Remove Element from Array in Place with Two Pointers

✅ Just solved LeetCode 27 — Remove Element! Small problem. Big lesson. 🧠 The task: remove all occurrences of a value from an array in-place and return the count of remaining elements. No extra array. No shortcuts. Just pointer logic. Here's the two-pointer approach I used: def removeElement(nums, val):   k = 0   for i in range(len(nums)):     if nums[i] != val:       nums[k] = nums[i]       k += 1   return k 🔍 The insight: → One pointer (i) scans every element → Another pointer (k) tracks valid write position → Overwrite only what doesn't match val → Result? O(n) time, O(1) space ✨ What's your go-to pattern for in-place array problems? Drop it below 👇 #DSA #LeetCode #CodingInterview #Programming #TwoPointers #Python #SoftwareEngineering #100DaysOfCode

  • No alternative text description for this image

To view or add a comment, sign in

Explore content categories