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: 5
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)
- 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.
- The Mth element of the array after K left rotations is
{ (K + M – 1) % N }th
element in the original array.
Efficient Approach: To optimize the problem, observe the following points:
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> 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 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 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# 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 |
<script> // Javascript program for the above approach // Function to return Mth element of
// array after k left rotations
function 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
var index = (K + M - 1) % N;
var result = a[index];
// Return the result
return result;
}
// Driver code
// Array initialization
var a = [ 3, 4, 5, 23 ];
// Size of the array
var N = a.length;
// Given K rotation and Mth element
// to be found after K rotation
var K = 2, M = 1;
// Function call
document.write(getFirstElement(a, N, K, M));
// This code contributed by gauravrajput1 </script> |
5
Time complexity: O(1)
Auxiliary Space: O(1)