# C++ Program to Find the Mth element of the Array after K left rotations

• Last Updated : 14 Oct, 2022

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;  ` `}  `

Output:

`5`

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

Please refer complete article on Find the Mth element of the Array after K left rotations for more details!

My Personal Notes arrow_drop_up
Related Articles