Coding Interview Prep— Contains Duplicate with 4 Solutions

Prompt

Solution 1 — Brute Force using nested for loops

/*
Brute Force method
Time: O(n^2)
Space: O(1)
*/
var containsDuplicate_nestedForLoops = function(nums){
    for (let i=0; i<nums.length; i++){
        for (let j=i+1; j<nums.length; j++){
            if (nums[i]==nums[j]){
                return true;
            }
        }
    }
    return false;
}

Solution 2 — Sorting Method and check the adjacent element

Sort the array and check if the next element is a duplicate of the current element.

/*
Sorting Method: compare 2 neighbors
Time: O(nlogn) from sorting
Space: O(1)
*/
var containsDuplicate = function(nums){
    const sort = nums.sort((a,b)=>a-b);

    for (let i=0; i<sort.length; i++){
        if (i+1 < sort.length){
            if (sort[i]==sort[i+1]){
                return true;
            }
        }
    }
    return false;
}

Solution 3 — Hashing Method

/*
Hashing Method
Time Complexity: O(n)
Space Complexity: O(n)
 */
var containsDuplicate = function(nums) {
    let obj = {};
    for (let i=0; i<nums.length; i++){
        let num = nums[i];
        if (obj[num]){
            return true;
        } else {
            obj[num] = true;
        }
    }
    return false;
}

Solution 4 — Compare Lengths of Array vs. Set

/*
Compare length of array vs. set. The set will remove the unique elements.
Time Complexity: O(n) to create the set
Space Complexity: O(n)
*/
var containsDuplicatex = function(nums) {
    return nums.length !== new Set(nums).size;
};