JavaScript interview questions (Count Boomerangs In Array )

JavaScript interview questions 🔥 Count Boomerangs In Array #reactinterviewquestions #javascriptinterviewquestions #reactjs #programming #coding
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[@g.pavani5275]

Very nice explanation bro it's really awesome ....waiting for regular videos to get uploaded in this interview series of JS

[@itscodingdoctor]

Thanks 😊

[@coder0015]

Started learning JS recently.
loved to watch this video, I literally want more of these!!

[@itscodingdoctor]

Great. You can find 23 more questions in this series

[@nitinsoni9956]

I love this series but I guess the difficulty level was not hard as it was kind of easy question. I solved the question with the exact solution that was shown in the video.

[@itscodingdoctor]

Happy to hear that you could solve it yourself.
It's a hard question though. But anyone having good knowledge of javascript can solve this.

[@itscodingdoctor]

I'll keep on increasing the difficultly level

[@nittasreenivas4043]

please solve this question alternating positive and negative number in js input:[9,4,-2,-1,5,0,-5,-3,2] output:[9,-2,4,-1,5,-5,0,-3,2]

[@itscodingdoctor]

Sure. Will create a video

[@nitinsoni9956]

At first the solution seems very simple but when I tried it damn it was tricky. Solved it by first finding indices of negative and positive numbers than using simple loop and finding the lower bounds length and upper bounds of respected arrays assigned the indices to new array and lasted compared which array is bigger and assigned remaining indices from that array.

[@nitinsoni9956]

let arr = [9, 4, -2, -1, 5, 0, -5, -3, 2,-2,2,3,3];
// pArr => positive indices nArr=> negative indices
let pArr = arr.map((el, i) => (el >= 0 ? i : " ")).filter((el) => el !== " ");
let nArr = arr.map((el, i) => (el < 0 ? i : " ")).filter((el) => el !== " ");
// to store the new alternate numbers
let alArr = [];
// bb & lb => higher bound & lower as one of two arrays can be bigger hence we have to loop till bigger array length
let bb = pArr.length > nArr.length ? pArr.length : nArr.length;
let lb = pArr.length < nArr.length ? pArr.length : nArr.length;
let pi=0,ni=0;
for (let i = 0; i < bb; i++) {
// as till lower bound both positive and negative indices array would have same number of elements
if(lb > i){
alArr.push(arr[pArr[pi]]);
alArr.push(arr[nArr[ni]]);
pi++;ni++;
}
else{
// as positive array is bigger remaing numbers would be added from this array
if(pArr.length > lb){
alArr.push(arr[pArr[pi]]);
pi++;
}
// as negative array is bigger remaing numbers would be added from this array
else{
alArr.push(arr[nArr[ni]]);
ni++;
}
}
}
console.log(alArr);

[@nittasreenivas4043]

@@nitinsoni9956 thankyou bro for helping out i just tried like this
let arr = [9, 4, -2, -1, 5, 0, -5, -3, 2,-2,2,3,3]
let positive = arr.filter((num) => {
return num >= 0;
})
let negative = arr.filter((num) => {
return num < 0;
})
let result = [];
for(let i=0,j=0; i

[@nitinsoni9956]

@@nittasreenivas4043 You're welcome and there was always room for improvement in my code. I'm looking forward how codingDoctor solves the given problem.

[@sakthiii711]

Nice video for new topic bro, Please can you make video for [1,2,3,4,5] this will come to X pattern cross method .this was my interview question i don't know how to solve this problem.

[@itscodingdoctor]

Thanks

[@itscodingdoctor]

I couldn't understand this properly

[@itscodingdoctor]

Can you give me the input and expected output? I'll try to solve and create a video too