# Convert an array into another by repeatedly removing the last element and placing it at any arbitrary index

• Last Updated : 16 Apr, 2021

Given two arrays A[] and B[], both consisting of a permutation of first N natural numbers, the task is to count the minimum number of times the last array element is required to be shifted to any arbitrary position in the array A[] to make both the arrays A[] and B[] equal.

Examples:

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.

Approach: The given problem can be solved by finding the first i consecutive elements of the first permutation which is the same as the subsequence of the second permutation, then the count of operations must be less at least (N – I), since the last (N – i) elements can be selected optimally and inserted at required indices. Therefore, (N – i) is the minimum number of steps required for the conversion of the array A[] to B[].

Below is the implementation of the above approach:

## 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

 ``
Output:
`1`

Time Complexity: O(N)
Auxiliary Space: O(1)

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