Find the Mth element of the Array after K left rotations

• Difficulty Level : Easy
• Last Updated : 29 Apr, 2021

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}
st 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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