Leetcode Solution - 1.0 Two Sum | Javascript

This is a detail Solution to the Leetcode question Two Sum written in Javascript.

[shrfu]

This is super helpful as a bootcamp student. I also appreciate seeing the errors happen.

[humancyyborg]

you are amazing, the best leetcode walkthrough so far... so natural, not just pounding straight into problems

[sreekumarmenon]

if the array is sorted, we can use 2 pointers approach as well -

function twoSum(nums, target){
let left =0;
let right = nums.length-1;
while(left < right){
const sum = nums[left] + nums[right];
if(sum == target){
return [left, right];
}
else if(sum > target){
right--;
}
else {
left++;
}
}

return [-1, -1];
}

[Onomandah]

Another solution, perhaps a much easier one, would be to have 2 pointers on either end of the array. But this solution was new for me. Thank you.

[Amar]

Thanks for the video!
I am a beginner, Learned a lot for you.

[narayanareddy]

Thanks for the detailed explanation.
Please continue this series 🔥

[hongkaihuang]

best explanation... You are a monster.

[islamibrahim]

This was really helpful.
(My attempt at interpreting the solution provided)
Halfway through your introduction of the solution I thought you were going to do a first loop where each difference is stored in the map and then a second one where we check each element against the map.
This is an interesting optimisation of this that uses the property that if two numbers add to a third, then removing one number will yield the other (which, based on the question, should be in the list. And no repetition so just one pass was sufficient).

Good stuff!

[altayezekariyas]

const twoSum = (nums, target) => {
for(i of nums ){
a = target - i
if (a in nums){
return [nums.indexOf(i), nums.indexOf(a)]
}
}
};

[ДениИмагожев]

That's perfect, very detailed and helpful👍🏻👍🏻👍🏻 thanks

[TheSaltyCook]

this was really great. thank you for walking through each step of the logic.

[javascript_developer]

I find this same approach on other videos too. but thanks your explaination did helped me understand.

[PavanKumar-mrtd]

Wonderful explanation i loved it. KEEP DOING..

[beluga]

another way if you dont want like this

method :-

const nums = [1, 2, 3, 4, 5, 6, 7, 8, 9];
let target = 5;
var countElements = function(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];
}
}
}
}
console.log(countElements(nums, target));

[SuperKojota]

Hi.Why you dont use classic if-else in this challenge? I am used to work with if(statement ) then { return something} then else{ return something}.

[saymemore]

Thank you for content and understanding;)

[nehaprakash]

This is Great!! Subscribed.. Thanks to you for this!

[simamorpho]

Simpler to understand working answer:

nums = [9, 0, 7, 2];
target = 9;

function twoSum(array, target) {
answer = [];
for (let i = 0; i < array.length; i++) {
for (let j = i + 1; j < array.length; j++) {
let sum = array[i] + array[j];
if (sum === target) {
if (array[i] === array[j]) continue;
else {
answer.push(i);
answer.push(j);
return answer;
}
}
}
}
}

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

[gearsighted]

Great stuff, I'm still a little confused but significantly less so! Subscribed! :D

[JustThink]

Definitely not the most efficient solution but this was nice to see as an addnl way to solve it.