通过重复删除最后一个元素并将其放置在任意索引处，将数组转换为另一个数组

给定两个数组A []和B [] ，它们都由前N个自然数组成，则任务是计算最后一个数组元素需要移到数组A []中任意位置的最小次数。 ]使数组A []和B []相等。

例子：

Input: A[] = {1, 2, 3, 4, 5}, B[] = {1, 5, 2, 3, 4}
Output:1
Explanation:
Initially, the array A[] is {1, 2, 3, 4, 5}. After moving the last array element, i.e. 5, and placing them between arr[0] (= 1) and arr[1](= 2) modifies the array to {1, 5, 2, 3, 4}, which is the same as the array B[].
Therefore, the minimum number of operations required to convert the array A[] to B[] is 1.

Input: A[] = {3, 2, 1}, B[] = {1, 2, 3}
Output: 2
Explanation:
Initially, the array A[] is {3, 2, 1}.
Operation 1: After moving the last array element, i.e. 1, to the beginning of the array, modifies the array to {1, 3, 2}.
Operation 2: After moving the last element of the array, i.e. 2 and placing them between the elements arr[0] (= 1) and arr[1] (= 3) modifies the array to {1, 2, 3}, which is the same as the array B[].
Therefore, the minimum number of operations required to convert the array A[] to B[] is 2.

方法：给定的问题可以通过找到第一个排列的前i个连续元素与第二个排列的子序列相同来解决，因此自最后一个运算起，操作数必须至少少(N – I )个(N – i)个元素可以最佳选择，并以所需的索引插入。因此， (N – i)是将数组A []转换为B []所需的最小步数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the minimum number
// of operations required to convert
// the array A[] into array B[]
int minCount(int A[], int B[], int N)
{
    // Stores the index in the first
    // permutation A[] which is same
    // as the subsequence in B[]
    int i = 0;
 
    // Find the first i elements in A[]
    // which is a subsequence in B[]
    for (int j = 0; j < N; j++) {
 
        // If element A[i]
        // is same as B[j]
        if (A[i] == B[j]) {
            i++;
        }
    }
 
    // Return the count of
    // operations required
    return N - i;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 3, 4, 5 };
    int B[] = { 1, 5, 2, 3, 4 };
 
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << minCount(A, B, N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
     
// Function to count the minimum number
// of operations required to convert
// the array A[] into array B[]
static int minCount(int A[], int B[], int N)
{
     
    // Stores the index in the first
    // permutation A[] which is same
    // as the subsequence in B[]
    int i = 0;
 
    // Find the first i elements in A[]
    // which is a subsequence in B[]
    for(int j = 0; j < N; j++)
    {
         
        // If element A[i]
        // is same as B[j]
        if (A[i] == B[j])
        {
            i++;
        }
    }
 
    // Return the count of
    // operations required
    return N - i;
}
 
// Driver Code
public static void main (String[] args)
{
    int A[] = { 1, 2, 3, 4, 5 };
    int B[] = { 1, 5, 2, 3, 4 };
    int N = A.length;
 
    System.out.println(minCount(A, B, N));
}
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
 
# Function to count the minimum number
# of operations required to convert
# the array A[] into array B[]
def minCount(A, B, N):
     
    # Stores the index in the first
    # permutation A[] which is same
    # as the subsequence in B[]
    i = 0
 
    # Find the first i elements in A[]
    # which is a subsequence in B[]
    for j in range(N):
         
        # If element A[i]
        # is same as B[j]
        if (A[i] == B[j]):
            i += 1
 
    # Return the count of
    # operations required
    return N - i
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 1, 2, 3, 4, 5 ]
    B = [ 1, 5, 2, 3, 4 ]
 
    N = len(A)
 
    print(minCount(A, B, N))
     
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to count the minimum number
// of operations required to convert
// the array A[] into array B[]
static int minCount(int[] A, int[] B, int N)
{
     
    // Stores the index in the first
    // permutation A[] which is same
    // as the subsequence in B[]
    int i = 0;
 
    // Find the first i elements in A[]
    // which is a subsequence in B[]
    for(int j = 0; j < N; j++)
    {
         
        // If element A[i]
        // is same as B[j]
        if (A[i] == B[j])
        {
            i++;
        }
    }
 
    // Return the count of
    // operations required
    return N - i;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] A = { 1, 2, 3, 4, 5 };
    int[] B = { 1, 5, 2, 3, 4 };
    int N = A.Length;
 
    Console.WriteLine(minCount(A, B, N));
}
}
 
// This code is contributed by ukasp

Javascript

 
// JavaScript program for the above approach
 
// Function to count the minimum number
// of operations required to convert
// the array A[] into array B[]
function minCount(A, B, N){
     
    // Stores the index in the first
    // permutation A[] which is same
    // as the subsequence in B[]
    var i = 0
 
    // Find the first i elements in A[]
    // which is a subsequence in B[]
    for(let j = 0; j < N; j++){
         
        // If element A[i]
        // is same as B[j]
        if (A[i] == B[j])
            i += 1
      }
    // Return the count of
    // operations required
    return N - i
     
}
 
// Driver Code
     
var A = [ 1, 2, 3, 4, 5 ]
var B = [ 1, 5, 2, 3, 4 ]
 
N = A.length
 
document.write(minCount(A, B, N))
     
// This code is contributed by AnkThon
 

输出：

1

时间复杂度： O(N)
辅助空间： O(1)