Solving MakeMyTrip Frontend Interview Question | Count Numbers | JavaScript

In this video, we will solve a front-end interview question from top companies like MakeMyTrip. We will implement a function that returns the count of numbers inside an array.
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[@DevtoolsTech]

Try question here: devtools.tech/questions/s/write-a-function-to-return-the-count-of-numbers-in-an-array-or-makemytrip-frontend-interview-question---qid---cjzjsfJUP0zHb2lgPJVy
Other useful links:
topmate.io/yomeshgupta
devtools.tech/questions/all
devtools.tech/lists/all

[@anujtiwari7218]

Tried a question got an approach .The way you write a code it is worth learning . Thanks sir ..

[@DevtoolsTech]

Glad to hear that! 🙌🏼

[@trivedi2690]

hi bhaiya do you have 30 min in this week would love to book a topmate session regarding a switch .

[@DevtoolsTech]

Yes sure. Please book through topmate.io/yomeshgupta

[@1-_-I]

Hey bro, how to tackle hacker Rank front end challenges, is there any other site where we can practice?

[@DevtoolsTech]

I haven't tried other platforms honestly in terms of solving more questions. Only one I can recommend is devtools.tech only.

[@1-_-I]

@@DevtoolsTech haha nice bhai, ofcourse devTools is there

[@DevtoolsTech]

Learnersbucket you can try too!

[@allpeace1]

More Video on Interview ques, Thanks in advance

[@DevtoolsTech]

More coming soon!

[@trivedi2690]

const isNumber =(nums )=>{
let count=0
for (let items of nums){
if(typeof(items)=== "number"){
count++;
}
if(Array.isArray(items)){
count+= isNumber(items)
}
}
return count;
}
const res=isNumber(input)
console.log(res)

[@gantavyasaraswat2130]

Simple solution without using Recursion
function countNumbers(collection) {
let count = 0;
let newArr = collection.flat(Infinity);
for(let i =0;i

[@DevtoolsTech]

Yes, it is one of the ways do it. Thanks for sharing!

[@manojpathak2500]

const countNumbers = (arr) => {
return arr.reduce((totalCount, el) => {
if (typeof el === 'number') totalCount++;
else if (Array.isArray(el)) totalCount += countNumbers(el);
return totalCount;
}, 0);
};