Problem

https://leetcode.com/problems/contains-duplicate/

Problem Summary

You are given an array of numbers.

You need to check:

• Does any number repeat at least once?
• If yes → return true
• If no → return false

Core Idea (easy explanation)

As you go through the array:

• Keep a set to store numbers you have already seen.

• For each number:

  • If it’s already in the set, then it's a duplicate → return true
  • Otherwise, add it to the set and continue

If you finish the loop and found no duplicates → return false

Why a Set?

A set ignores duplicates and checking if something exists in a set is very fast (O(1)).

Step-by-Step Explanation

Let’s take: nums = [1, 2, 3, 1]

1. Start with an empty set: {}
2. Look at 1 → not in set → add it → set becomes {1}
3. Look at 2 → not in set → add it → {1, 2}
4. Look at 3 → not in set → add it → {1, 2, 3}
5. Look at 1 again → this time 1 is already in the set
6. So return true

🧠 Time and Space Complexity

• Time: O(n)
• Space: O(n)

📝 Pseudocode

create empty set seen

for each number in nums:
    if number is in seen:
        return true
    add number to seen

return false

🐍 Python Solution

class Solution:
    def containsDuplicate(self, nums):
        seen = set()
        for num in nums:
            if num in seen:
                return True
            seen.add(num)
        return False

☕ Java Solution

import java.util.HashSet;
import java.util.Set;

class Solution {
    public boolean containsDuplicate(int[] nums) {
        Set<Integer> seen = new HashSet<>();

        for (int num : nums) {
            if (seen.contains(num)) {
                return true;
            }
            seen.add(num);
        }

        return false;
    }
}