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C++ Program to 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:Â
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: 5Â
Explanation:Â
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. Â
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:Â

5

Â

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!