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📜  将数组A []分成子集,它们的总和和大小等于数组B []的元素

📅  最后修改于: 2021-05-19 18:07:45             🧑  作者: Mango

给定一个由N个整数组成的数组A [] ,任务是将数组A []拆分为多个子集,这些子集的总和相等且长度等于数组B []中的元素。

例子:

方法:请按照以下步骤解决问题:

  • 由于已经给出了子集的数量,因此请计算每个子集的总和。
  • 遍历数组B []的每个元素,找到长度B [i]的每种可能的组合,并检查组合的总和是否等于所需的总和。
  • 对每个数组元素B [i]重复上述步骤,并打印最终答案

下面是上述方法的实现:

Python
# Python Program to implement
# the above approach
from itertools import combinations
  
# Function to split elements of first
# array into subsets of size given as
# elements in the second array
def main(A, B):
  
    # Required sum of subsets
    req_sum = sum(A) // len(B)
  
    # Stores the subsets
    final = []
  
    # Iterate the array B[]
    for i in B:
  
        # Generate all possible
        # combination of length B[i]
        temp = list(combinations(A, i))
  
    # Iterate over all the combinations
        for j in temp:
  
            # If the sum is equal to the 
            # required sum of subsets
            if(sum(j) == req_sum):
  
                # Store the subset
                temp = list(j)
                final.append(temp)
  
                for k in temp:
  
                    # Removing the subset
                    # from the array
                    A.remove(k)
                break
  
    # Printing the final result
    print(final)
  
  
# Driver Code
if __name__ == '__main__':
    # Value of array A and B
    A = [1, 2, 3, 4, 5, 6]
    B = [2, 2, 2]
  
    # Function call
    main(A, B)


输出:
[[1, 6], [2, 5], [3, 4]]

时间复杂度: O(N 3 )
辅助空间: O(2 maxm ),其中maxm是数组B []中的最大元素