Number Of Unique Flavors After Sharing K Candies - Leetcode Solution

Problem Statement:

You have K candies and N children. You want to give each child at least one candy. Given an integer array of length N where candies[i] is the number of candies that child i wants, return the number of unique ways to distribute the candies.

Example:

Input: candies = [1,1,2,2,3,3], k = 7 Output: 29 Explanation: You can give each child 1 candy and give the extra 2 candies to child 1 and child 2. There are 7C2 = 21 ways to distribute the remaining 2 candies. So, the total number of unique ways to distribute the candies is 1 + 21 = 22. Also, there are 2C1 * 2C1 * 3C1 = 12 ways to choose the number of candies for each child. Therefore, the total number of unique ways to distribute the candies is 22 * 12 = 264.

Solution:

To solve this problem, we can use combinatorics. First, we need to distribute one candy to each child. So, we have K-N candies left. Then, we need to distribute these candies among the N children in such a way that each child gets at least one candy.

Let's consider the simplest case where all children want one candy each. In this case, we just need to choose N candies out of K-N candies. We can do this in (K-N)C(N) ways.

Now, let's consider the general case where each child can demand any number of candies. To distribute the remaining candies, we can use the stars and bars technique. We can imagine the K-N candies as stars and the N-1 gaps between the children as bars. The number of ways to distribute the candies is then (K-N+N-1)C(N-1).

Finally, we need to take into account the number of ways we can assign the number of candies for each child. This can be done using the multinomial coefficients. The number of ways to distribute the candies among the children is then (K-N+N-1)C(N-1)*(C(p1,q1)C(p2,q2)...*C(pk,qk)), where p1,q1,p2,q2,...,pk,qk are the number of candies demanded and distributed to each child.

In Python, we can implement the solution as follows:

def distributeCandies(candies, k): n = len(candies)

# distribute one candy to each child
k -= n

# number of ways to distribute remaining candies
ways = math.comb(k+n-1, n-1)

# number of ways to assign candy count to each child
count = collections.Counter(candies)
p = len(count)
q = list(count.values())
assign = 1
for i in range(p):
    assign *= math.comb(k, q[i])
    k -= q[i]
return ways*assign

Time Complexity:

The time complexity of this solution is O(N log K), where K is the total number of candies and N is the number of children. This is because we need to sort the candy count and iterate over all unique counts, which takes O(N log N) time. For each unique count, we need to compute the number of ways to distribute those candies, which takes O(log K) time using math.comb. Therefore, the total time complexity is O(N log N + N log K) = O(N log K).

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