Sort 0s, 1s, and 2s

Problem You’re given an array of integers and need to sort an array containing only 0s, 1s, and 2s. Strategy At first, it feels like you can just sort the array or count frequencies. But the follow-up asks for a one-pass solution with constant space. Instead, I thought about it differently: At each position, I asked— Where should this element go? So I used three pointers: low for placing 0s mid for traversal high for placing 2s So for every element: If it’s 0 → move it to the front If it’s 1 → leave it where it is If it’s 2 → move it to the end This way, everything gets sorted in a single pass. Code class Solution: def sort012(self, arr): low = 0 mid = 0 high = len(arr) - 1 while mid <= high: if arr[mid] == 0: arr[low], arr[mid] = arr[mid], arr[low] low += 1 mid += 1 elif arr[mid] == 1: mid += 1 else: arr[mid], arr[high] = arr[high], arr[mid] high -= 1 Key Lines Explained if arr[mid] == 0: Move 0 to the front by swapping with the low pointer. elif arr[mid] == 1: Already in the corre