Divide Chocolate

To solve this problem, we can use binary search along with a greedy approach. Here’s the step-by-step guide:

  1. Define the Search Range: Since we want to maximize the total sweetness of the piece you can get, we’ll search for the result within the range from 1 to the sum of all sweetness values in the sweetness array.
  2. Binary Search: Perform a binary search to find the maximum total sweetness that can be obtained.
    • The mid-point in the binary search represents the minimum sweetness that you are trying to achieve for your piece.
    • For each mid-point, iterate through the sweetness array and try to divide the chocolate bar into pieces such that each piece has at least the mid-point sweetness.
    • Check whether it is possible to divide the chocolate into k+1 pieces, each with at least the mid-point sweetness. If it’s possible, the mid-point can be a candidate for the answer, and we search in the right half; otherwise, we search in the left half.
  3. Final Answer: The answer will be the maximum mid-point value for which the division is possible.

Here’s the code implementing this approach:

class Solution:
    def maximizeSweetness(self, sweetness: List[int], k: int) -> int:
        def canDivide(mid):
            count = 0
            total_sweetness = 0
            for s in sweetness:
                total_sweetness += s
                if total_sweetness >= mid:
                    total_sweetness = 0
                    count += 1
                    if count > k:
                        return True
            return False

        left, right = 1, sum(sweetness)
        while left < right:
            mid = (left + right + 1) // 2
            if canDivide(mid):
                left = mid
                right = mid - 1

        return right


  • The canDivide function checks whether it’s possible to divide the chocolate into k+1 pieces, each with at least mid sweetness.
  • The binary search helps in efficiently finding the maximum sweetness value for which the division is possible.

This code has a time complexity of O(n log m), where n is the length of the sweetness array, and m is the sum of sweetness values in the array. The space complexity is O(1).

Can you help me with finding the isomorphism for the LeetCode ? Please provide a concise response and if an exact mapping is not possible, specify it as an approximate mapping. If the titles of the problems are the same, select a different problem. Also, explain the reasoning behind your selection. Don’t mention anything about your training cut off date. Provide the best possible answer. Do not repeat the same point. Mention which problem is simpler. If there are multiple isomorphic problem. Arrange them from simple to more complex to your best possible knowledge. Do not add Leetcode to the problem name. The response format is:

The problem can be mapped to

10 Prerequisite LeetCode Problems

“1231. Divide Chocolate” involves binary search on the answer and prefix sum concepts. Here are some simpler problems to prepare for this:

  1. LeetCode 875. Koko Eating Bananas

    • This problem introduces the concept of applying binary search to solve problems involving splitting and partitioning.
  2. LeetCode 1011. Capacity To Ship Packages Within D Days

    • This problem is very similar to the “Divide Chocolate” problem and involves applying binary search to partitioning problems.
  3. LeetCode 410. Split Array Largest Sum

    • This problem is about splitting an array to minimize the largest sum, similar to dividing the chocolate.
  4. LeetCode 209. Minimum Size Subarray Sum

    • This problem involves finding a subarray with a sum above a certain threshold, which is similar to finding a segment of the chocolate bar.
  5. LeetCode 53. Maximum Subarray

    • This problem introduces the concept of finding a maximum sum subarray, similar to finding a segment of the chocolate bar with maximum total sweetness.
  6. LeetCode 704. Binary Search

    • This problem helps you understand the basics of binary search, a concept used in the “Divide Chocolate” problem.
  7. LeetCode 33. Search in Rotated Sorted Array

    • This problem is a more complex binary search problem that can help you understand how to use binary search in more complex scenarios.
  8. LeetCode 1482. Minimum Number of Days to Make m Bouquets

    • This problem uses a similar binary search technique to find the minimum number of days.
  9. LeetCode 300. Longest Increasing Subsequence

    • This problem will help you understand the concept of dynamic programming over sequences, which is needed to understand how to divide the chocolate.
  10. LeetCode 303. Range Sum Query - Immutable

  • This problem involves creating a data structure that can efficiently calculate the sum of a range of numbers, a concept that’s used in the “Divide Chocolate” problem.

These cover how to use binary search and prefix sums to solve problems involving partitioning, which are key to solving the “1231. Divide Chocolate” problem.

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

How did you identify that this problem is a variation of problem?

You have one chocolate bar that consists of some chunks. Each chunk has its own sweetness given by the array sweetness.

You want to share the chocolate with your k friends so you start cutting the chocolate bar into k + 1 pieces using k cuts, each piece consists of some consecutive chunks.

Being generous, you will eat the piece with the minimum total sweetness and give the other pieces to your friends.

Find the maximum total sweetness of the piece you can get by cutting the chocolate bar optimally.

Example 1:

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]

Example 2:

Input: sweetness = [5,6,7,8,9,1,2,3,4], k = 8 Output: 1 Explanation: There is only one way to cut the bar into 9 pieces.

Example 3:

Input: sweetness = [1,2,2,1,2,2,1,2,2], k = 2 Output: 5 Explanation: You can divide the chocolate to [1,2,2], [1,2,2], [1,2,2]


0 <= k < sweetness.length <= 104 1 <= sweetness[i] <= 105