Divide Chocolate

Problem Statement

In this scenario, you possess a chocolate bar that is segmented into multiple chunks, each characterized by an individual sweetness value provided by an array named sweetness. Your objective is to share this chocolate with k friends. The task involves making k cuts on the chocolate, thereby splitting it into k + 1 segments, ensuring each segment comprises consecutive chunks from the original array.

Your role as the sharer dictates that you will end up consuming the segment that has the lowest total sweetness out of the k + 1 segments, keeping generosity in mind. The problem asks to determine the strategy for cutting the bar in such a way that the segment you consume has the maximum sweetness possible, relative to all potential ways of segmenting the bar.

This involves finding an optimal cutting strategy that maximizes the minimum sweetness segment that you will eat.

Examples

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]

Constraints

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

Approach and Intuition

The problem can be considered as a form of the "Partition to K Equal Sum Subsets" optimization problem, where the goal is to partition an array in a way that the smallest sum of the partitions is as large as possible. The solution entails a mix of binary search and greedy approaches to effectively find an optimal cut:

1. Greedy Partitioning: Implementing a greedy algorithm helps determine whether it is possible to partition the chocolate bar such that each segment has at least x sweetness. This involves iterating over the sweetness array and partitioning it into segments where each segment at least has x total sweetness until we either successfully create k+1 segments or fail.

2. Binary Search: The key insight for maximizing the minimum sweetness of the consumed segment is utilizing binary search across possible sweetness values (ranging from the minimal sweetness chunk to possibly the sum of all chunks divided by k+1).

   • Initialize the boundaries: Start with low as the smallest sweetness value in the array and high as the sum of all sweetness values divided by k+1. This high boundary ensures that even in the most skewed partition, each part can at least potentially hold this value.
   • Iterate and adjust boundaries: For each midpoint value in the current low to high range, use the greedy algorithm to check if it's possible to partition the bar such that each piece has at least the midpoint sweetness. Adjust the search space based on whether the partitioning is possible:
     • If partitioning is feasible, it implies that a larger minimum segment sweetness might be achievable, so adjust the lower boundary up.
     • If it's not feasible, reduce the upper boundary.

This approach is guided by the constraints and examples given, ensuring computational efficiency given the upper limits of sweetness.length and individual sweetness[i] values. The binary search narrows down the potential values for maximum minimal sweetness efficiently, while the greedy check offers a systematic way to validate each potential cut strategy.

Solutions

class Solution {
public:
    int maximizeSweetness(vector<int>& sweets, int people) {
        int indiv = people + 1;
        int low = *min_element(sweets.begin(), sweets.end());
        int high = accumulate(sweets.begin(), sweets.end(), 0) / indiv;
    
        while (low < high) {
            int median = (low + high + 1) / 2;
            int currentSweet = 0;
            int count = 0;
    
            for (int sweet : sweets) {
                currentSweet += sweet;
                if (currentSweet >= median) {
                    count += 1;
                    currentSweet = 0;
                }
            }
    
            if (count >= indiv) {
                low = median;
            } else {
                high = median - 1;
            }
        }
    
        return high;
    }
};