Solve the Easy-level Contains Duplicate coding interview problem online in Python & JavaScript. Run your code against real test cases in the browser and study worked examples. It covers arrays and hashmap. Read the step-by-step solution and complexity analysis. Java starter code is also included for reference (no in-browser JVM yet, so it can't be run here).

Contains Duplicate – Solve Online in Python & JavaScript

Maximum Subarray

FizzBuzz

Determine whether any integer appears more than once in an array.

Return `true` when any value occurs at least twice in `nums`; otherwise return `false`.

**Input / output**
- Input: `nums: integer[]`
- Output: boolean

**Examples**
1. `[1,2,3,1]` returns `true`.
2. `[1,2,3,4]` returns `false`.

**Constraints**
- `0 <= nums.length <= 100,000`
- `-1,000,000,000 <= nums[i] <= 1,000,000,000`

**Follow-up**
Compare a hash set with sorting when memory is constrained or modifying the input is allowed.

- We need to decide whether any value appears more than once.
- The exact duplicate value does not need to be returned, only `True` or `False`.
- An empty or fully distinct list should return `False`.

Understand the Problem

- A direct approach compares every pair of numbers.
- If any pair has the same value, we found a duplicate.
- This works, but checking all pairs takes O(n²) time.

Consider the Brute Force Idea

- A set can tell us in average O(1) time whether we have seen a number before.
- Scan the array once from left to right.
- When a number is already in the set, return `True` immediately.

Use a Set for Fast Lookups

- Keep a `seen` set.
- For each number, check membership before adding it.
- If the loop finishes, every number was distinct.

Build the Check Incrementally

- The final solution returns as soon as a duplicate is found.
- Time complexity is O(n) because each element is processed once.
- Space complexity is O(n) in the worst case for the set of seen values.