Segregate even and odd numbers using Lomuto’s Partition Scheme

Given an array arr[] of integers, segregate even and odd numbers in the array such that all the even numbers should be present first, and then the odd numbers.

Examples: 

Input: arr[] = {7, 2, 9, 4, 6, 1, 3, 8, 5}
Output: 2 4 6 8 7 9 1 3 5

Input: arr[] = {1, 3, 2, 4, 7, 6, 9, 10}
Output:  2 4 6 10 7 1 9 3

We have discussed two different approaches in Segregate Even and Odd numbers

Here we are going to discuss Lomuto’s Partition Scheme to segregate even and odd.

• Start with an array of numbers.

For the sake of illustration, let's consider the following array: [7, 2, 9, 4, 6, 1, 3, 8, 5].

• Choose a pivot element from the array.

In this case, let's choose the last element, which is 5, as the pivot.

• Initialize two pointers, i and j.
  • The pointer i will keep track of the boundary between the even and odd numbers, and
  • the pointer j will iterate through the array.
• Iterate through the array using the pointer j.
  • Compare each element with the pivot and perform the following steps:
    • If the element is even, swap it with the element at index i and increment i.
    • If the element is odd, do nothing and continue.
• Finally, swap the pivot element with the element at index I.

Updated Array: 2 4 6 8 5 1 3 7 9 

Illustration

Let's go through the steps:

• Initially, i = 0 and j = 0.
• The first element is 7 (odd),
  • so we do nothing and move j to the next element.
• The second element is 2 (even),
  • so we swap it with the element at index i (which is also 2) and increment i to 1.
• The third element is 9 (odd),
  • so we do nothing and move j to the next element.
• The fourth element is 4 (even),
  • so we swap it with the element at index i (which is 9) and increment i to 2.
• The fifth element is 6 (even),
  • so we swap it with the element at index i (which is 9) and increment i to 3.
• The sixth element is 1 (odd),
  • so we do nothing and move j to the next element.
• The seventh element is 3 (odd),
  • so we do nothing and move j to the next element.
• The eighth element is 8 (even),
  • so we swap it with the element at index i (which is 1) and increment i to 4.
• The ninth element is 5 (odd),
  • so we do nothing and move j to the next element.

• After iterating through the array, we partitioned the even and odd numbers.

The array now looks like this: [2, 4, 6, 8, 5, 1, 3, 9, 7].

• Finally, swap the pivot element with the element at index I.

In this case, we swap 5 with 7. This step ensures that the pivot element is in its final sorted position.

The array after the final swap will be: [2, 4, 6, 8, 7, 1, 3, 9, 5].

Time Complexity: O(n) 
Auxiliary Space: O(1), since no extra space has been taken.