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Sri Mahalakshmi
Sri Mahalakshmi

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Sorting 0s, 1s, and 2s Using the Counting Method in Python

#python #beginners #programming

Problem Explanation

You are given an array that contains only 0s, 1s, and 2s.
Your task is to sort the array in ascending order without using any built-in sort function.

Example:

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

Idea Behind the Solution

Instead of comparing and swapping elements, we do something smarter:

  1. Count how many 0s, 1s, and 2s are present
  2. Rewrite the array using those counts

This avoids unnecessary operations and keeps the logic simple.

Python Code with Explanation 

class Solution:
    def sort012(self, arr):

        count0 = 0
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We create a variable count0 to store how many times 0 appears in the array.

        count1 = 0
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This variable stores how many times 1 appears.

        count2 = 0
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This variable stores how many times 2 appears.

        for num in arr:
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We loop through each element in the array one by one.

            if num == 0:
                count0 += 1
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If the element is 0, we increase the count of 0s.

            elif num == 1:
                count1 += 1
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If the element is 1, we increase the count of 1s.

            else:
                count2 += 1
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If it's not 0 or 1, it must be 2, so we increase the count of 2s.

Now we know how many 0s, 1s, and 2s are present.

        index = 0
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We start filling the array from index 0.

        for i in range(count0):
            arr[index] = 0
            index += 1
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We fill the array with all the 0s first.
Each time we place a 0, we move to the next index.

        for i in range(count1):
            arr[index] = 1
            index += 1
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Next, we fill all the 1s.

        for i in range(count2):
            arr[index] = 2
            index += 1
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Finally, we fill all the 2s.

        return arr
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We return the sorted array.

Complete Code (Final) 

class Solution:
    def sort012(self, arr):
        count0 = 0
        count1 = 0
        count2 = 0

        for num in arr:
            if num == 0:
                count0 += 1
            elif num == 1:
                count1 += 1
            else:
                count2 += 1

        index = 0

        for i in range(count0):
            arr[index] = 0
            index += 1

        for i in range(count1):
            arr[index] = 1
            index += 1

        for i in range(count2):
            arr[index] = 2
            index += 1

        return arr
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Step-by-Step Example

Input:

arr = [0, 1, 2, 0, 1, 2]
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Step 1: Count

  • count0 = 2
  • count1 = 2
  • count2 = 2

Step 2: Rewrite array

  • First 2 positions → 0
  • Next 2 positions → 1
  • Last 2 positions → 2

Output:

[0, 0, 1, 1, 2, 2]
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Time and Space Complexity

  • Time Complexity: O(n) (one pass to count + one pass to fill)
  • Space Complexity: O(1) (no extra space used)

Key Takeaway

This method is one of the easiest ways to solve the problem because:

  • No swapping is required
  • No complex logic
  • Just counting and rewriting

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