2226. Maximum Candies Allocated to K Children

Problem Description

In this problem, you have an array candies, where each element represents a pile of candies (with index starting from 0). Each pile contains a certain number of candies denoted by candies[i]. The array gives you the flexibility to split any pile into smaller sub-piles but does not allow you to combine two different piles into one.

Additionally, you are given an integer k, which represents the number of children to whom the candies must be distributed. The goal is to distribute the candies to k children in such a way that each child receives exactly the same number of candies. However, each child can receive at most one pile (which can potentially be a sub-pile of one of the original piles), and it's possible that not all piles will be distributed.

The objective is to find the maximum number of candies that you can give to each child under these conditions.

Intuition

To solve this problem, we need to determine the maximum number of candies each child can get. The key here is understanding that if we can give x candies to each child, then surely we can give any amount less than x candies as well (by dividing the piles into smaller ones). However, we cannot always give more than x candies. So, our solution will likely involve finding the optimal x.

One way to approach this is through binary search. The minimum number of candies a child could get is 0 (if k is larger than the total number of candies available) and the maximum is the size of the largest pile in candies.

Using binary search, we can find the largest possible value for x in the range [0, max(candies)] such that we can still distribute the candies to k children with each child receiving x candies. We consider a middle value mid within this range and check if it is possible to divide the piles in such a way that at least k children can receive mid candies each. If it is possible, then we know we can try to increase x and search in the upper half of the range; otherwise, we search in the lower half of the range.

The provided code implements this binary search approach to efficiently determine the maximum amount of candies that can be distributed to each child such that every child receives the same amount.

Learn more about Binary Search patterns.

Solution Approach

The solution provided is a binary search approach, which solves the problem by iterating over a possible range of candies that can be distributed to each child and narrowing down that range to find an optimal solution.

Here's a step-by-step explanation of the implementation:

We initialize two pointers, left and right. left represents the lower bound (minimum number of candies we can give per child), set to 0, since it's possible that we cannot give any candies to the children if there are fewer candies than k. right represents the upper bound (maximum number of candies we can give to each child), which would be the largest pile in candies, since no child can receive more than the size of the largest pile.
```
1left, right = 0, max(candies)
```
We perform the binary search by checking the middle value between the left and right pointers. This middle value, mid, posits a hypothetical maximum number of candies to be distributed per child.
```
1mid = (left + right + 1) >> 1
```
For the current middle value mid, we calculate whether it is possible to achieve this distribution by summing up the number of children we can satisfy if each were to be given mid candies. This is done by dividing each pile size by mid and summing those values across all piles.
```
1cnt = sum(v // mid for v in candies)
```
If cnt (the number of potential children receiving mid candies) is greater than or equal to k, then distributing mid candies to each child is possible, and we can potentially give more, so we adjust the left pointer to mid.
```
1if cnt >= k:
2    left = mid
```
If cnt is less than k, then we cannot give out mid candies to each child, and we might need to reduce the number of candies we give to each child. Therefore, we adjust the right pointer to mid - 1.
```
1else:
2    right = mid - 1
```
This search process continues until left and right converge, which happens when left == right. When the binary search terminates, left will hold the maximum number of candies that we can distribute to each child.
Finally, the function returns the value of left as the answer.
```
1return left
```

The algorithm effectively uses binary search on a sorted range of possible candy distributions, utilizing the idea that if a certain amount x is doable, so is any amount less than x. We use division to check the feasibility of each mid point, counting the number of children that can be satisfied with mid candies and adjusting our search space accordingly until we find the maximum feasible distribution.

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Example Walkthrough

Let's take an example to illustrate the solution approach. Suppose we have an array of piles of candies given as candies = [4, 7, 12, 19], and the number of children k = 3.

We want to distribute these candies such that each of the 3 children gets the same number of candies, and we are tasked with finding the maximum number of candies each child can get.

We begin by initializing left = 0 and right = 19 (19 being the largest pile of candies).
We perform binary search. Initially, mid = (0 + 19 + 1) >> 1 is 10. We check if we can distribute 10 candies to each child.
To check the feasibility, we sum up how many children can receive 10 candies:
- Pile 1 can be divided into 4 // 10 children, which is 0.
- Pile 2 can be divided into 7 // 10 children, which is 0.
- Pile 3 can be divided into 12 // 10 children, which is 1.
- Pile 4 can be divided into 19 // 10 children, which is 1.
The total number of children who can receive candies is 0 + 0 + 1 + 1 = 2.
Since we can only satisfy 2 children, but we need to satisfy 3, 10 candies per child is too high. So, we set right = mid - 1, which is 9.
We repeat the binary search. Now, mid = (0 + 9 + 1) >> 1 is 5. We calculate:
- Pile 1 can be divided into 4 // 5 children, which is 0.
- Pile 2 can be divided into 7 // 5 children, which is 1.
- Pile 3 can be divided into 12 // 5 children, which is 2.
- Pile 4 can be divided into 19 // 5 children, which is 3.
Now, the total is 0 + 1 + 2 + 3 = 6.
Since the total number of children who can receive candies is now 6, and we only need to satisfy 3, we have found that giving 5 candies per child is possible. We could potentially give more, so we adjust left to mid, which is 5.
We continue this process, but now as left and right converge, both will be equal to 5, and the loop ends.
The binary search concludes with left = 5, which is the maximum number of candies we can distribute to each child.

Each child can receive 5 candies, which can be distributed as follows:

Child 1 receives 5 candies from pile 2.
Child 2 receives 5 candies from pile 3.
Child 3 receives 5 candies from pile 4.

2 candies will remain in pile 3, and 14 candies will remain in pile 4, but since we can't combine them and we can only give one pile to each child, this is the most we can distribute evenly.

Solution Implementation

1class Solution:
2    def maximumCandies(self, candies: List[int], k: int) -> int:
3        # Initialize left to 0 (minimum possible result) and 
4        # right to the maximum number of candies in any pile
5        left, right = 0, max(candies)
6
7        # Binary search for the highest number of candies per child
8        while left < right:
9            # Mid is the number of candies to distribute per child; add 1 and 
10            # shift right by 1 to get the upper middle for even numbers
11            mid = (left + right + 1) // 2
12          
13            # Count how many children can receive 'mid' number of candies
14            count = sum(candy // mid for candy in candies)
15          
16            # If the number of children that can receive 'mid' candies is at least k,
17            # mid can be a potential answer, so we move left pointer up to mid
18            if count >= k:
19                left = mid
20            # Otherwise, we decrease the right pointer, as we need less candies per child
21            else:
22                right = mid - 1
23      
24        # left is now the maximum number of candies per child that we can distribute
25        return left
26

1class Solution {
2    public int maximumCandies(int[] candies, long k) {
3        // Initialize the search range for the maximum number of candies per child.
4        int low = 0;
5        int high = (int) 1e7;
6      
7        // Use a binary search to find the maximum number of candies 
8        // that can be distributed to `k` children evenly.
9        while (low < high) {
10            // Calculate the middle point of the current search range.
11            // Use `+ 1` to ensure the `mid` is biased towards the higher part of the range.
12            // The bitwise right shift operator `>> 1` effectively divides the sum of `low` and `high` by 2.
13            int mid = (low + high + 1) >> 1;
14          
15            // Count the total number of children that can receive `mid` candies.
16            long count = 0;
17            for (int candy : candies) {
18                count += candy / mid;
19            }
20          
21            // If the total number of children that can receive `mid` candies is at least `k`,
22            // then we know that it's possible to give out at least `mid` candies to each child.
23            // So we update `low` to `mid` to search in the higher half of the range next.
24            if (count >= k) {
25                low = mid;
26            }
27            // If not, we need to search in the lower half of the range, so we adjust `high`.
28            else {
29                high = mid - 1;
30            }
31        }
32      
33        // `low` will represent the highest number of candies that can be distributed evenly to `k` children.
34        return low;
35    }
36}
37

1class Solution {
2public:
3    int maximumCandies(vector<int>& candies, long long totalChildren) {
4        // Initialize the search range
5        int minCandies = 0, maxCandies = 1e7;
6
7        // Perform a binary search to find the maximum number of candies
8        // that can be distributed to each child
9        while (minCandies < maxCandies) {
10            // Calculate the mid-point for the current search range
11            int mid = (minCandies + maxCandies + 1) >> 1;
12
13            // Counter to record the total number of children we can distribute
14            // candies to with the current 'mid' number of candies per child
15            long long childCount = 0;
16
17            // Iterate through each pile of candies
18            for (int candiesInPile : candies) {
19                // Increment the count by the number of children we can satisfy
20                // with the current pile using 'mid' candies per child
21                childCount += candiesInPile / mid;
22            }
23
24            // If we can give at least 'mid' candies to 'totalChildren' or more,
25            // adjust the lower bound of the search range
26            if (childCount >= totalChildren) {
27                minCandies = mid;
28            } else {
29                // Otherwise, reduce the upper bound of the search range
30                maxCandies = mid - 1;
31            }
32        }
33
34        // After narrowing down the search range, 'minCandies' will hold the
35        // maximum number of candies that can be distributed to each child
36        return minCandies;
37    }
38};
39

1// Function to calculate the maximum number of candies
2// that can be distributed to each child
3function maximumCandies(candies: number[], totalChildren: number): number {
4    // Initialize the search range
5    let minCandies: number = 0;
6    let maxCandies: number = 1e7;
7
8    // Perform a binary search to find the maximum number of candies
9    // that can be distributed to each child
10    while (minCandies < maxCandies) {
11        // Calculate the mid-point for the current search range
12        let mid: number = Math.floor((minCandies + maxCandies + 1) / 2);
13
14        // Counter to record the total number of children we can distribute
15        // candies to with the current 'mid' number of candies per child
16        let childCount: number = 0;
17
18        // Iterate through each pile of candies
19        for (let candiesInPile of candies) {
20            // Increment the count by the number of children we can satisfy
21            // with the current pile using 'mid' candies per child
22            childCount += Math.floor(candiesInPile / mid);
23        }
24
25        // If we can give at least 'mid' candies to 'totalChildren' or more,
26        // adjust the lower bound of the search range
27        if (childCount >= totalChildren) {
28            minCandies = mid;
29        } else {
30            // Otherwise, reduce the upper bound of the search range
31            maxCandies = mid - 1;
32        }
33    }
34
35    // After narrowing down the search range, 'minCandies' will hold the
36    // maximum number of candies that can be distributed to each child
37    return minCandies;
38}
39

Time and Space Complexity

Time Complexity

The time complexity of the given code is mainly determined by the while loop and the sum operation within that loop. The while loop uses a binary search pattern which runs until left is less than right. In terms of the binary search, this takes O(log(max(candies))) time, because we are repeatedly halving the search space that starts with the maximum value in candies.

Inside this loop, there is a sum operation which includes a list comprehension that iterates over all elements in the candies list. This operation has a time complexity of O(n) where n is the number of elements in candies, because it must visit each element once.

Since the list comprehension is done every time the while loop runs, the combined time complexity is the product of the two, resulting in O(n * log(max(candies))).

Space Complexity

The space complexity of the given code is O(1). Despite the list comprehension inside the sum function, it does not create a new list due to the generator expression; instead, it calculates the sum on the fly. Thus, the only additional space used is for a few variables (left, right, mid, cnt), which is constant and does not depend on the input size. Therefore, the space used is constant and does not grow with the size of the input.

Learn more about how to find time and space complexity quickly using problem constraints.

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