Overview

Given an array of integers, determine whether any value appears at least twice. Return true if any duplicate exists, false if every element is distinct.

Implement hasDuplicates(nums).

Constraints

• Input is always an array of numbers (may be empty).
• An empty array and a single-element array return false.
• Must handle negative numbers and zero.
• Aim for $O(n)$ time complexity.

Examples

hasDuplicates([1, 2, 3, 1]);        // true  — 1 appears twice
hasDuplicates([1, 2, 3, 4]);        // false — all distinct
hasDuplicates([]);                   // false
hasDuplicates([1, 1, 1, 3, 3, 4]);  // true

Notes

• A Set-based approach gives $O(n)$ time: iterate and check membership before inserting.
• Alternatively, sort first ($O(n \log n)$) and scan for adjacent duplicates — but this mutates the input.
• Watch for the early-exit optimization: return true as soon as you find the first duplicate.

Solution

function hasDuplicates(nums) {
  const seen = new Set();
  for (const n of nums) {
    if (seen.has(n)) return true;
    seen.add(n);
  }
  return false;
}

Resources

• Contains Duplicate — LeetCode
• Set — MDN