Given a permutation arrays A[] consisting of N numbers in range [1, N], the task is to left rotate all the even numbers and right rotate all the odd numbers of the permutation and print the updated permutation.
Note: N is always even.
Examples:
Input: A = {1, 2, 3, 4, 5, 6, 7, 8}
Output: {7, 4, 1, 6, 3, 8, 5, 2}
Explanation:
Even element = {2, 4, 6, 8}
Odd element = {1, 3, 5, 7}
Left rotate of even number = {4, 6, 8, 2}
Right rotate of odd number = {7, 1, 3, 5}
Combining Both odd and even number alternatively.
Input: A = {1, 2, 3, 4, 5, 6}
Output: {5, 4, 1, 6, 3, 2}
Approach:
- It is clear that the odd elements are always on even index and even elements are always laying on odd index.
- To do left rotation of even number we choose only odd indices.
- To do right rotation of odd number we choose only even indices.
- Print the updated array.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include<bits/stdc++.h> using namespace std; // function to left rotate void left_rotate(int arr[]) { int last = arr[1]; for (int i = 3; i < 6; i = i + 2) { arr[i - 2] = arr[i]; } arr[6 - 1] = last; } // function to right rotate void right_rotate(int arr[]) { int start = arr[6 - 2]; for (int i = 6- 4; i >= 0; i = i - 2) { arr[i + 2] = arr[i]; } arr[0] = start; } // Function to rotate the array void rotate(int arr[]) { left_rotate(arr); right_rotate(arr); for (int i = 0; i < 6; i++) { cout << (arr[i]) << " "; } } // Driver code int main() { int arr[] = { 1, 2, 3, 4, 5, 6 }; rotate(arr); } // This code is contributed by rock_cool |
5 4 1 6 3 2
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Rotate all odd numbers right and all even numbers left in an Array of 1 to N for more details!
