Code Challenge: Contains Duplicate

Working through these challenges on LeetCode is an interesting experience. I try to solve the problems or at least think an approach through prior to hitting the docs and as a last result searching for solution specifics. This is to avoid getting overly frustrated since the purpose is to try and stretch my knowledge while working through the challenges.

This was a fun one as the difficulty wasn't too frustrating and provided a few different possible solutions.

The challenge this time is:

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 :

• Input: nums = [1,2,3,1]
• Output: true

Constraints:

• 1 <= nums.length <= 10^5
• -10^9 <= nums[i] <= 10^9

I love the MDN docs, I knew of indexOf and I was curious if there were any parameters I could use to solve this challenge. What I discovered was lastIndexOf. Using these two methods, finding duplicates is only a matter of comparing the indices of the first and last instances of the value.

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var containsDuplicate = function(nums) {
    for(let i=0;i<nums.length;i++){
        let current= nums[i];
        let firstIndex= nums.indexOf(current);
        let lastIndex= nums.lastIndexOf(current);
        if(firstIndex!=lastIndex){
            return true;
        }
    }
    return false
};

Alternatives

I was happy to see my solution was passing all tests but while lastIndexOf was useful it seemed like there were likely other more standard solutions.

Using Slice & Includes

I decided to try a combination of using slice and includes

• Use slice to create a new array with only the values after the current value
• Use includes to check for the current value inside the new array.

Initially I was concerned about duplicates present before the current position. Using the same logic of slice and includes, I checked both the array of values before after the current position. This failed the challenges time/performance constraint. Then I realized it wasn't even a necessary concern, since I was iterating through the whole array the before values has already been checked.

When run without the before values checked this solution also was accepted.

var containsDuplicate = function(nums) {
    for(let i=0;i<nums.length;i++){
        let current= nums[i];
        let arrayAfter= nums.slice(i+1, nums.length);
        if(arrayAfter.includes(current)){
            return true;
        }
    }
    return false
};

Using a Set

I spent a little time after searching for other solutions to see how I felt mine fit in. I found this post by Will Harris and found his solution using the javascript Set object. I've looked at sets before but haven't had much chance to use them, since each value in a set need to be unique, its going to filter out any duplicates in the array. The solution worked and looked very clean.

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