Python Solution
HardLeetCode PremiumArrayBinary Search
- Problem
- #1231
- Pattern
- Monotonic Stack
- Reading time
- 3 min
- Source
- leetcode.com
The problem
You have one chocolate bar that consists of some chunks. Each chunk has its own sweetness given by the array sweetness.
Example
- Input
- sweetness = [1,2,3,4,5,6,7,8,9], k = 5
- Output
- 6
- Explanation
- You can divide the chocolate to [1,2,3], [4,5], [6], [7], [8], [9]
Python solution
Python
class Solution:
def maximizeSweetness(self, sweetness: List[int], k: int) -> int:
def check(x: int) -> bool:
s = cnt = 0
for v in sweetness:
s += v
if s >= x:
s = 0
cnt += 1
return cnt > k
l, r = 0, sum(sweetness)
while l < r:
mid = (l + r + 1) >> 1
if check(mid):
l = mid
else:
r = mid - 1
return lComplexity
| Measure | Complexity |
|---|---|
| Time | O(n \times \log \sum_{i=0}^{n-1} sweetness[i]) |
| Space | O(1) auxiliary |
Pattern: Monotonic Stack
Answer "what is the next greater element" for every position in one pass. LeetCode 1231. Divide Chocolate is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.
The monotonic stack guide has the Python template for the pattern and the 225 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1231. Divide Chocolate?
- LeetCode 1231. Divide Chocolate is rated Hard on LeetCode.
- What is the time complexity of LeetCode 1231. Divide Chocolate?
- The Python solution on this page runs in O(n \times \log \sum_{i=0}^{n-1} sweetness[i]).
- What is the space complexity of LeetCode 1231. Divide Chocolate?
- The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 1231. Divide Chocolate cover?
- LeetCode 1231. Divide Chocolate is tagged Array and Binary Search on LeetCode.
- Is LeetCode 1231. Divide Chocolate a premium problem?
- Yes. LeetCode 1231. Divide Chocolate is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.