How I Sorted an Array of 0s, 1s, and 2s in Python Without Using sort()

 Introduction:

While practicing Python problems, I came across an interesting question where I had to sort an array that contains only 0s, 1s, and 2s.

The twist was that I was not allowed to use the built-in sort() function. At first, I didn’t know how to approach it, but then I learned a simple method using counting.

In this blog, I’ll explain how I solved it step by step.

Problem Statement:

We are given an array like this:

2 0 1 2 0 1 0

We need to sort it in ascending order:

0 0 0 1 1 2 2

My Idea:

Instead of sorting directly, I thought:

• Why not count how many 0s, 1s, and 2s are there?

Then I can just rebuild the array using those counts.

Code I Used:

arr = list(map(int, input("Enter elements (0s,1s,2s): ").split()))
count0 = count1 = count2 = 0
for num in arr:
    if num == 0:
        count0 += 1
    elif num == 1:
        count1 += 1
    else:
        count2 += 1
i = 0
for i in range(count0):
    arr[i] = 0
    i += 1
for i in range(count1):
    arr[i] = 1
    i += 1
for i in range(count2):
    arr[i] = 2
    i += 1
print(arr)

Explanation:

• First, I take input from the user

• Then I count:

  • how many 0s

  • how many 1s

  • how many 2s

• After that, I overwrite the array:

  • fill 0s first

  • then 1s

  • then 2s

This way, the array becomes sorted without using sort().

Example Run:

Input: arr[] = [0, 1, 2, 0, 1, 2]
Output: [0, 0, 1, 1, 2, 2]

What I Learned:

• Even without sort(), we can solve problems using logic

• Counting technique is very useful

• This problem improved my thinking skills

Conclusion:

This was a simple but interesting problem for me as a beginner. It helped me understand how to approach problems in different ways instead of always depending on built-in functions.