Demo preview: These solutions pass the automated test suite but have not been human-reviewed. Treat trade-offs and prose as draft.

Easyhashing

Contains Duplicate

Problem (restated)

Return true if any value appears at least twice in the array.

Intuition

A set remembers seen values; a second sighting is a duplicate.

Approaches

Hash set

Unverified

Time O(n)Space O(n)

Idea. Insert each number into a set; if already present, return true.

Walkthrough. [1,2,3,1] → see 1 again → true.

Trade-offs. Sorting is O(n log n) with O(1) extra space if allowed to mutate.

Solution

export function containsDuplicate(nums: number[]): boolean {
  const seen = new Set<number>();
  for (const x of nums) {
    if (seen.has(x)) return true;
    seen.add(x);
  }
  return false;
}

export function containsDuplicate(nums: number[]): boolean {
  const seen = new Set<number>();
  for (const x of nums) {
    if (seen.has(x)) return true;
    seen.add(x);
  }
  return false;
}

Template connection

Hash-set “seen before?” from the Hashing template.

Reflection

“At least twice” is a set. Sorting and comparing neighbors is O(n log n).
Are the empty array and a single element both false?
What happens if you insert first and only then ask whether the value was already present?