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2226. Maximum Candies Allocated to K Children

Description

You are given a 0-indexed integer array candies. Each element in the array denotes a pile of candies of size candies[i]. You can divide each pile into any number of sub piles, but you cannot merge two piles together.

You are also given an integer k. You should allocate piles of candies to k children such that each child gets the same number of candies. Each child can take at most one pile of candies and some piles of candies may go unused.

Return the maximum number of candies each child can get.

 

Example 1:

Input: candies = [5,8,6], k = 3
Output: 5
Explanation: We can divide candies[1] into 2 piles of size 5 and 3, and candies[2] into 2 piles of size 5 and 1. We now have five piles of candies of sizes 5, 5, 3, 5, and 1. We can allocate the 3 piles of size 5 to 3 children. It can be proven that each child cannot receive more than 5 candies.

Example 2:

Input: candies = [2,5], k = 11
Output: 0
Explanation: There are 11 children but only 7 candies in total, so it is impossible to ensure each child receives at least one candy. Thus, each child gets no candy and the answer is 0.

 

Constraints:

  • 1 <= candies.length <= 105
  • 1 <= candies[i] <= 107
  • 1 <= k <= 1012

Solutions

We notice that if each child can receive $v$ candies, then for any $v' \lt v$, each child can also receive $v'$ candies. Therefore, we can use binary search to find the maximum $v$ such that each child can receive $v$ candies.

We define the left boundary of the binary search as $l = 0$ and the right boundary as $r = \max(\text{candies})$, where $\max(\text{candies})$ represents the maximum value in the array $\text{candies}$. During the binary search, we take the middle value $v = \left\lfloor \frac{l + r + 1}{2} \right\rfloor$ each time, and then calculate the total number of candies each child can receive. If the total is greater than or equal to $k$, it means each child can receive $v$ candies, so we update the left boundary $l = v$. Otherwise, we update the right boundary $r = v - 1$. Finally, when $l = r$, we have found the maximum $v$.

The time complexity is $O(n \times \log M)$, where $n$ is the length of the array $\text{candies}$, and $M$ is the maximum value in the array $\text{candies}$. The space complexity is $O(1)$.

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class Solution:
    def maximumCandies(self, candies: List[int], k: int) -> int:
        l, r = 0, max(candies)
        while l < r:
            mid = (l + r + 1) >> 1
            if sum(x // mid for x in candies) >= k:
                l = mid
            else:
                r = mid - 1
        return l

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

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class Solution {
public:
    int maximumCandies(vector<int>& candies, long long k) {
        int l = 0, r = ranges::max(candies);
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            long long cnt = 0;
            for (int x : candies) {
                cnt += x / mid;
            }
            if (cnt >= k) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
};

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func maximumCandies(candies []int, k int64) int {
    return sort.Search(1e7, func(v int) bool {
        v++
        var cnt int64
        for _, x := range candies {
            cnt += int64(x / v)
        }
        return cnt < k
    })
}

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function maximumCandies(candies: number[], k: number): number {
    let [l, r] = [0, Math.max(...candies)];
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        const cnt = candies.reduce((acc, cur) => acc + Math.floor(cur / mid), 0);
        if (cnt >= k) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l;
}

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