Problem Requirements You are given a list of whole numbers called nums. Your goal is to check if any number appears more than once in this list. * If any number shows up at least twice,

return `tr

Problem Requirements You are given a list of whole numbers called nums. Your goal is to check if any number appears more than once in this list. * If any number shows up at least twice,

return true. * If every number in the list is unique,

return false.

Sample Cases

Case 1: *

Input: nums = [1, 2, 3, 3] *

Result: true

Case 2: *

Input: nums = [1, 2, 3, 4] *

Result: false ***

Solution Strategy To solve this efficiently, we can use a

HashSet. A HashSet is a tool that allows us to quickly check if we have seen a number before. Here is the step-by-step logic: 1. Create an empty

HashSet. 2. Loop through each number in the nums array. 3. For every number, check if it is already inside the

HashSet. *

If yes: This means we have seen it before. We found a duplicate, so we return true. *

If no: We add the number to the

HashSet and check the next one. 4. If we check all numbers and find no duplicates,

return false. This approach is fast because checking a

HashSet takes

O(1) time on average. # Code Implementation

java class Solution { public boolean containsDuplicate(int[] nums) { // Create a HashSet to store numbers we have seen HashSet<Integer> seen = new HashSet<>(); // Loop through the array for (int num : nums) { // If the number is already in the set, we found a duplicate if (seen.contains(num)) {

**return** true; } // Otherwise, add the number to the set seen.add(num); } // If no duplicates are found after checking all numbers return false; } }

Complexity Analysis *

Time Complexity:

O(n) * We go through the list of numbers exactly once. *

Space Complexity:

O(n) * In the worst case (no duplicates), we store every number in the

HashSet.