# Minimum count of elements required to obtain the given Array by repeated mirror operations

Given an array arr[] consisting of N integers, the task is to find the array K[] of minimum possible length such that after performing multiple mirror operations on K[], the given array arr[] can be obtained.

Mirror Operation: Appending all the array elements to the original array in reverse order.
Illustration:
arr[] = {1, 2, 3}
After 1st mirror operation, arr[] modifies to {1, 2, 3, 3, 2, 1}
After 2nd mirror operation, arr[] modifies to {1, 2, 3, 3, 2, 1, 1, 2, 3, 3, 2, 1}

Examples:

Input: N = 6, arr[] = { 1, 2, 3, 3, 2, 1 }
Output:
Explanation:
Subarrays {1, 2, 3} and {3, 2, 1} are mirror images of each other.
Single mirror operation on {1, 2, 3} obtains the given array.
Therefore, the minimum number of elements required is 3.
Input: N = 8, arr[] = { 1, 2, 2, 1, 1, 2, 2, 1 }
Output:
Explanation:
Subarrays {1, 2, 2, 1} and {1, 2, 2, 1} are mirror images of each other.
Subarray {1, 2} and {2, 1} are mirror images of each other.
{1, 2} -> {1, 2, 2, 1} -> {1, 2, 2, 1, 1, 2, 2, 1}
Therefore, the minimum number of elements required is 2.

Naive Approach:
The simplest approach to solve the problem is to generate all the possible subarrays from the given array of size less than equal to N/2 and, for each subarray, check if performing mirror operation gives the array arr[] or not. Print the minimum length subarray satisfying the condition. If no subarray is found to be satisfying, print No
Time Complexity: O(N3
Auxiliary Space: O(N)
Efficient Approach:
The above approach can be further optimized using Divide and Conquer technique. Follow the steps below to solve the problem:

• Initialize a variable K = N and then, check whether the prefix of A[] of length K is a palindrome or not.
• If the prefix of length K is a palindrome then divide K by 2 and perform the above checking.
• If the prefix is not a palindrome then the answer is the current value of K.
• Keep checking while K > 0 until K is odd.
• If K is odd, then print the current value of K.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to find minimum number` `// of elements required to form A[]` `// by performing mirroring operation` `int` `minimumrequired(``int` `A[], ``int` `N)` `{` `    ``// Initialize K` `    ``int` `K = N;`   `    ``int` `ans;`   `    ``while` `(K > 0) {`   `        ``// Odd length array` `        ``// cannot be formed by` `        ``// mirror operation` `        ``if` `(K % 2 == 1) {` `            ``ans = K;` `            ``break``;` `        ``}`   `        ``bool` `ispalindrome = 1;`   `        ``// Check if prefix of` `        ``// length K is palindrome` `        ``for` `(``int` `i = 0; i < K / 2; i++) {`   `            ``// Check if not a palindrome` `            ``if` `(A[i] != A[K - 1 - i])`   `                ``ispalindrome = 0;` `        ``}`   `        ``// If found to be palindrome` `        ``if` `(ispalindrome) {` `            ``ans = K / 2;` `            ``K /= 2;` `        ``}`   `        ``// Otherwise` `        ``else` `{` `            ``ans = K;` `            ``break``;` `        ``}` `    ``}`   `    ``// Return the final answer` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `a[] = { 1, 2, 2, 1, 1, 2, 2, 1 };` `    ``int` `N = ``sizeof` `a / ``sizeof` `a[0];`   `    ``cout << minimumrequired(a, N);` `    ``return` `0;` `}`

## Java

 `// Java Program to implement` `// the above approach` `class` `GFG{` `  `  `// Function to find minimum number` `// of elements required to form A[]` `// by performing mirroring operation` `static` `int` `minimumrequired(``int` `A[], ``int` `N)` `{` `    ``// Initialize K` `    ``int` `K = N;` ` `  `    ``int` `ans=``0``;` ` `  `    ``while` `(K > ``0``)` `    ``{` ` `  `        ``// Odd length array` `        ``// cannot be formed by` `        ``// mirror operation` `        ``if` `(K % ``2` `== ``1``) ` `        ``{` `            ``ans = K;` `            ``break``;` `        ``}` ` `  `        ``int` `ispalindrome = ``1``;` ` `  `        ``// Check if prefix of` `        ``// length K is palindrome` `        ``for` `(``int` `i = ``0``; i < K / ``2``; i++)` `        ``{` ` `  `            ``// Check if not a palindrome` `            ``if` `(A[i] != A[K - ``1` `- i])` ` `  `                ``ispalindrome = ``0``;` `        ``}` ` `  `        ``// If found to be palindrome` `        ``if` `(ispalindrome == ``1``)` `        ``{` `            ``ans = K / ``2``;` `            ``K /= ``2``;` `        ``}` ` `  `        ``// Otherwise` `        ``else` `        ``{` `            ``ans = K;` `            ``break``;` `        ``}` `    ``}` ` `  `    ``// Return the final answer` `    ``return` `ans;` `}` ` `  `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `a[] = { ``1``, ``2``, ``2``, ``1``, ``1``, ``2``, ``2``, ``1` `};` `    ``int` `N = a.length;` ` `  `    ``System.out.println(minimumrequired(a, N));` `}` `}`   `// This code is contributed by rock_cool`

## Python3

 `# Python3 program to implement ` `# the above approach `   `# Function to find minimum number ` `# of elements required to form A[] ` `# by performing mirroring operation ` `def` `minimumrequired(A, N):` `    `  `    ``# Initialize K ` `    ``K ``=` `N` `    `  `    ``while` `(K > ``0``):` `        `  `        ``# Odd length array ` `        ``# cannot be formed by ` `        ``# mirror operation` `        ``if` `(K ``%` `2``) ``=``=` `1``:` `            ``ans ``=` `K` `            ``break` `        `  `        ``ispalindrome ``=` `1` `        `  `        ``# Check if prefix of ` `        ``# length K is palindrome ` `        ``for` `i ``in` `range``(``0``, K ``/``/` `2``):` `            `  `            ``# Check if not a palindrome` `            ``if` `(A[i] !``=` `A[K ``-` `1` `-` `i]):` `                ``ispalindrome ``=` `0` `                `  `        ``# If found to be palindrome ` `        ``if` `(ispalindrome ``=``=` `1``):` `            ``ans ``=` `K ``/``/` `2` `            ``K ``=` `K ``/``/` `2` `            `  `        ``# Otherwise` `        ``else``:` `            ``ans ``=` `K` `            ``break` `    `  `    ``# Return the final answer ` `    ``return` `ans`   `# Driver code` `A ``=` `[ ``1``, ``2``, ``2``, ``1``, ``1``, ``2``, ``2``, ``1` `]` `N ``=` `len``(A)`   `print``(minimumrequired(A, N))` `        `  `# This code is contributed by VirusBuddah_`

## C#

 `// C# program to implement` `// the above approach` `using` `System;`   `class` `GFG{`   `// Function to find minimum number` `// of elements required to form []A` `// by performing mirroring operation` `static` `int` `minimumrequired(``int``[] A, ``int` `N)` `{` `    `  `    ``// Initialize K` `    ``int` `K = N;`   `    ``int` `ans = 0;`   `    ``while` `(K > 0) ` `    ``{`   `        ``// Odd length array` `        ``// cannot be formed by` `        ``// mirror operation` `        ``if` `(K % 2 == 1) ` `        ``{` `            ``ans = K;` `            ``break``;` `        ``}`   `        ``int` `ispalindrome = 1;`   `        ``// Check if prefix of` `        ``// length K is palindrome` `        ``for``(``int` `i = 0; i < K / 2; i++)` `        ``{`   `            ``// Check if not a palindrome` `            ``if` `(A[i] != A[K - 1 - i])` `                ``ispalindrome = 0;` `        ``}`   `        ``// If found to be palindrome` `        ``if` `(ispalindrome == 1)` `        ``{` `            ``ans = K / 2;` `            ``K /= 2;` `        ``}`   `        ``// Otherwise` `        ``else` `        ``{` `            ``ans = K;` `            ``break``;` `        ``}` `    ``}`   `    ``// Return the readonly answer` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int``[] a = { 1, 2, 2, 1, 1, 2, 2, 1 };` `    ``int` `N = a.Length;`   `    ``Console.WriteLine(minimumrequired(a, N));` `}` `}`   `// This code is contributed by amal kumar choubey`

## Javascript

 ``

Output:

`2`

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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