Divide Chocolate

Divide Chocolate - 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.

Input & Output

Example 1 — Basic Case

$ Input: sweetness = [1,2,3,4,5,6,7,8,9], k = 2

› Output: 14

💡 Note: Optimal cuts create pieces [1,2,3,4,5] (sum=15), [6,7] (sum=13), [8,9] (sum=17). You get the minimum piece with sweetness 13. But better cuts [1,2,3,4] (sum=10), [5,6,7] (sum=18), [8,9] (sum=17) give minimum 10. The optimal is actually [1,2,3,4,5,6] (sum=21), [7] (sum=7), [8,9] (sum=17) → min=7. After trying all possibilities, the maximum minimum is 14.

Example 2 — Small Array

$ Input: sweetness = [5,6,7,8,9,1,2,3,4], k = 8

› Output: 1

💡 Note: With k=8 cuts on 9 elements, we create 9 pieces of 1 element each. The minimum will be the smallest element, which is 1.

Example 3 — No Cuts Possible

$ Input: sweetness = [1,2,2,1,2,2,1,2,2], k = 2

› Output: 5

💡 Note: Best cuts create pieces like [1,2,2] (sum=5), [1,2,2] (sum=5), [1,2,2] (sum=5). Each piece has sweetness 5, so minimum is 5.

Constraints

1 ≤ sweetness.length ≤ 10⁴
1 ≤ sweetness[i] ≤ 10⁵
0 ≤ k ≤ sweetness.length - 1

Visualization

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Asked in

G Google 12 f Facebook 8 a Amazon 6

The key insight is to use binary search on the answer space rather than trying all cut combinations. We search for the maximum possible minimum sweetness by testing if each candidate value is achievable through greedy partitioning. Best approach is Binary Search on Answer. Time: O(n × log(sum)), Space: O(1)

Common Approaches

✓ Two Pass Hash

⏱️ Time: N/A Space: N/A

Bfs

⏱️ Time: N/A Space: N/A

Binary Search on Answer

⏱️ Time: O(n × log(sum)) Space: O(1)

Instead of trying all cut combinations, we binary search on the answer space. We guess a minimum sweetness value and check if it's possible to make k cuts such that every piece has at least that sweetness.

Brute Force - Try All Cut Combinations

⏱️ Time: O(C(n,k) × n) Space: O(k)

We try all possible combinations of k cut positions among n-1 possible positions. For each combination, we calculate the sweetness of each piece and find the minimum. We track the maximum of all such minimums.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

💡 Explanation

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