# Find the Mth element of the Array after K left rotations

Given non-negative integers K, M, and an array arr[] with N elements find the Mth element of the array after K left rotations.

Examples:

Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output: 5
Explanation:
The array after first left rotation a1[ ] = {4, 5, 23, 3}
The array after second left rotation a2[ ] = {5, 23, 3, 4}
1st element after 2 left rotations is 5.

Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2
Output:
Explanation:
The array after 3 left rotation has 5 at its second position.

Naive Approach: The idea is to Perform Left rotation operation K times and then find the Mth element of the final array.

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

Efficient Approach: To optimize the problem, observe the following points:

1. If the array is rotated N times it returns the initial array again.

For example, a[ ] = {1, 2, 3, 4, 5}, K=5 then the array after 5 left rotation a5[ ] = {1, 2, 3, 4, 5}.

Therefore, the elements in the array after Kth rotation is the same as the element at index K%N in the original array.

2. The Mth element of the array after K left rotations is

{ (K + M – 1) % N }th

element in the original array.

3.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach  ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to return Mth element of  ` `// array after k left rotations  ` `int` `getFirstElement(``int` `a[], ``int` `N,  ` `                    ``int` `K, ``int` `M)  ` `{  ` `    ``// The array comes to original state  ` `    ``// after N rotations  ` `    ``K %= N;  ` ` `  `    ``// Mth element after k left rotations  ` `    ``// is (K+M-1)%N th element of the  ` `    ``// original array  ` `    ``int` `index = (K + M - 1) % N;  ` ` `  `    ``int` `result = a[index];  ` ` `  `    ``// Return the result  ` `    ``return` `result;  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` `    ``// Array initialization  ` `    ``int` `a[] = { 3, 4, 5, 23 };  ` ` `  `    ``// Size of the array  ` `    ``int` `N = ``sizeof``(a) / ``sizeof``(a[0]);  ` ` `  `    ``// Given K rotation and Mth element  ` `    ``// to be found after K rotation  ` `    ``int` `K = 2, M = 1;  ` ` `  `    ``// Function call  ` `    ``cout << getFirstElement(a, N, K, M);  ` `    ``return` `0;  ` `}  `

## Java

 `// Java program for the above approach  ` `import` `java.util.*;  ` `class` `GFG{  ` `     `  `// Function to return Mth element of  ` `// array after k left rotations  ` `public` `static` `int` `getFirstElement(``int``[] a, ``int` `N,  ` `                                  ``int` `K, ``int` `M)  ` `{  ` `     `  `    ``// The array comes to original state  ` `    ``// after N rotations  ` `    ``K %= N;  ` ` `  `    ``// Mth element after k left rotations  ` `    ``// is (K+M-1)%N th element of the  ` `    ``// original array  ` `    ``int` `index = (K + M - ``1``) % N;  ` ` `  `    ``int` `result = a[index];  ` ` `  `    ``// Return the result  ` `    ``return` `result;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `     `  `    ``// Array initialization  ` `    ``int` `a[] = { ``3``, ``4``, ``5``, ``23` `};  ` ` `  `    ``// Size of the array  ` `    ``int` `N = a.length;  ` ` `  `    ``// Given K rotation and Mth element  ` `    ``// to be found after K rotation  ` `    ``int` `K = ``2``, M = ``1``;  ` ` `  `    ``// Function call  ` `    ``System.out.println(getFirstElement(a, N, K, M));  ` `}  ` `}  ` ` `  `// This code is contributed by divyeshrabadiya07  `

## Python3

 `# Python3 program for the above approach  ` ` `  `# Function to return Mth element of  ` `# array after k left rotations  ` `def` `getFirstElement(a, N, K, M):  ` `     `  `    ``# The array comes to original state  ` `    ``# after N rotations  ` `    ``K ``%``=` `N  ` ` `  `    ``# Mth element after k left rotations  ` `    ``# is (K+M-1)%N th element of the  ` `    ``# original array  ` `    ``index ``=` `(K ``+` `M ``-` `1``) ``%` `N  ` ` `  `    ``result ``=` `a[index]  ` ` `  `    ``# Return the result  ` `    ``return` `result  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `'__main__'``:  ` `     `  `    ``# Array initialization  ` `    ``a ``=` `[ ``3``, ``4``, ``5``, ``23` `]  ` ` `  `    ``# Size of the array  ` `    ``N ``=` `len``(a)  ` ` `  `    ``# Given K rotation and Mth element  ` `    ``# to be found after K rotation  ` `    ``K ``=` `2` `    ``M ``=` `1` ` `  `    ``# Function call  ` `    ``print``(getFirstElement(a, N, K, M))  ` ` `  `# This code is contributed by mohit kumar 29  `

## C#

 `// C# program for the above approach  ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `// Function to return Mth element of  ` `// array after k left rotations  ` `public` `static` `int` `getFirstElement(``int``[] a, ``int` `N,  ` `                                  ``int` `K, ``int` `M)  ` `{  ` `     `  `    ``// The array comes to original state  ` `    ``// after N rotations  ` `    ``K %= N;  ` ` `  `    ``// Mth element after k left rotations  ` `    ``// is (K+M-1)%N th element of the  ` `    ``// original array  ` `    ``int` `index = (K + M - 1) % N;  ` ` `  `    ``int` `result = a[index];  ` ` `  `    ``// Return the result  ` `    ``return` `result;  ` `}  ` ` `  `// Driver code ` `public` `static` `void` `Main(``string``[] args)  ` `{ ` `     `  `    ``// Array initialization  ` `    ``int` `[]a = { 3, 4, 5, 23 };  ` ` `  `    ``// Size of the array  ` `    ``int` `N = a.Length;  ` ` `  `    ``// Given K rotation and Mth element  ` `    ``// to be found after K rotation  ` `    ``int` `K = 2, M = 1;  ` ` `  `    ``// Function call  ` `    ``Console.Write(getFirstElement(a, N, K, M)); ` `} ` `} ` ` `  `// This code is contributed by rutvik_56 `

## Javascript

 ` `

Output:

`5`

Time complexity: O(1)
Auxiliary Space: O(1)

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