Question

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

Example 1:

Input: nums = [1,2,3,1]

Output: true

Explanation:

The element 1 occurs at the indices 0 and 3.

Example 2:

Input: nums = [1,2,3,4]

Output: false

Explanation:

All elements are distinct.

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]

Output: true

Constraints:

• 1 <= nums.length <= 105
• 109 <= nums[i] <= 109

My Solution

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var containsDuplicate = function(nums) {
    let map = new Map();
    for (let i = 0; i < nums.length; i++){
        if (map.has(nums[i])) {
            return true
        }
        map.set(nums[i], 1)
    }
    return false
};

Key Takeaway

I used a hash map to efficiently check if the array contains duplicates. By leveraging the hash map's constant time complexity for insertion and lookup O(1), I was able to iterate through the array in linear time O(n).

Here’s how it works:

1. As I iterate through the array, I check if the current value already exists in the hash map.
2. If it does, the function immediately returns true, indicating a duplicate is found.
3. If not, I add the current value as a key in the hash map and continue.

This approach is both time-efficient and simple to implement. While the space complexity is O(n) due to the hash map, it's a reasonable trade-off for the performance gain, especially compared to less efficient O(n^2) brute-force methods.