Divide Chocolate - Leetcode Solution

The Divide Chocolate problem on LeetCode asks us to find the maximum total sweetness of the given chocolate bar, which we can divide up into k groups. The sweetness of a group is defined as the sum of the sweetness values of all the pieces in that group. We need to minimize the minimum total sweetness of any group.

Solution:

To solve this problem, we can use a binary search algorithm. We know that the minimum possible sweetness value of any group is 1, and the maximum possible sweetness value is the sum of all the pieces in the chocolate bar. So, we can perform a binary search on this range to find the maximum total sweetness that is possible while minimizing the minimum total sweetness of any group.

For each middle value of the binary search range, we check if it is possible to divide the chocolate bar into k groups such that the minimum total sweetness of any group is less than or equal to the current middle value. To check this, we iterate through the pieces of the chocolate bar one by one and keep adding them to the current group's sweetness total until it exceeds the middle value. When this happens, we start a new group and continue iterating through the remaining pieces. If we are able to create k groups without the minimum sweetness value exceeding the middle value, then we know that we can still increase the minimum sweetness value and search for a greater total sweetness value, so we move the left search pointer to the middle value + 1. If we are not able to create k groups, then we know that we need to reduce the sweetness value, so we move the right search pointer to the middle value - 1.

Finally, the maximum total sweetness value that we have found is the answer to the problem.

Time Complexity:

The binary search algorithm has a time complexity of O(log(sum of all pieces)). Within the binary search, we iterate through all the pieces of the chocolate bar, which takes O(n) time, where n is the number of pieces in the chocolate bar. Therefore, the overall time complexity of the algorithm is O(n log(sum of all pieces)).

Space Complexity:

The algorithm uses constant extra space, so the space complexity is O(1).

Here is the Python code for the solution:

class Solution:
    def maximizeSweetness(self, sweetness: List[int], k: int) -> int:
        left, right = 1, sum(sweetness)
        while left < right:
            mid = (left + right + 1) // 2
            curr_total = 0
            groups = 0
            for s in sweetness:
                curr_total += s
                if curr_total >= mid:
                    groups += 1
                    curr_total = 0
            if groups >= k:
                left = mid
            else:
                right = mid - 1
        return left

In the code, we perform a binary search on the range from 1 to the sum of all the elements in the sweetness list. We first initialize the left and right pointers, then we run the binary search loop until the left pointer is greater than or equal to the right pointer. We calculate the midpoint using integer division and check if it is possible to divide the chocolate bar into k groups such that the minimum total sweetness of any group is less than or equal to the middle value. We iterate through all the pieces of the chocolate bar, keeping track of the current group's total sweetness and when it reaches or exceeds the middle value, we start a new group and reset the current group's total sweetness. If we are able to create k groups with a minimum total sweetness value less than or equal to the middle value, then we move the left pointer to the middle value (since we're looking for the maximum total sweetness value). Otherwise, we move the right pointer to the middle value - 1 (since we need to reduce the minimum total sweetness value). At the end of the binary search, we return the left pointer, which will contain the maximum total sweetness value that satisfies the conditions of the problem.

1