# C++ Program to Rotate all odd numbers right and all even numbers left in an Array of 1 to N

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.Â
Note: N is always even.
Examples:Â

Input: A = {1, 2, 3, 4, 5, 6, 7, 8}Â
Output: {7, 4, 1, 6, 3, 8, 5, 2}Â
Explanation:Â
Even element = {2, 4, 6, 8}Â
Odd element = {1, 3, 5, 7}Â
Left rotate of even number = {4, 6, 8, 2}Â
Right rotate of odd number = {7, 1, 3, 5}Â
Combining Both odd and even number alternatively.
Input: A = {1, 2, 3, 4, 5, 6}Â
Output: {5, 4, 1, 6, 3, 2}Â
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Approach:

1. It is clear that the odd elements are always on even index and even elements are always laying on odd index.
2. To do left rotation of even number we choose only odd indices.
3. To do right rotation of odd number we choose only even indices.
4. Print the updated array.

Below is the implementation of the above approach:

## C++

 // C++ program to implement // the above approach #include 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

Output:
5 4 1 6 3 2

Time Complexity: O(N)Â
Auxiliary Space: O(1)
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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!

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