JavaScript Algorithms - 9 - Prime Number

[@xitri1068]

Wow the optimize way of doing it really humbled me down. I learnt that there's always more optimize and better way of writing code.

[@ajayraja6636]

Could you please Zoom in replit for next video? My eye needs to zoom in for that theme

[@ogucharles2800]

Thanks a lot for the lessons. Anticipating for the next class.

[@arasefe]

Very good example of O(n) analysis

[@khusnijafar587]

Great explanation, thank you sir.

[@gabrieludo3436]

Thanks, sir, you deserve 1Million plus subscribers

[@veryhightower]

Hello. Thank you very much for your videos. I learn algorithms with your playlist. And also recording videos with this progress and with my solves.

[@Adarshabinash]

Hey I have a little confusion in the code..When inside the for loop, I use the if condition for n%i===0 and then inside the for loop only I use the return true in an else condition, it returns wrong values for numbers like 2 and 55..but works perfect if I do as the tutorial...any idea where I am doing wrong?

[@veryhightower]

@@Adarshabinash could you share you code for review?

[@ebrubendeliyeva]

Thank you)

[@abhisheksharma10600]

function isPrime(n) { if (n === 2) { return true; } for (let i = 2; i < n; i++) { if (n % i === 0) { return false; } } return true; }

[@oscargm1979]

Careful, if the user enters 1 it won't enter through the loop at all and will go for the final return true. The first guard clausule should be as in the video if(n < 2)

[@CarlosRodriguez-pn7fe]

Amazing xplanation.

[@jmeahra]

amazing, keep uploading

[@HodaElsyed]

Love it 🌹

[@Adarshabinash]

Hey I have a little confusion in the code..When inside the for loop, I use the if condition for n%i===0 and then inside the for loop only I use the return true in an else condition, it returns wrong values for numbers like 2 and 55..but works perfect if I do as the tutorial...any idea where I am doing wrong?

[@xraven3387]

When you return true in the for loop from the else statement when n%i !== 0 it automatically returns true and stops the for loop, therefore stops the function before the completion of the for loop.

[@manishaggarwal5643]

can't i do in for loop i < (n / 2 + 1); this will reduce our steps

[@CrouTonG]

function isPrime(n) { if (n < 2) return false else if (n % 2 !== 0 || n === 2) return true return false }

[@khairiyusoff5040]

but if n=9, it return true.

[@bzchii7474]

121 will return true here, which is wrong as 121 is not a prime number :)

[@NaveenKumar-ne4yf]

This method checks if reminder is 0(zero), but for 9,15,27.. the reminder will not be zero(0) and it will return them as Prime numbers.

[@francismpangirwa1856]

function prime(n) { let result = true; for (let i = 2; i < n; i++) { if (n % i === 0) { result = false; } } return result; } console.log(prime(5)); console.log(prime(4));

[@gamersruin]

The square root trick is interesting, but does it actually optimize the function? The first instance where n % i === 0 will trigger the return statement, so all future loops will not run.

[@antony2894]

If the instance (n % i === 0) is true it means that there is other number other than n and 1 that can divide n giving result 0, therefore it is not prime

[@gamersruin]

@@antony2894 Thank you for the explanation

[@nicolau7k]

function isPrime(n) { let cont = 0 for (let i=1; i

[@demno2258]

Amazing tutorial as always expect ; thanks a lot but I have bug in prime(2)

[@shirefhamam4457]

if you mean it returns false, you need to add if statement if n=2 it shall returns true, otherwise it checks for the remaining lines

[@davidpokrajac557]

What if for example we put 1 as the parameter...first if statement is success and we return false...would it be then consant O(1) time complexity?

[@damo190]

Sir, the value of 'this' is lexical(static) scoped or dynamic scoped? I have a confusion. I think it's dynamic scoped because it gets its value through execution context. But I have already learned, JavaScript only uses lexical scope

[@-xyzan-]

when Global Execution Context get created then we will get 'this' and it is an object provided by the browser and it belongs to lexical scope

[@nothing_7774]

yeah and 'window' too can be usefull

[@yasserelbatal153]

good work go Keeping

[@piotrekjazz1287]

Second solution would be more optimal if the loop wasn't stopped by return, otherway math function only adds extra calculations, both solutions are at most O(sqrt(n))

[@yusufalp1786]

Your statement is correct only for non-prime numbers. Think of a prime number, say 37. without Math.sqrt, loop goes for 35 times, with it it goes for 6 times. Imagine a bigger prime number.

[@AnilRoxsta]

​@@yusufalp1786thank you 🙏

[@oscargm1979]

Calculate primes in a given range: const extractPrimes = range => { return range.filter((n) => isPrime(n)); } console.log(extractPrimes(Array.from(Array(100).keys())))

[@ojooladimeji5177]

function primeNumber(n){ for (i = 2; i < n; i++){ if (n % i || n===2){ return ' Prime Number' } else { return ' Not a Prime Number' } } }

[@DantesSagan]

Also might be work as well in that way: function IsPrime(n){ // for(let i = 2; i < n; i += 1){ // if(n % i === 0){ // return false; // } // } if(n === 1){ // return false // } // return true; let i = 2; while(i < n){ if(n % i === 0) { return false } i += 1; } if(n === 1){ return false; } return true } let result = IsPrime(1); let resultTwo = IsPrime(4); let resultThree = IsPrime(5); let resultFour = IsPrime(37); let resultFive = IsPrime(97); console.log(result, resultTwo, resultThree, resultFour, resultFive)

[@HuntsDown]

What does it meant to trace the function execution? So many terms and phrases are used but nobody seems to explains what it means or what the process is of doing so. Can someone point me in the right direction?

[@ronaldngarombo1026]

Same question am asking?

[@armoredchimp]

I believe he just means to go through the code step by step, writing it down on a piece of paper. It's not a specific technical term related to coding, he just means you should take the time to understand exactly what the function is doing each step of the way.

[@HuntsDown]

@armoredchimp thanks bro! I got this already, but hopefully your comment will help someone else. :)

[@ibearua]

how do come up with the math principles, i 'm pretty bad with remembering math concepts

[@saifhamdare72]

i have the most popular yet boring solution to it.. PRACTICE🙌🙌

[@srustithakkar1335]

Since we already know almost all numbers are divided by 2,3,5,7. I came up with a solution that can optimize it to O(logn): if(input < 2){ return false }else if(input % 2 === 0 || input % 3 === 0 || input%5 ===0 || input % 7 ===0){ if( input === 2 ||input === 3 || input === 5 ||input === 7 ){ return true } return false }else { for(let i = 2; i < input; i++){ if(i % input === 0){ return false } } return true } comments appreciated!

[@zisgeim09]

omg what a bullshit

[@pronoychowdhurypothik3586]

Sir , if i want to be a expert at this kind of math,which subject or topics of the math ,i should study on? Sir ,Please answer my this question,because i wanted to study cse,but somehow i'm out of the opportunity now ,But still i'm trying to learn everything covered in a normal cse graduation even if it is not academically.

[@Alegend13]

function isPrime(n) { if(n == 2 || n == 3){ return true } else if(n < 2 || n % 2 == 0 || n % 3 == 0) { return false } return true }

[@charanraj.g6560]

how does this code works for 2 even its remainder is zero

[@blessingalfred]

Solution for prime numbers (will only run a maximum of nine times) const findPrime = n => { if (n < 2) return false let isPrime = true for (let i = 2; i < Math.min(9, n); i++){ if ((n % i === 0)){ isPrime = false; break } } return isPrime }

[@jeremiahojo2048]

😂Never seen a more complicated definition of prime numbers

[@Abdelrahman24434]

complicated?

[@rituraj9552]

what is the Big O of this program: function isPrime(n) { if (n % 2 != 0 && n > 2) { return true } else { return false } } console.log(isPrime(1))

[@m_97__]

Brother In if statement why n%2 ?

[@oussamahamdi6993]

And i think we can check if n > 2 && isEven then we return false because all Prime numbers except for 2 are Odd.

[@PawanChauhan-nd6il]

What about this?Is below mention code is wrong.. const prime = (n) => { const val = n % 2; let result = false; if (val == 1 && n != 1) { result = true; } return result; } console.log(prime(5)); console.log(prime(4)); console.log(prime(990));

[@reactworld8323]

second solution does not work like if you put 9 then it will check only up to 3 and 3 is prime number and 9 is not prime number

[@jagadeeshm.v.1887]

Inside the for loop he is checking 9 % 3 === 0 which will be true and it will return false. So 9 is not a prime number

[@elias-soykat]

if we use Math.sqrt(n) it will give us the wrong output some of the time. for 25 we know it's not a prime number, but when we use Math.sqrt(25) it will say it's a prime number. If I have said anything wrong, please correct me.

[@angelmanuelmayhuagutierrez8103]

Hi Elias. Notice that in the for loop we will pass also the square root because of the

[@s.pavankumar9671]

// Better optimum solution 0(sqrt(n)) function isPrime(number) { if (number

[@sandeepshorton]

function isPrime(number) { let count = 0 for (let i = 0; i

[@deidomandoura]

function primeNumber(n) { if (n == 2 || n == 3 || n == 5 || n == 7 || n == 11) return true; if (n % 2 == 0 || n % 3 == 0 || n % 5 == 0 || n % 7 == 0 || n % 11 == 0) return false; return true; } this is my code and it works for me, but i'm not sure, can any one validate it is it true or not ?

[@raunakgiri21]

221 is a composite number. But your code will return it as a prime number.

[@aravinthrs998]

function isPrimes(n){ if(n%2 === 0 || n%3 === 0 || n%5 === 0){ return false }else if(n

[@aizealawin4788]

While I understand your logic, there are still numbers that are not divisible by 2, 3, or 5 and have other factors. The first that comes to my mind is 121. The logic in the video above encompasses all possible numbers because of the inherent property of how there will always be a factor smaller than the square root of a number.

[@rsrinivas6025]

169 what will reault will your code works

[@christocdas1146]

💯💯

[@CoderSineNomine]

how about this; const prime = (n) => { const result = n < 2 ? false : n >= 2 && n % 2 === 0 ? false : true; return result; };

[@mounirhabbas7969]

Math.sqrt(n) return true with 9 but 9 not a prime number

[@pawanshimpi100]

Root 9 is 3 You have to check the modulus of nine till 3 and not less than 3 Your condition should be something like for(i = 2; i

[@abhigyankhaund6830]

Please rate my algorithm:- This is my logic for prime number with time complexity o(1) where any input you provide the result will be same let prime=(n)=>{ if(n>3) { if((n%2==0)||(n%3==0)||(n%Math.sqrt(n)==0)){ return `${n} is Composite`; } else{ return `${n} is prime`; } } else if(n==1){ return `${n} is neither prime nor composite`; } else{ return `${n} is prime`; } } let n=289; console.log(prime(n)); //logic is any number is considered composite if the number is greater than 3 and is either divisible by 2 or by 3 or by its square root

[@prachi_arora]

function isPrime(n){ if(n

[@anaselmakhloufi2180]

return n % 2 == 0 ? false : n < 2 ? false : true

[@me-wtf]

How to chance formik initialState values based on stateChanges inside useEffect? (multiple values at onces)

[@bonshippedit]

function primeNumber(n) { if (n < 2) { return false; } for (let i = 2; i < n; i++) { if (n % 2 === 0) { return false; } } return true; } console.log(primeNumber(9)); this return true mean while 9 is not a prime number

[@bonshippedit]

function primeNumber(n) { if (n < 1) { return `${n} is not a prime number`; } if (n === 2) { return `${n} is a prime number`; } if (n % 2 === 0) { return `${n} is not a prime number`; } const sqrt = Math.sqrt(n); for (let i = 3; i

[@metaltank2540]

Sir, I have tried a different solution and it works with Big-O - O(n). function primeNumber(n){ if(n > 1){ let result = [null] for (let i = 1; i el) if(filteredResult.length === 2){ console.log(`${n} is a Prime number`) } else { console.log(`${n} is not a Prime Number`) } } else { console.log("The given number must be greater than 1") } } primeNumber(5)

[@pawanshimpi100]

The if condition is not right as for bigger numbers such as n = 100 , you will have filteredResult.length more than 2

[@sathishkumar-sn1gu]

function primeNumber(n){ if(n < 2) return false; if(n % 2 === 0) return false; else return true; } console.log(primeNumber(1)); cant you use like this to optimize the code then here the time complexity will be O(1)

[@vaibhavsharma4743]

function isPrime(n){ if(n

[@nassergamal6041]

i think this solution more easy can any one give me feedback is it true ? function isPrime(num) { if (num < 2 || num % 2 == 0) { return false; } return true; } why you used for loop ?

[@AnilRoxsta]

Consider 9 as input, you'll see the need of `for` loop now

[@manishaggarwal5643]

haha you are best.... sorry for that

[@ankitbinjola]

But 9 is not a prime number

[@NaveenKumar-ne4yf]

Yup same

[@أميرالمؤمنينبنمحمدأكرم]

// this code will make the O constant function isPrime(n) { if(n < 2) return false; if(n % 2 == 0 && n != 2) return false; if(n > 7){ if(n % 3 == 0) return false; if(n % 5 == 0) return false; if(n % 7 == 0) return false; } return true }

[@desertedislander]

function isPrime(n){ if(n

[@jeff1571]

I thought all prime numbers are odd except 2, so then can't we just check if the given n is odd or 2 and return that they are prime numbers instead of putting them in a loop?

[@Ashokm1690]

All odd numbers are not prime numbers

[@jeff1571]

@@Ashokm1690 Yeah I realized it later on hahaha. Thanks for the reply tho appreciate it

[@hulihulayy7917]

if (n < 2){ return false; } else if( n === 2){ return true; } else { if(n % 2 === 0){ return false; }else{ return true; } } thoughts below?

[@zzzzüp]

doesn't this solution work? function isPrime(n) { if (n%2 === 0 || n%3 === 0) return false; return true; }

[@zzzzüp]

okay, it doesn't if n=55 for example

[@codelikealpha]

Just imagine if it did, you could have been called a geniuse

[@falcon04v8]

here, this will work for every numbers function prime(n){ if(n < 2){ return false } else if(n === 2 || n === 3 || n === 5){ return true } else if(n % 2 === 0 || n % 3 === 0 || n % 5 === 0){ return false } else{ return true } }

[@ezzinesmichi]

with your programm the number 25 will be prime!!!!!!!

[@denizdeniz4948]

This is my solution function isPrime(n) { if (n == 2) { return true; } else if (n > 2 && n%2 !== 0) { return true; } else { return false } };

[@alhaquekhan938]

What if we pass isPrime(2) i think there is little bug in this code

[@mohamedgamal-cg7dz]

no the code is right because for loop body doesn't execute in case of 2

[@Farduswahid31]

2 is prime.. your code mark 2 as non prime

[@unqcreation3098]

function prime (n) { if(n>0){ return n%2 === 0 || n === 1 ? false : true } } console.log(prime(1)) console.log(prime(5)) console.log(prime(4)) More reliable solution for this Problem, Done by me // Big-O = O(1)

[@nicolau7k]

the code don't work with 2, but seems nice

[@MarzoPlays]

Consider 9 as input, you'll see the need of `for` loop now

[@sreerags5818]

There is no need of loop

[@NoumanKhan-ck7vx]

function prime(n) { if (n % 2 === 0) { console.log(" no prime"); } else { console.log(" prime"); } } prime(7);

[@ezzinesmichi]

with your program the number 9 will be prime!!!!!!!!!!

[@aravinthrs998]

inside if condition add if(n%2 === 0 || n%3 === 0) like this, here also one problem is there console.log(55) means its show true so if(n%2 === 0 || n%3 === 0 || n%5 === 0)

[@bzchii7474]

@@aravinthrs998 urs will return true if n=121 XD which is not a prime number

[@mustafahassan9383]

bool isPrime(int n) { bool isPrime = false; if((n > 1) && (n % 2 != 0)){ isPrime = true; } return isPrime; }