2021

// Function to check for duplicates using sorting
function containsDuplicate(nums) {
  // Sort the array to bring duplicates together
  nums.sort((a, b) => a - b);

  // Compare each element with its previous
  for (let i = 1; i < nums.length; i++) {
    if (nums[i] === nums[i - 1]) {
      return true; // Duplicate found
    }
  }

  return false; // No duplicates
}

// Sample input
const nums = [1, 2, 3, 1];

// Call the function and print result
console.log(containsDuplicate(nums) ? "true" : "false");

Complexity Analysis

Time Complexity: O(n log n), due to the sorting step.

Space Complexity: O(1), assuming in-place sorting is used.

Optimal Approach

Algorithm

Intuition: A set only stores unique elements, so inserting all elements of the array into a set will automatically remove duplicates. If the size of the set is smaller than the size of the array, it means duplicates were present.

Create an empty unordered set.
Iterate through the array and insert each element into the set.
After insertion, compare the size of the set with the size of the original array.
If the set size is smaller, return true indicating duplicates exist.
Otherwise, return false.

Code