The Most Asked Coding Interview Question By Big Tech!!

Solution to a very popular coding interview question asked recently by Google - 2 Sum.

#softwareengineering #javascript #shorts

[AlgoJS]

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[VedankPande]

There's no way google asks twosum in interviews

[yyxzzx]

Maybe they'd give you a complement for finding this.

[KebabTM]

To everyone saying Google wouldn’t ask it, I’ve done multiple coding interviews for tech companies and the first question is always something easy like this. They want to know that you have a basic grounding in programming (plenty of uni graduates still haven’t learnt anything yet since they search answers or cheat), they want to know how you approach problems in general and they want to test your knowledge of time complexity. An easy question like this lets a recruiter instantly fail your application quickly if you couldn’t do it to certain standard (without the recruiter needing to be extremely technical), the harder questions that come afterwards test knowledge of algorithms and data structures and let companies deny you for a certain role, or categorise you for different teams.

[nagyzoli]

When you build the map, it will take you "n" steps to add an element (inserting a new element into a binary search tree, which is what a map is at base is worst case N). So every time you invoke the map.set(..) option, you have a "hidden cost" of N, making the total complexity O(N^2) in case no elements are found.

[evanfreethy]

I suggest you experiment with providing the intuition behind the solution up front - then walk through the solution. It makes it easier to learn in my opinion.

[nitishkumarprusty]

function solution(nums, target){
let start = 0;
let end = nums.length - 1;
let res = [];
while(start <= end){
if(nums[start] + nums[end] == target){
res.put(nums[start], nums[end]);
start++;
end- -;
}else if(nums[start] + nums[end] > target){
end- -;
}else{
start++;
}
}
return res;
}

Time complexity- O(n)

[classicjonesy]

One way to solve this is to calculate number of possible combos then use a while loop to exhaust every combo in parallel

[gavindeane]

The singular of "indices" is "index"!

[SirBearingtonSupporter]

You forgot the other part of this question and that is the follow up. How do you handle large data sets that won’t fit into a single array // handle multiple data streams to allow them to run in parallel.

[meow]

function solution(nums, target) {
return [0, 1]
}

complexity O(1)
you're welcome

[thecorleone]

we need subtitles for your videos, much love for you <3

[FeuervogelIra]

No, its not O(n). Its O(n²) what you did there. Maps use Hashmaps which have O(n) worst case complexity. Yielding O(n²) overall complexity. It is only O(n) _average_ complexity. Important difference.

[mlguy]

You know you grind leetcode when you know the url for this problem and level by heart.

[supergamerstv]

function solution(target, nums) {
const list = [];
for (let i = 0; i < nums.length; ++i) {
const computation = target - nums[i];
if (computation <= 0 || !nums.includes(computation)) {
continue;
}
list.push([nums.indexOf(computation), i]);
}
return list;
}

[roman-romadin]

Please write ✍️ condition at the top - it is convenient to have a condition in the form of text 📝 before the task. Thank you for this great content.

[alexanderten]

Spelling without jumping to different parts of the code, amazing, especially for the speaker

[zahoorahmedsayyad]

function solution (nums, target) {
for (i = 0; i < nums.length; i++) {
for (j = i+1; j < nums.length; j++) {
if (nums[i]+nums[j] == target)
return [i, j];
}
}
return null;
}

[parisgiannoukos]

simple staff hahaha

const nums = [-2, -5, 1, 15, 2, 7];
const target = 9;

function solution(nums, target) {
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] === target) return [i, j];
}
}

return [];
}

console.log(solution(nums, target));

[sidharthshanu]

i think doing it binary search style is better left and right if their sum> num right-- or sum==num left and right or left-- should be more efficient