Two Pointers — the interview technique that wins interviews

You have two indices. Each moves according to a rule. The input is often sorted (or you sort it first). This one pattern collapses many O(n²) brute-force ideas into O(n).

Three shapes

Opposite ends (converging)

Start at lo = 0, hi = n - 1. Move inward based on a condition.

• Two-sum on sorted array
• Palindrome check
• Container with most water
• 3-sum (fix one, two-pointer on the rest)

Same direction (fast + slow)

Both start at the beginning. One moves faster (or one moves conditionally).

• In-place dedup, in-place remove
• Cycle detection (Floyd's tortoise and hare)
• Find middle of linked list

Two arrays (merge-like)

One pointer per array. Advance the one whose element is smaller.

• Merge two sorted arrays
• Intersection of two sorted arrays

When you can't use it

• Unsorted numeric data where sorting would break the required order → often a hash-map approach instead.
• When you need to remember every past element — that's a stack or a hash map.

Interview trap

Interviewers love: "same problem, but the array is not sorted." Sort it first (O(n log n)) then two-pointer (O(n)). Total: O(n log n) — still often better than the O(n²) brute force.

Try it

1. Write is_palindrome(s: str) -> bool using two pointers.
2. Given nums = [0, 1, 0, 2, 0, 3], move all zeros to the end in-place using same-direction two pointers.