Odd Even Linked List | Two Chains, One Pass | LeetCode 75 #30

Odd Even Linked List (LeetCode 328, Medium) — explained in Hindi, from absolute zero. Group all odd-index nodes first, then even-index nodes — in one pass, O(1) space, by rewiring arrows in place. The card-deal intuition: two players (odd and even pointers), every node dealt to exactly one chain, no node ever dropped. Full animated dry run on [2,1,3,5,6,4,7], line-by-line C++ walkthrough (why every line exists), and three hard parts opened slowly: (1) why a skipped node is never lost — the held-by-the-other-chain invariant, (2) the final join via evenHead plus the cycle trap if the even tail is not null, (3) why the loop needs BOTH even and even→next checks — one check crashes on each parity. Correctness proof via loop invariant, O(n) time / O(1) space reasoning, 8 edge cases, and interview Q&A. Problem: https://leetcode.com/problems/odd-eve... Timestamps: 00:00 Intro — Two Chains, One Pass 01:17 Problem — Odd Positions vs Even Values 04:21 Brute Force — Copy Nodes (and why it fails) 06:04 The Idea — Card Dealing 09:07 Hard Part 1 — Skipped Node Never Lost 13:24 Hard Part 2 — Final Join and Cycle Trap 16:59 Hard Part 3 — Loop Condition (Both Checks) 19:56 Full Dry Run — [2,1,3,5,6,4,7] 24:06 C++ Code — Line by Line (Why) 28:18 Edge Cases — 8 Tests 30:21 Correctness Proof — Loop Invariant 32:23 Interview Q&A 33:42 Outro — Recap #LeetCode #DSAinHindi #CodingInterview #Placements #Cpp #LeetCode75 #LinkedList #OddEvenLinkedList