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}

Asked in: Microsoft

Method 1 (Using Partition)

Partition the input array such that all odd elements are moved to left and all even elements on right. This step takes O(n).
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 swaped 
        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 swaped      
        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 

Output:

7 5 3 1 2 4

Time complexity: O(n log n)
Space complexity: O(1)
Method 2 (Using negative multiplication) :

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

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 
?> 

Output:

7 5 3 1 2 4

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

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.
When both the elements are odd: The larger element must appear on left of the smaller element.
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 

Thanks to Amandeep Singh for suggesting this solution.
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