Sort even-placed elements in increasing and odd-placed in decreasing order

1.7

We are given an array of n distinct numbers, the task is to sort all even-placed numbers in increasing and odd-place numbers in decreasing order. The modified array should contain all sorted even-placed numbers followed by reverse sorted odd-placed numbers.

Note that the first element is considered as even because of its index 0.

Examples:

Input:  arr[] = {0, 1, 2, 3, 4, 5, 6, 7}
Output: arr[] = {0, 2, 4, 6, 7, 5, 3, 1}
Even-place elements : 0, 2, 4, 6
Odd-place elements : 1, 3, 5, 7
Even-place elements in increasing order : 
0, 2, 4, 6
Odd-Place elements in decreasing order : 
7, 5, 3, 1

Input: arr[] = {3, 1, 2, 4, 5, 9, 13, 14, 12}
Output: {2, 3, 5, 12, 13, 14, 9, 4, 1}
Even-place elements : 3, 2, 5, 13, 12
Odd-place elements : 1, 4, 9, 14
Even-place elements in increasing order : 
2, 3, 5, 12, 13
Odd-Place elements in decreasing order : 
14, 9, 4, 1

The idea is simple, we create two auxiliary arrays evenArr[] and oddArr[] respectively. We traverse input array and put all even-placed elements in evenArr[] and odd placed elements in oddArr[]. Then we sort evenArr[] in ascending and oddArr[] in descending order. Finally copy evenArr[] and oddArr[] to get the required result.

// Program to separately sort even-placed and odd
// placed numbers and place them together in sorted
// array.
#include <bits/stdc++.h>
using namespace std;

void bitonicGenerator(int  arr[], int n)
{
    // create evenArr[] and oddArr[]
    vector<int> evenArr;
    vector<int> oddArr;

    // Put elements in oddArr[] and evenArr[] as
    // per their position
    for (int i = 0; i < n; i++)
    {
        if (!( i % 2 ))
            evenArr.push_back(arr[i]);
        else
            oddArr.push_back(arr[i]);
    }

    // sort evenArr[] in ascending order
    // sort oddArr[] in descending order
    sort(evenArr.begin(), evenArr.end() );
    sort(oddArr.begin(), oddArr.end(), greater<int>());

    int i = 0;
    for (int j=0; j<evenArr.size(); j++)
        arr[i++] = evenArr[j];
    for (int j=0; j<oddArr.size(); j++)
        arr[i++] = oddArr[j];
}

//Driver Program
int main()
{
    int arr[] = { 1, 5, 8, 9, 6, 7, 3, 4, 2, 0 };
    int n = sizeof(arr)/sizeof(arr[0]);
    bitonicGenerator(arr, n);
    for (int i=0; i<n; i++)
       cout << arr[i] << " ";
    return 0;
}

Output:

1 2 3 6 8 9 7 5 4 0

Time Complexity : O(n Log n)
Space Complexity : O(n)

The above problem can also be solve without use of Auxiliary space. The idea is to swaps first half odd index positions with second half even index positions and then sort the first half array in increasing order and second half array in decreasing order. Thanks to SWARUPANANDA DHUA for suggesting this.

// Program to sort even-placed elements in increasing and 
// odd-placed in decreasing order with constant space complexity

#include <bits/stdc++.h>
using namespace std;

void bitonicGenerator(int arr[], int n)
{
    // first odd index
    int i = 1;
    
    // last index
    int j = n-1;
    
    // if last index is odd
    if(j%2 != 0)
        // decrement j to even index
        j--;
    
    // swapping till half of array
    while(i < j)
    {
        swap(arr[i], arr[j]);
        i += 2; 
        j -= 2;
    }
    
    // Sort first half in increasing
    sort(arr, arr + (n+1)/2);
    
    // Sort second half in decreasing
    sort(arr + (n+1)/2, arr + n, greater<int>());
    
}

//Driver Program
int main()
{
    int arr[] = {1, 5, 8, 9, 6, 7, 3, 4, 2, 0};
    int n = sizeof(arr)/sizeof(arr[0]);
    bitonicGenerator(arr, n);
    for (int i=0; i<n; i++)
    cout << arr[i] << " ";
    return 0;
}
// This code is contributed by SWARUPANANDA DHUA

Output:

1 2 3 6 8 9 7 5 4 0

Time Complexity : O(n Log n)
Space Complexity : O(1)

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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