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Divide Chocolate

Number: 1192

Difficulty: Hard

Paid? Yes

Companies: Google

Problem Description

You are given a chocolate bar represented as an array of positive integers where each integer denotes the sweetness of a chunk. You need to make k cuts to split the chocolate into k+1 contiguous pieces. Being generous, you will eat the piece with the minimum total sweetness. The goal is to maximize the total sweetness of the piece you get by strategically choosing where to cut the chocolate.

Key Insights

  • The problem asks for maximizing the minimum sum among the contiguous pieces.
  • A binary search over possible minimum sweetness values can be used to solve the problem efficiently.
  • Greedily try forming pieces by summing consecutive chunks and cut whenever the sum reaches or exceeds the candidate value.
  • If you can form at least k+1 pieces with a candidate value, it means that candidate is achievable.
  • The final answer is the largest candidate for which the above condition holds.

Space and Time Complexity

Time Complexity: O(n * log(S)) where n is the length of the sweetness array and S is the total sum (or maximum possible sum) considered in binary search. Space Complexity: O(1) extra space (not counting the input array).

Solution

The solution uses a binary search to determine the maximum possible minimum sum (sweetness) you can achieve. The binary search is applied over the potential sweetness values. For each candidate value mid, we traverse the sweetness array, summing the chunks until we get a sum greater than or equal to mid. If a valid segment is found, we increment our count of segments and reset the sum. If we can form at least k+1 segments, then mid is feasible. Otherwise, mid is too high. We adjust the binary search boundaries accordingly until the optimal value is found.

Code Solutions

# Python solution with line-by-line comments
class Solution:
    def maximizeSweetness(self, sweetness, k):
        # Function to check if we can cut the chocolate into at least k+1 pieces with each piece >= minSweetness
        def canDivide(minSweetness):
            current_sum = 0
            pieces = 0
            # Iterate over each chunk in the chocolate bar
            for chunk in sweetness:
                current_sum += chunk  # Add current chunk's sweetness to current sum
                # If the current sum is large enough, a piece is formed
                if current_sum >= minSweetness:
                    pieces += 1  # Increase the count of pieces
                    current_sum = 0  # Reset current sum for the next piece
            return pieces >= k + 1  # Check if we have at least k+1 pieces
        
        # Initialize binary search boundaries:
        # low is the smallest possible sweetness (minimum element in sweetness array)
        # high is the total sum of sweetness divided by (k+1) as an upper bound
        low, high = min(sweetness), sum(sweetness) // (k+1)
        
        # Binary search to find the maximum minimum sweetness that can be achieved
        while low < high:
            mid = (low + high + 1) // 2  #Choose mid value, biased to the upper half
            if canDivide(mid):  # If cutting is possible with candidate mid
                low = mid  # Try for a larger minimum sweetness
            else:
                high = mid - 1  # Otherwise reduce the candidate sweetness
        return low

# Example usage:
# sol = Solution()
# print(sol.maximizeSweetness([1,2,3,4,5,6,7,8,9], 5))  # Output should be 6
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