Sort all even numbers in ascending order and then sort all odd numbers in descending order

Given an array of integers (both odd and even), sort them in such a way that the first part of the array contains odd numbers sorted in descending order, rest portion contains even numbers sorted in ascending order.
Examples:

Input: arr[] = {1, 2, 3, 5, 4, 7, 10}
Output: arr[] = {7, 5, 3, 1, 2, 4, 10}

Input: arr[] = {0, 4, 5, 3, 7, 2, 1}
Output: arr[] = {7, 5, 3, 1, 0, 2, 4} 

Method 1 (Using Partition)

1. Partition the input array such that all odd elements are moved to the left and all even elements on right. This step takes O(n).
2. Once the array is partitioned, sort left and right parts individually. This step takes O(n Log n).

Below is the implementation of the above idea.  

C++

// C++ program sort array in even and odd manner.
// The odd numbers are to be sorted in descending
// order and the even numbers in ascending order
#include <bits/stdc++.h>
using namespace std;

// To do two way sort. First sort even numbers in
// ascending order, then odd numbers in descending
// order.
void twoWaySort(int arr[], int n)
{

    // Current indexes from left and right
    int l = 0, r = n - 1;

    // Count of odd numbers
    int k = 0;

    while (l < r)
    {

        // Find first even number
        // from left side.
        while (arr[l] % 2 != 0)
        {
            l++;
            k++;
        }

        // Find first odd number
        // from right side.
        while (arr[r] % 2 == 0 && l < r)
            r--;

        // Swap even number present on left and odd
        // number right.
        if (l < r)
            swap(arr[l], arr[r]);
    }

    // Sort odd number in descending order
    sort(arr, arr + k, greater<int>());

    // Sort even number in ascending order
    sort(arr + k, arr + n);
}

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

Java

// Java program sort array in even and odd manner.
// The odd numbers are to be sorted in descending
// order and the even numbers in ascending order

import java.util.Arrays;
import java.util.Collections;

public class GFG {
    // To do two way sort. First sort even numbers in
    // ascending order, then odd numbers in descending
    // order.
    static void twoWaySort(Integer arr[], int n)
    {
        // Current indexes from left and right
        int l = 0, r = n - 1;

        // Count of odd numbers
        int k = 0;

        while (l < r)
        {

            // Find first even number from left side.
            while (arr[l] % 2 != 0)
            {
                l++;
                k++;
            }

            // Find first odd number from right side.
            while (arr[r] % 2 == 0 && l < r)
                r--;

            // Swap even number present on left and odd
            // number right.
            if (l < r)
            {

                // swap arr[l] arr[r]
                int temp = arr[l];
                arr[l] = arr[r];
                arr[r] = temp;
            }
        }

        // Sort odd number in descending order
        Arrays.sort(arr, 0, k, Collections.
                                  reverseOrder());

        // Sort even number in ascending order
        Arrays.sort(arr, k, n);
    }

    // Driver Method
    public static void main(String[] args)
    {
        Integer arr[] = { 1, 3, 2, 7, 5, 4 };

        twoWaySort(arr, arr.length);

        System.out.println(Arrays.toString(arr));
    }
}

Python

# Python program to sort array
# in even and odd manner
# The odd numbers are to be
# sorted in descending order
# and the even numbers in
# ascending order

# To do two way sort. First
# sort even numbers in ascending
# order, then odd numbers in
# descending order.
def two_way_sort(arr, arr_len):

    # Current indexes l->left
    # and r->right
    l, r = 0, arr_len - 1

    # Count of number of
    # odd numbers, used in
    # slicing the array later.
    k = 0

    # Run till left(l) < right(r)
    while(l < r):

        # While left(l) is odd, if yes
        # increment the left(l) plus
        # odd count(k) if not break the
        # while for even number found
        # here to be swapped
        while(arr[l] % 2 != 0):
            l += 1
            k += 1

        # While right(r) is even,
        # if yes decrement right(r)
        # if not break the while for
        # odd number found here to
        # be swapped
        while(arr[r] % 2 == 0 and l < r):
            r -= 1

        # Swap the left(l) and right(r),
        # which is even and odd numbers
        # encountered in above loops
        if(l < r):
            arr[l], arr[r] = arr[r], arr[l]

    # Slice the number on the
    # basis of odd count(k)
    odd = arr[:k]
    even = arr[k:]

    # Sort the odd and
    # even array accordingly
    odd.sort(reverse = True)
    even.sort()

    # Extend the odd array with
    # even values and return it.
    odd.extend(even)

    return odd

# Driver code
arr_len = 6
arr = [1, 3, 2, 7, 5, 4]
result = two_way_sort(arr, arr_len)
for i in result:
    print(str(i) + " "),

# This code is contributed
# by JaySiyaRam

C#

// C# program sort array in even and odd manner.
// The odd numbers are to be sorted in descending
// order and the even numbers in ascending order
using System;
using System.Linq;

class GFG {
    // To do two way sort. First sort even numbers in
    // ascending order, then odd numbers in descending
    // order.
    static void twoWaySort(int[] arr, int n)
    {
        // Current indexes from left and right
        int l = 0, r = n - 1;

        // Count of odd numbers
        int k = 0;

        while (l < r)
        {

            // Find first even number
            // from left side.
            while (arr[l] % 2 != 0)
            {
                l++;
                k++;
            }

            // Find first odd number from right side.
            while (arr[r] % 2 == 0 && l < r)
                r--;

            // Swap even number present
            // on left and odd
            // number right.
            if (l < r) {
                // swap arr[l] arr[r]
                int temp = arr[l];
                arr[l] = arr[r];
                arr[r] = temp;
            }
        }

        // Sort odd number in descending order
        Array.Sort(arr, 0, k);
        Array.Reverse(arr, 0, k);

        // Sort even number in ascending order
        Array.Sort(arr, k, n - k);
    }

    // Driver Method
    public static void Main(String[] args)
    {
        int[] arr = { 1, 3, 2, 7, 5, 4 };

        twoWaySort(arr, arr.Length);

        Console.WriteLine(String.Join(" ", arr));
    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// Javascript program sort array
// in even and odd manner.
// The odd numbers are to be
// sorted in descending
// order and the even numbers in
// ascending order

    // To do two way sort.
    // First sort even numbers in
    // ascending order, then odd
    // numbers in descending
    // order.
    function twoWaySort(arr,n)
    {
        // Current indexes from
        // left and right
        let l = 0, r = n - 1;

        // Count of odd numbers
        let k = 0;

        while (l < r)
        {

            // Find first even number
            // from left side.
            while (arr[l] % 2 != 0)
            {
                l++;
                k++;
            }

            // Find first odd number
            // from right side.
            while (arr[r] % 2 == 0 && l < r)
                r--;

            // Swap even number present
            // on left and odd
            // number right.
            if (l < r)
            {

                // swap arr[l] arr[r]
                let temp = arr[l];
                arr[l] = arr[r];
                arr[r] = temp;
            }
        }

        let odd=new Array(k);
        for(let i=0;i<k;i++)
        {
            odd[i]=arr[i];
        }
        let even=new Array(n-k);
        for(let i=0;i<n-k;i++)
        {
            even[i]=arr[k+i];
        }
        // Sort odd number in descending order
        odd.sort(function(a,b){return b-a;});
        // Sort even number in ascending order
        even.sort(function(a,b){return a-b;});

        return odd.concat(even);

    }
    // Driver Method
    let arr=[1, 3, 2, 7, 5, 4 ];
    let ans=twoWaySort(arr, arr.length);
    for(let i=0;i<ans.length;i++)
    {
        document.write(ans[i]+" ");
    }


    // This code is contributed by rag2127

</script>

Output

7 5 3 1 2 4 

Time complexity: O(n log n) 
Auxiliary Space: O(1)

Method 2 (Using negative multiplication) :

1. Make all odd numbers negative.
2. Sort all numbers.
3. Revert the changes made in step 1 to get original elements back.

Implementation:

C++

// C++ program sort array in even and odd manner.
// The odd numbers are to be sorted in descending
// order and the even numbers in ascending order
#include <bits/stdc++.h>
using namespace std;

// To do two way sort. First sort even numbers in
// ascending order, then odd numbers in descending
// order.
void twoWaySort(int arr[], int n)
{
    // Make all odd numbers negative
    for (int i = 0; i < n; i++)
        if (arr[i] & 1) // Check for odd
            arr[i] *= -1;

    // Sort all numbers
    sort(arr, arr + n);

    // Retaining original array
    for (int i = 0; i < n; i++)
        if (arr[i] & 1)
            arr[i] *= -1;
}

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

Java

// Java program sort array in even and odd manner.
// The odd numbers are to be sorted in descending
// order and the even numbers in ascending order

import java.util.Arrays;

public class GFG
{
    // To do two way sort. First sort even numbers in
    // ascending order, then odd numbers in descending
    // order.
    static void twoWaySort(int arr[], int n)
    {
        // Make all odd numbers negative
        for (int i = 0; i < n; i++)
            if ((arr[i] & 1) != 0) // Check for odd
                arr[i] *= -1;

        // Sort all numbers
        Arrays.sort(arr);

        // Retaining original array
        for (int i = 0; i < n; i++)
            if ((arr[i] & 1) != 0)
                arr[i] *= -1;
    }

    // Driver Method
    public static void main(String[] args)
    {
        int arr[] = { 1, 3, 2, 7, 5, 4 };

        twoWaySort(arr, arr.length);

        System.out.println(Arrays.toString(arr));
    }
}

Python3

# Python 3 program to sort array in
# even and odd manner. The odd
# numkbers are to be sorted in
# descending order and the even
# numbers in ascending order

# To do two way sort. First sort
# even numbers in ascending order,
# then odd numbers in descending order.
def twoWaySort(arr, n):

    # Make all odd numbers negative
    for i in range(0, n):

        # Check for odd
        if (arr[i] & 1):
            arr[i] *= -1

    # Sort all numbers
    arr.sort()

    # Retaining original array
    for i in range(0, n):
        if (arr[i] & 1):
            arr[i] *= -1

# Driver code
arr = [1, 3, 2, 7, 5, 4]
n = len(arr)
twoWaySort(arr, n);
for i in range(0, n):
    print(arr[i], end = " ")

# This code is contributed by Smitha Dinesh Semwal

C#

// Java program sort array in even and
// odd manner. The odd numbers are to
// be sorted in descending order and
// the even numbers in ascending order
using System;

public class GFG
{

    // To do two way sort. First sort
    // even numbers in ascending order,
    // then odd numbers in descending
    // order.
    static void twoWaySort(int[] arr, int n)
    {

        // Make all odd numbers negative
        for (int i = 0; i < n; i++)

            // Check for odd
            if ((arr[i] & 1) != 0)
                arr[i] *= -1;

        // Sort all numbers
        Array.Sort(arr);

        // Retaining original array
        for (int i = 0; i < n; i++)
            if ((arr[i] & 1) != 0)
                arr[i] *= -1;
    }

    // Driver Method
    public static void Main()
    {
        int[] arr = { 1, 3, 2, 7, 5, 4 };

        twoWaySort(arr, arr.Length);

        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
}

// This code is contributed by Smitha

PHP

<?php
// PHP program sort array in even and odd manner.
// The odd numbers are to be sorted in descending
// order and the even numbers in ascending order

// To do two way sort. First sort even numbers in
// ascending order, then odd numbers in descending
// order.
function twoWaySort(&$arr, $n)
{
    // Make all odd numbers negative
    for ($i = 0 ; $i < $n; $i++)
        if ($arr[$i] & 1) // Check for odd
            $arr[$i] *= -1;

    // Sort all numbers
    sort($arr);

    // Retaining original array
    for ($i = 0 ; $i < $n; $i++)
        if ($arr[$i] & 1)
            $arr[$i] *= -1;
}

// Driver code
$arr = array(1, 3, 2, 7, 5, 4);
$n = sizeof($arr);
twoWaySort($arr, $n);
for ($i = 0; $i < $n; $i++)
    echo $arr[$i] . " ";

// This code is contributed by ita_c
?>

Javascript

<script>

// Javascript program sort array in even and odd manner.
// The odd numbers are to be sorted in descending
// order and the even numbers in ascending order

// To do two way sort. First sort even numbers in
// ascending order, then odd numbers in descending
// order.
function twoWaySort(arr, n)
{
    // Make all odd numbers negative
    for (let i = 0; i < n; i++)
        if (arr[i] & 1) // Check for odd
            arr[i] *= -1;

    // Sort all numbers
    arr.sort((a,b) => a-b);

    // Retaining original array
    for (let i = 0; i < n; i++)
        if (arr[i] & 1)
            arr[i] *= -1;
}

// Driver code

    let arr = [ 1, 3, 2, 7, 5, 4 ];
    let n = arr.length;
    twoWaySort(arr, n);
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");

//This code is contributed by Mayank Tyagi
</script>

Output

7 5 3 1 2 4 

Time complexity: O(n log n) 
Auxiliary Space: O(1)

This method may not work when input array contains negative numbers. However, there is a way to handle this. We count the positive odd integers in the input array then sort again. Readers may refer this for implementation.

Method 3 (Using comparator):
This problem can be easily solved by using the inbuilt sort function with a custom compare method. On comparing any two elements there will be three cases: 

1. When both the elements are even: In this case, the smaller element must appear in the left of the larger element in the sorted array.
2. When both the elements are odd: The larger element must appear on left of the smaller element.
3. One is odd and the other is even: The element which is odd must appear on the left of the even element.

Below is the implementation of the above approach: 

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to print
// the contents of the array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// To do two way sort. Make comparator function
// for the inbuilt sort function of c++ such that
// odd numbers are placed before even in descending
// and ascending order respectively
bool compare(int a, int b)
{

    // If both numbers are even,
    // smaller number should
    // be placed at lower index
    if (a % 2 == 0 && b % 2 == 0)
        return a < b;

    // If both numbers are odd larger number
    // should be placed at lower index
    if (a % 2 != 0 && b % 2 != 0)
        return b < a;

    // If a is odd and b is even,
    // a should be placed before b
    if (a % 2 != 0)
        return true;

    // If b is odd and a is even,
    // b should be placed before a
    return false;
}

// Driver code
int main()
{
    int arr[] = { 1, 3, 2, 7, 5, 4 };
    int n = sizeof(arr) / sizeof(int);

    // Sort the array
    sort(arr, arr + n, compare);

    // Print the sorted array
    printArr(arr, n);

    return 0;
}

// This code is contributed by Nikhil Yadav

Java

// Java implementation of the approach
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

class GFG
{

    // Utility function to print
    // the contents of the array
    static void printArr(ArrayList<Integer> arr, int n) {
        for (int i = 0; i < n; i++)
            System.out.print(arr.get(i) + " ");
    }

    // Driver code
    public static void main(String args[]) {
        ArrayList<Integer> arr = new ArrayList<Integer>();
        arr.add(1);
        arr.add(3);
        arr.add(2);
        arr.add(7);
        arr.add(5);
        arr.add(4);
        int n = arr.size();


        // Sort the array
        Collections.sort(arr, new Comparator<Integer>() {
            @Override
            public int compare(Integer a, Integer b) {

                // If both numbers are even,
                // smaller number should
                // be placed at lower index
                if (a % 2 == 0 && b % 2 == 0)
                    return (a - b);

                // If both numbers are odd larger number
                // should be placed at lower index
                if (a % 2 != 0 && b % 2 != 0)
                    return (b - a);

                // If a is odd and b is even,
                // a should be placed before b
                if (a % 2 != 0)
                    return -1;

                // If b is odd and a is even,
                // b should be placed before a
                return 0;
            }
        });

        // Print the sorted array
        printArr(arr, n);
    }
}

// This code is contributed by Saurabh Jaiswal

Python3

# Python3 implementation of the approach

# Utility function to print
# the contents of the array
from functools import cmp_to_key

def printArr(arr, n):

    for i in range(n):
        print(arr[i], end = " ")

        # To do two way sort. Make comparator function
        # for the inbuilt sort function of c++ such that
        # odd numbers are placed before even in descending
        # and ascending order respectively
def compare(a, b):

    # If both numbers are even,
    # smaller number should
    # be placed at lower index
    if (a % 2 == 0 and b % 2 == 0 and a < b):
        return -1

    # If both numbers are odd larger number
    # should be placed at lower index
    if (a % 2 != 0 and b % 2 != 0 and b > a):
        return 1

    # If a is odd and b is even,
    # a should be placed before b
    if (a % 2 != 0):
        return -1

    # If b is odd and a is even,
    # b should be placed before a
    return 1

# Driver code
arr = [1, 3, 2, 7, 5, 4]
n = len(arr)

# Sort the array
arr.sort(key = cmp_to_key(compare))

# Print the sorted array
printArr(arr, n)

# This code is contributed by shinjanpatra

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

public class GFG
{

  // Utility function to print
  // the contents of the array
  static void printArr(List<int> arr, int n) {
    for (int i = 0; i < n; i++)
      Console.Write(arr[i] + " ");
  }
  private static int Compare(int a, int b)
  {

    // If both numbers are even,
    // smaller number should
    // be placed at lower index
    if (a % 2 == 0 && b % 2 == 0 && a<b)
      return -1;

    // If both numbers are odd larger number
    // should be placed at lower index
    if (a % 2 != 0 && b % 2 != 0 && b>a)
      return 1;

    // If a is odd and b is even,
    // a should be placed before b
    if (a % 2 != 0)
      return -1;

    // If b is odd and a is even,
    // b should be placed before a
    return 1;
  }

  // Driver code
  public static void Main(String []args) {
    List<int> arr = new List<int>();
    arr.Add(1);
    arr.Add(3);
    arr.Add(2);
    arr.Add(7);
    arr.Add(5);
    arr.Add(4);
    int n = arr.Count;

    // Sort the array
    arr.Sort(Compare);

    // Print the sorted array
    printArr(arr, n);
  }
}


// This code is contributed by 29AjayKumar

Javascript

<script>
        // JavaScript implementation of the approach

        // Utility function to print
        // the contents of the array
        function printArr(arr, n)
        {
            for (let i = 0; i < n; i++)
                document.write(arr[i] + " ");
        }

        // To do two way sort. Make comparator function
        // for the inbuilt sort function of c++ such that
        // odd numbers are placed before even in descending
        // and ascending order respectively
        function compare(a, b) {

            // If both numbers are even,
            // smaller number should
            // be placed at lower index
            if (a % 2 == 0 && b % 2 == 0 && a < b)
                return -1;

            // If both numbers are odd larger number
            // should be placed at lower index
            if (a % 2 != 0 && b % 2 != 0 && b > a)
                return 1;

            // If a is odd and b is even,
            // a should be placed before b
            if (a % 2 != 0)
                return -1;

            // If b is odd and a is even,
            // b should be placed before a
            return 1;
        }

        // Driver code
        var arr = [1, 3, 2, 7, 5, 4];
        var n = arr.length;

        // Sort the array
        arr.sort(compare);

        // Print the sorted array
        printArr(arr, n);

// This code is contributed by Potta Lokesh
    </script>

Output

7 5 3 1 2 4 

Time complexity : O(n*logn) [sort function has average time complexity n*logn]
Auxiliary Space : O(1)

Thanks to Amandeep Singh for suggesting this solution.

This article is contributed by DANISH_RAZA. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.