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:
- 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++
#include<bits/stdc++.h>
using namespace std;
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;
}
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;
}
void rotate( int arr[])
{
left_rotate(arr);
right_rotate(arr);
for ( int i = 0; i < 6; i++)
{
cout << (arr[i]) << " " ;
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
rotate(arr);
}
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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!
Last Updated :
27 Jan, 2022
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