巧克力分配问题的Python程序

给定一个包含 n 个整数的数组，其中每个值代表一包中巧克力的数量。每个包可以有不定数量的巧克力。有 m 个学生，任务是分发巧克力包，这样：

1. 每个学生得到一包。
2. 给学生的巧克力包中的巧克力数量与巧克力数量之差最小。

例子：

Input : arr[] = {7, 3, 2, 4, 9, 12, 56} , m = 3 
Output: Minimum Difference is 2 Explanation:
We have seven packets of chocolates and 
we need to pick three packets for 3 students 
If we pick 2, 3 and 4, we get the minimum 
difference between maximum and minimum packet 
sizes.

Input : arr[] = {3, 4, 1, 9, 56, 7, 9, 12} , m = 5 
Output: Minimum Difference is 6 
Explanation:
The set goes like 3,4,7,9,9 and the output 
is 9-3 = 6

Input : arr[] = {12, 4, 7, 9, 2, 23, 25, 41, 
30, 40, 28, 42, 30, 44, 48, 
43, 50} , m = 7 
Output: Minimum Difference is 10 
Explanation:
We need to pick 7 packets. We pick 40, 41, 
42, 44, 48, 43 and 50 to minimize difference 
between maximum and minimum. 

来源：Flipkart 面试经历

一个简单的解决方案是生成 arr[0..n-1] 的所有大小为 m 的子集。对于每个子集，找出其中最大和最小元素之间的差异。最后，返回最小差值。
一个有效的解决方案是基于这样的观察，即为了最小化差异，我们必须从排序的数据包中选择连续的元素。我们首先对数组 arr[0..n-1] 进行排序，然后找到大小为 m 的子数组，其中最后一个元素和第一个元素之间的差异最小。

下图是上述方法的试运行：

下面是上述方法的实现：

Python3

# Python3 program to solve chocolate
# distribution problem
  
# arr[0..n-1] represents sizes of packets
# m is number of students.
# Returns minimum difference between maximum
# and minimum values of distribution.
def findMinDiff(arr, n, m):
  
    # if there are no chocolates or number
    # of students is 0
    if (m==0 or n==0):
        return 0
  
    # Sort the given packets
    arr.sort()
  
    # Number of students cannot be more 
    # than number of packets
    if (n < m):
        return -1
  
    # Largest number of chocolates
    min_diff = arr[n-1] - arr[0]
  
    # Find the subarray of size m such that
    # difference between last (maximum in case
    # of sorted) and first (minimum in case of
    # sorted) elements of subarray is minimum.
    for i in range(len(arr) - m + 1):
        min_diff = min(min_diff ,  
                       arr[i + m - 1] - arr[i])    
          
    return min_diff
  
# Driver Code
if __name__ == "__main__":
    arr = [12, 4, 7, 9, 2, 23, 25, 41,
           30, 40, 28, 42, 30, 44, 48, 
           43, 50]
  
    # Number of students
    m = 7 
    n = len(arr)
    print("Minimum difference is", 
           findMinDiff(arr, n, m))
# This code is contributed by Smitha

输出：

Minimum difference is 10

时间复杂度： O(n Log n)，因为我们在子数组搜索之前应用排序。

请参阅有关巧克力分配问题的完整文章以获取更多详细信息！