巧克力分配问题的 Javascript 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：

Javascript

// Javascript Code For 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.
function findMinDiff(arr, n, m)
{    
    // If there are no chocolates or
    // number of students is 0
    if (m == 0 || n == 0)
        return 0;
   
    // Sort the given packets
    arr.sort(function(a, b){return a - b});
   
    // Number of students cannot be
    // more than number of packets
    if (n < m)
        return -1;
   
    // Largest number of chocolates
    let min_diff = Number.MAX_VALUE;
   
    // 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(let i = 0; i + m - 1 < n; i++)
    {
        let diff = arr[i + m - 1] - arr[i];
           
        if (diff < min_diff)
            min_diff = diff;
    }
    return min_diff;
}
  
// Driver code
let arr = [12, 4, 7, 9, 2, 23, 25, 
           41, 30, 40, 28, 42, 30, 
           44, 48, 43, 50];
              
// Number of students
let m = 7; 
let n = arr.length;
  
document.write("Minimum difference is " + 
                findMinDiff(arr, n, m));
// This code is contributed by divyesh072019

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！