通过完全删除第(arr [i] + arr [i + 1])个元素来修改数组

给定的阵列ARR []由第一N的自然数，其中ARR [I] = I(基于1的索引)和整数K，一个正的，则任务是打印阵列ARR []除去每隔(ARR [之后所获得的我] + ARR [I + 1])^个从阵列元件中的每^第i^个动作恰好K倍。

例子：

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}, K = 2
Output: 1 2 4 5 7
Explanation:
Initially, the array arr[] is {1, 2, 3, 4, 5, 6, 7, 8}.
Operation 1: Delete every (A[1] + A[2])^th element, i.e., every 3^rd element from the array arr[]. Now the array modifies to {1, 2, 4, 5, 7, 8}.
Operation 2: Delete every (A[2] + A[3])^th element, i.e., every 6^th element from the array arr[]. Now the array modifies to {1, 2, 4, 5, 7}.
After performing the above operations, the array obtained is {1, 2, 4, 5, 7}.

Input: N = 10, K = 3
Output: 1 2 4 5 7 10

方法：给定的问题可以通过删去每(ARR [I] + ARR第[i + 1])^个术语从阵列中的每个^第i^个操作执行给定操作恰好K次可以解决。请按照以下步骤解决问题：

• 使用变量i遍历[0，K – 1]范围，并执行以下步骤：
  • 在每次删除操作之后，初始化辅助数组B []以存储数组arr []的元素。
  • 遍历给定数组arr [] ，如果当前索引不能被值(arr [i] + arr [i + 1])整除，则将该元素插入数组B []中。
  • 完成上述步骤后，将数组B []的所有元素插入数组arr []中。
• 完成上述步骤后，将数组arr []打印为结果数组。

下面是上述方法的实现：

Python3

# Python approach for the above approach
  
# Function to modify array by removing
# every K-th element from the array
def removeEveryKth(l, k):
    
    for i in range(0, len(l)):
        
        # Check if current element
        # is the k-th element
        if i % k == 0:
            l[i] = 0
  
    # Stores the elements after
    # removing every kth element
    arr = [0]
      
    for i in range(1, len(l)):
          
        # Append the current element
        # if it is not k-th element
        if l[i] != 0:
            arr.append(l[i])
  
    # Return the new array after
    # removing every k-th element
    return arr
  
# Function to print the array
def printArray(l):
    
    # Traverse the array l[]
    for i in range(1, len(l)):
        print(l[i], end =" ")
  
    print()
  
# Function to print the array after
# performing the given operations
# exactly k times
def printSequence(n, k):
  
    # Store first N natural numbers
    l = [int(i) for i in range(0, n + 1)]
  
    x = 1
  
    # Iterate over the range [0, k-1]
    for i in range(0, k):
  
        # Store sums of the two
        # consecutive terms
        p = l[x] + l[x + 1]
  
        # Remove every p-th
        # element from the array
        l = removeEveryKth(l, p)
          
        # Increment x by 1 for
        # the next iteration
        x += 1
      
    # Print the resultant array
    printArray(l)
  
# Driver Code
N = 8
K = 2
  
# Function Call
printSequence(N, K)

输出：

1 2 4 5 7

时间复杂度： O(N * K)
辅助空间： O(N)