LeetCode 27 | Remove Element | O(n) Time | O(1) Space | Two Pointer Solution

Description: Remove Element is a classic beginner-friendly LeetCode problem that helps improve your understanding of arrays, in-place modification, and the two-pointer technique. You are given an integer array nums and an integer val. Remove all occurrences of val in-place and return the number of remaining elements. The order of the elements may be changed, and you must use only constant extra space. ✅ Approach The idea is simple and efficient: Use a pointer to track the position where the next valid element should be placed Traverse the array from left to right If the current element is not equal to val, place it at the valid position Increment the valid position pointer Skip elements equal to val After processing all elements, the pointer value represents the new length of the array This approach modifies the array in-place and is perfect for learning the two-pointer technique. ✅ Time Complexity O(n) → We traverse the array only once ✅ Space Complexity O(1) → No extra data structures are used 🔗 LeetCode https://leetcode.com/u/TR0kHGhQN9/ 🔗 LinkedIn   / dhruvi-patel-bb2b9239a   📝 Notes A classic in-place array modification problem Excellent introduction to the two-pointer technique No additional array or data structure is required Frequently asked in coding interviews and assessments Helps strengthen array traversal and pointer manipulation skills Demonstrates how to optimize space usage effectively #leetcode #dsa #coding #programming #java #python #cplusplus #arrays #twopointers #removeelement #leetcodeeasy #codinginterview #algorithms #softwareengineer #computerscience #datastructures #arraymanipulation #inplacealgorithm #logicbuilding