Divide Chocolate

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;
    }
};

This solution addresses the problem of dividing chocolate into the maximum possible minimum sweetness that each person can receive. The provided code is written in C++ and focuses on a binary search method to optimize the problem of distributing sweetness evenly among a specified number of people.

Explanation of the Code:

• The function maximizeSweetness accepts a vector sweets, which contains the sweetness levels of various chocolate pieces, and an integer people, representing the number of people sharing these sweets.

• The goal is to make divisions such that at least people + 1 individuals get at least a certain level of sweetness that is maximized.

• int indiv = people + 1; declares the total individual divisions needed which is one more than the number of people.

• The chocolate pieces' minimum sweetness is stored in low, while high is initialized to the average total sweetness per individual achieved by summing the entire sweetness and dividing it by indiv.

• The main logic involves a binary search within the while loop where low is less than high. The median point of current low and high values determines the trial division of sweetness.

• If the sum of chocolates currentSweet reaches or exceeds this median sweetness during distribution (for loop iterating over sweets), the count of such successful divisions increments. This distribution resets currentSweet to zero every time the division condition is satisfied.

• Based on the successful divisions (count), if they meet or exceed the required individuals (indiv), the search boundary adjusts to potentially increase the minimum sweetness. Otherwise, it decreases to find the closest possible fit under the median condition.

• The process iterates until the optimal division point of sweetness (high) is found.

In summary, the algorithm efficiently ensures that the distribution allows for the most equitable distribution of sweetness to just over the number of people required, using a binary search mechanism to find the maximum of the minimum possible distribution of sweetness per person.

class Solution {
    public int maxSweetnessDistribution(int[] sweetnessLevels, int k) {
        int peopleCount = k + 1;
        int low = Arrays.stream(sweetnessLevels).min().getAsInt();
        int high = Arrays.stream(sweetnessLevels).sum() / peopleCount;

        while (low < high) {
            int middle = (low + high + 1) / 2;
            int currentSweetness = 0;
            int peopleAssigned = 0;
            
            for (int current : sweetnessLevels) {
                currentSweetness += current;
                
                if (currentSweetness >= middle) {
                    peopleAssigned += 1;
                    currentSweetness = 0;
                }
            }
            
            if (peopleAssigned >= peopleCount) {
                low = middle;
            } else {
                high = middle - 1;
            }
        }
        
        return high;
    }
}

This Java solution effectively addresses the problem of distributing chocolate bars such that the minimum sweetness level among given parts is maximized, taking into consideration the number of parts (k + 1).

• The function maxSweetnessDistribution takes two parameters:

  • int[] sweetnessLevels: an array of integers representing the sweetness of each chocolate piece.
  • int k: the number of parts less than the total desired parts, as the pieces will be divided into k + 1 parts.

• The calculation for the minimum possible sweetness (low) is derived from the minimum value in the sweetnessLevels array.

• The maximum possible sweetness (high) is calculated by averaging the sum of all the sweet values in the sweetnessLevels over k + 1.

• Use a binary search technique to find the highest minimum sweetness possible:

  1. Determine the middle point sweetness between low and high.
  2. Count how many groups can be formed where each group has a sweetness at least equal to the middle sweetness.
  3. If the number of groups formed is at least equal to peopleCount, adjust the low to middle.
  4. Otherwise, decrease high to middle - 1.

• The loop continues until low is less than high, gradually refining the possible richest minimum sweetness that can be distributed equally among k + 1 persons.

• Finally, the function returns the value of high, which represents the maximum possible minimum sweetness that can be achieved for the most equitable distribution.

This solution leverages efficient binary search principles and performs the calculations by continually adjusting the search space based on the divisibility of the sweetness levels into the required number of parts.

var calculateMaxSweetness = function(sweets, divisions) {
    let personCount = divisions + 1;
    let minimum = Math.min(...sweets);
    let maximum = Math.floor(sweets.reduce((acc, cur) => acc + cur) / personCount);
    
    while (minimum < maximum) {
        const middle = Math.floor((minimum + maximum + 1) / 2);
        let currentSweetness = 0;
        let peopleCounted = 0;
        
        for (const sweet of sweets) {
            currentSweetness += sweet;
            if (currentSweetness >= middle) {
                peopleCounted += 1;
                currentSweetness = 0;
            }
        }
        
        if (peopleCounted >= personCount) {
            minimum = middle;
        } else {
            maximum = middle - 1;
        }
    }
    
    return maximum;
};

The solution presents a JavaScript function named calculateMaxSweetness which aims to distribute a list of sweetness values among a specified number of people such that everyone gets a maximum and roughly equal amount of sweets. The function utilizes a binary search algorithm to efficiently determine the optimal division of sweets.

The JavaScript function works as follows:

1. The total number of persons is identified by adding one to the given number of divisions, acknowledging that the sweets need to be divided among these many individuals.
2. Initialize minimum as the smallest value in the sweets array, recognizing the least amount of sweetness that could potentially be received if divided equally.
3. Establish maximum by calculating the average sweetness per person after summing all the items in the sweets and dividing by personCount.
4. Employ a binary search between minimum and maximum:
   • Calculate the middle value of current minimum and maximum.
   • Traverse through the sweets array, aggregating sweetness until it matches or exceeds the middle value, which might be a potential maximum sweetness per person. If reached, reset the accumulated sweetness and increment a count of persons who have received their share.
   • Adjust minimum or maximum based on whether the number of persons who can get at least the middle-value amount of sweetness meets or exceeds personCount. Increase minimum to shrink the search space if the current middle-value can be distributed to sufficient people, otherwise decrease maximum.
5. The search terminates when minimum and maximum converge, resulting in maximum being the highest possible minimum sweetness each person can receive under optimal division.

This approach ensures that the solution is computationally efficient, handling the division even for large input data by leveraging the binary search paradigm. Moreover, it assures that every individual gets the fairest possible share of sweetness that the input constraints allow.

class Solution:
    def distributeChocolates(self, chocolates: List[int], k: int) -> int:
        num_people = k + 1
        low = min(chocolates)
        high = sum(chocolates) // num_people

        while low < high:
            mid = (low + high + 1) // 2
            current_sum = 0
            count_people = 0

            for choco in chocolates:
                current_sum += choco

                if current_sum >= mid:
                    count_people += 1
                    current_sum = 0

            if count_people >= num_people:
                low = mid
            else:
                high = mid - 1

        return high

In this Python solution, you find the maximum amount of chocolate that can be evenly distributed among a number of people so that each gets at least a certain amount k. The chocolate distribution is defined by a list where each element denotes a packaged amount of chocolates.

The solution involves following steps in a binary search manner:

1. Define the number of required distributions by adding one to k to factor in all people.
2. Set the lower boundary (low) as the minimum amount from the chocolate list and the upper boundary (high) as the total chocolate amount divided by the number of people (num_people).
3. Use a binary search mechanism where:
   • The middle value (mid) is calculated. This value represents a potential maximum amount of chocolate a single person could receive if chocolates are distributed fairly.
   • Traverse through each chocolate packet in the list and keep adding to current_sum until it at least matches mid. Each time it does, increment a counter for people who have received their share (count_people).
4. Adjust the low and high boundaries based on whether the number of people who got their share is at least as much as the number of people you need to distribute to.
5. Continue adjusting the boundaries until low is less than high.

The function returns the value of high which indicates the maximum minimum amount of chocolates that can be distributed to all people such that no one receives less than this number. This value ensures a fair distribution among the receivers ensuring that the number of people who at least get k amounts of chocolate is maximized within the given constraints.