最小化 K 个子集中存在的最大和最小元素之间的差异总和

给定一个大小为N的数组 arr[]和一个整数K ，任务是通过将数组拆分为 K个子集，使每个子集仅由唯一的数组元素组成，从而最小化每个子集的最大和最小元素之间的差值之和。

例子：

Input: arr[] = { 6, 3, 8, 1, 3, 1, 2, 2 }, K = 4
Output: 6
Explanation:
One of the optimal ways to split the array into K(= 4) subsets are { { 1, 2 }, { 2, 3 }, { 6, 8 }, { 1, 4 } }.
Sum of difference of maximum and minimum element present in each subset = { (2 – 1) + (3 – 2) + (8 – 6) + (3 – 1) } = 6.
Therefore, the required output is 6

Input: arr[] = { 2, 2, 1, 1 }, K = 1
Output: -1

方法：该问题可以使用带有位掩码的动态规划来解决。以下是递推关系：

mask: i^th bit of mask checks if array element is already selected in a subset or not.
l: index of last element selected in a subset.
j: index of current element selected in a subset.

If count of set bits in mask mod (N / K) == 1:
dp[mask][l] = min(dp[mask][l], dp[mask ^ (1 << l)][j])

Otherwise,
dp[mask][j] = min(dp[mask][j], dp[mask ^ (1 << j)][l] + nums[l] – nums[j])

请按照以下步骤解决问题：

• 迭代 mask 的所有可能值，即[0, 2 ^N – 1]
• 初始化一个数组，比如n_z_bits[] ，以存储子集中已经选择的元素的数量。
• 使用上面的递推关系并填充递推关系的所有可能的dp状态。
• 最后，打印dp[(1 << N ) – 1] 中的最小元素。

下面是上述方法的实现：

Python3

# Python program to implement
# the above approach
from itertools import permutations 
from itertools import combinations
  
# Function to minimize the sum of
# difference between maximums and
# minimums of K subsets of an array
def MinimizeSum(nums, k):
  
    # Stores count of elements
    # in an array
    n = len(nums)
      
    # Base Case
    if k == n:
        return 0
  
    # Initialize DP[][] array
    dp = [[float("inf")] * n for _ in range(1 << n)]
  
    # Sort the array
    nums.sort()
  
    # Mark i-th element
    # as not selected
    for i in range(n):
        dp[1 << i][i] = 0
  
    # Iterate over all possible
    # values of mask
    for mask in range(1 << n):
  
        # Store count of set bits
        # in mask
        n_z_bits = []
          
  
        # Store index of element which is 
        # already selected in a subset
        for p, c in enumerate(bin(mask)): 
            if c == "1":
                temp = len(bin(mask)) - p - 1
                n_z_bits.append(temp)
  
        # If count of set bits in mask
                # mod (n // k) equal to 1
        if len(n_z_bits) % (n//k) == 1:
            for j, l in permutations(n_z_bits, 2):
                temp = dp[mask ^ (1 << l)][j]
                dp[mask][l] = min(dp[mask][l], temp)
  
        else:
            for j, l in combinations(n_z_bits, 2):
                if nums[j] != nums[l]:
  
                    # Check if l-th element 
                    # is already selected or not
                    mask_t = mask ^ (1 << l)
  
                    temp = (dp[mask_t][j] + 
                             nums[j] - nums[l])
  
                    # Update dp[mask][l]        
                    dp[mask][l] = min(dp[mask][l],
                                            temp)
      
    # Return minimum element 
    # from dp[(1 << N) - 1]
    if min(dp[-1]) != float("inf"):
        return min(dp[-1])
      
    # If dp[-1] is inf then the 
    # partition is not possible 
    else: 
        return -1
      
# Driver Code
if __name__ == "__main__":
    # Given array
    arr = [ 6, 3, 8, 1, 3, 1, 2, 2 ]
    K = 4
  
    # Function call
    print(MinimizeSum(arr, K))

输出：

6

时间复杂度： O(N* 2 ^N )
辅助空间： O(N * 2 ^N )

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