Find the intersection node of two linked lists (if any) ListNode GetIntersectionNode(ListNode headA, ListNode headB) { if (headA == null || headB == null) return null; ListNode a = headA, b = headB; while (a != b) { a = (a == null) ?

Short answer: headB : a.next; b = (b == null) ? headA : b.next; return a; // either intersection or null Explanation: Two pointers traverse both lists; if no intersection, both will reach null simultaneously. headB : a.next; b = (b == null) ? headA : b.next; return a; // either intersection or null Explanation: Two pointers traverse both lists; if no intersection, both…… will reach null simultaneously. headB : a.next; b = (b ==…

Explain a bit more

null) ? headA : b.next; return a; // either intersection or null Explanation: Two pointers traverse both lists; if no intersection, both will reach null simultaneously. headB : a.next; b = (b == null) ? headA : b.next; return a; // either intersection or null Explanation: Two pointers traverse both lists; if no intersection, both will reach null simultaneously. headB : a.next; b = (b == null) ? headA : b.next; return a; // either intersection or null Explanation: Two pointers traverse both lists; if no intersection, both will reach null simultaneously. headB : a.next; b = (b == null) ? headA : b.next; return a; // either intersection…

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.