Solution Explanation: Maximum Candies Allocated to K Children

This problem asks to find the maximum number of candies each of k children can get, given an array of candy piles and the restriction that each child can take at most one pile. We can't merge piles, but we can split them.

The key observation is that the maximum number of candies each child can receive exhibits a monotonically decreasing property. If we can distribute x candies per child, we can also distribute any number less than x. This property allows us to efficiently use binary search.

Approach:

1. Binary Search: We perform a binary search on the possible range of candies per child. The search space is [0, max(candies)], where max(candies) is the largest pile of candies.

2. Check Feasibility: In each iteration of the binary search, we check if it's possible to distribute mid (the middle value in the search space) candies to each child. This is done by calculating the total number of children we can satisfy with mid candies per child:

   • For each pile candies[i], we calculate candies[i] / mid, which represents how many children can be satisfied from this pile.
   • We sum this up for all piles to get the total number of children that can be satisfied.
   • If this total is greater than or equal to k, it means we can distribute mid candies to each of k children.

3. Adjust Search Space:

   • If the total number of children satisfied is greater than or equal to k, it means mid is a feasible amount of candies per child, so we update the left boundary of the search space to mid. This is because we're trying to find the maximum possible amount.
   • Otherwise, mid is too high, and we update the right boundary to mid - 1.

4. Return Result: The binary search continues until the left and right boundaries converge (l == r). At this point, l (or r) represents the maximum number of candies each child can receive.

Time Complexity: O(N log M), where N is the number of candy piles and M is the maximum number of candies in a pile. The binary search takes O(log M) iterations, and each iteration requires O(N) time to check feasibility.

Space Complexity: O(1). The algorithm uses a constant amount of extra space.

Code Examples:

The code examples below demonstrate the solution in several programming languages. They all follow the same binary search algorithm outlined above.

Python3

def maximumCandies(candies, k):
    left, right = 0, max(candies)
    while left < right:
        mid = (left + right + 1) // 2  # Integer division for mid
        total_children = sum(candy // mid for candy in candies)
        if total_children >= k:
            left = mid
        else:
            right = mid - 1
    return left
 

Java

class Solution {
    public int maximumCandies(int[] candies, long k) {
        int left = 0, right = Arrays.stream(candies).max().getAsInt();
        while (left < right) {
            int mid = (left + right + 1) / 2;
            long totalChildren = 0;
            for (int candy : candies) {
                totalChildren += candy / mid;
            }
            if (totalChildren >= k) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
}

C++

#include <algorithm>
#include <numeric>
 
class Solution {
public:
    int maximumCandies(vector<int>& candies, long long k) {
        int left = 0, right = *max_element(candies.begin(), candies.end());
        while (left < right) {
            int mid = (left + right + 1) / 2;
            long long totalChildren = 0;
            for (int candy : candies) {
                totalChildren += candy / mid;
            }
            if (totalChildren >= k) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
};

Go

func maximumCandies(candies []int, k int64) int {
	left, right := 0, 0
	for _, c := range candies {
		if c > right {
			right = c
		}
	}
 
	for left < right {
		mid := (left + right + 1) / 2
		totalChildren := int64(0)
		for _, candy := range candies {
			totalChildren += int64(candy / mid)
		}
		if totalChildren >= k {
			left = mid
		} else {
			right = mid - 1
		}
	}
	return left
}

JavaScript

/**
 * @param {number[]} candies
 * @param {number} k
 * @return {number}
 */
var maximumCandies = function(candies, k) {
    let left = 0;
    let right = Math.max(...candies);
    while (left < right) {
        let mid = Math.floor((left + right + 1) / 2);
        let totalChildren = 0;
        for (let candy of candies) {
            totalChildren += Math.floor(candy / mid);
        }
        if (totalChildren >= k) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return left;
};

These codes all implement the same efficient binary search approach to solve the problem. Remember to choose the code that best suits your preferred programming language.