JavaScript LeetCode Contains Duplicate

Introduction

Continuing through the problems of LeetCode. I am not haphazardly selecting any questions. I am following along with this list for those that were curious:

techinterviewhandbook.org/best-practice-que..

Prompt

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

At first glance, the problem seems to be pretty simple. The idea here is to iterate over the array and find any duplicates. There’s a bit of a cheat code we can use with JavaScript in this one.

First Solution (cheating)

There is a standard built-in object for JavaScript called Set

But what does this object do?

Set objects are collections of values. You can iterate through the elements of a set in insertion order. A value in the Set may only occur once; it is unique in the Set's collection.

Knowing this, the only thing we need to do is convert our array into a Set and compare its length with the original number array length.

var containsDuplicate = function(nums) {
    const set = new Set([...nums]);
    return set.size != nums.length;
};

Converting an array into a Set is simple, just spread it into a new array in the Set constructor. It has a property to calculate the size (number of items). We just have to compare that with the length of the original array, nums.

return set.size != nums.length;

Second Solution

Much like our (first solution) we can create a hash table of our array as we’re iterating and evaluate it in place.

var containsDuplicate = function(nums) {
   const hashTable = new Map();

    for(let i = 0; i < nums.length; i++) {
        if(hashTable.has(nums[i])) return true;
        else hashTable.set(nums[i], true);

    }
    return false;
};

To break this down a little bit: we first iterate over the array of numbers. If the map already has the value then we return true. We check this with the .has property of map s

if(hashtable.has(nums[i]) return true;

Otherwise, we add it to the map and move on.

else hashTable.set(nums[i], true);

The value is pretty irrelevant as we don’t really care much for it. There probably is a data structure more suited here.

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