Today, many KZfaqrs are only focused on sharing more materials, without considering how companies are asking questions or how freshers can really face interviews. Great work by Striver and GFG for providing practical advice on these topics.

We need more such type of mock interviews

Here i can see Arjit is struggling to get approach in the very start that clearly shows a real feel to interview ...a nice video to learn a lot from thank you @striver for such gem content ❤️❤️.....I want to know how you choose interviewee?

Sieve of eratosthenes + dp on subsequences

I think what he was talking at the end about functions is "Pure Functions". Changing a global variable from within a function is a "Side effect" and it's a bad practice to do that unless absolutely necessary. Pure functions should not have any side effects and given a set of input you can be rest assured that the output will always be the same no matter how many times the function is called.

minimum coin required + seive

thanks! lot to learn here even if you get the question by yourself.

Definitely top notch content

Mixture of Sieve and Knapsack.

The very first intuition i got after seeing the question was that , first i will prepare M primes in some array/vector. Then i can relate this qn. to unbounded knapsack problem (coin change problem where I've to find minimum number of denominations to make Target price). Or target sum problem to find minimum number of elements to make target . Striver Dp series helped me alot to think out of the box also ❤️❤️. thanks @striver

Show us an interview as well where the candidate barely knows anything...!

Brute force of this solution that strike in my mind is making a vector to store all the m primes and after that use unbounded knapsack to find minimun number of prime to make n

Clearly understood 💯...How companies approach a fresher

First thought comes to mind after seeing this question is Coin Change. 1. Find first 3 primes(precompute primes if given lot of queries) 2. BFS to get minimum primes

So inspiring men striver you are really great!!!

Loved it Sir ❤❤❤

binary search lower bound 1st que

For brute force approach we can tell that storing ist M prime no in array then simple do combination sum where duplicacy allowed that is we can use same element again T.c : exponential S.c O(M)

very useful !

What about we use bfs algorithm? Like simply store first m primes in a queue and then just simply keep iterating until you get the value

i think quetion can be easily solved using lower bound concept. in nlogn complexity.

similar to combination sum

sieve of eratosthenes + dp on subsequences(coin change problem modified)

If they're asking this level of questions for an OA. Then I won't even make it to the interviews lol.

Can anyone give a test case where gready fail?

Million dollar DSA problem - Given k, X and a stream of numbers , determine the average of the last k numbers , excluding the largest X numbers

❤️❤️

Coin Change variation where coins are M, and target sum is N, and how come 5+3+2 is 11 ?

25:00

@striver@GeeksforGeeks if N

What if we find first M primes using a Sieve, and then use a recursion on that list with a constraint of M[prime] < N ? that would just take around O(n) time asymptotically? (I'm not sure Im solving it as I see the question 1st time so I'll update later). Lets take an example: Say I take N=10 and M=1. So my M list is: M[]={2} //since 1st prime is M Now of course here 10%2 is 0 so it works (we can do 10/2 to see how many 2's we need) For the next problem: N=11 M=3 So my M list is: M[]={2,3,5} //3 primes Here, we can see M[3] < N //and M is strictly increasing list so we don't need to check more We see 11%5 !=0 We store 5 in a list say K. So we subtract 5 from 11 i.e. 6 and call this same function on 6 with M[]={2,3}, and concatenate the output on K[]. so what I mean is we get 5 first, then we send 6, so we have M[]={2,3} so we check 6%3 and its 0 so we do 6/3 to see how many 3's we need. So finally we get {5,3,3} is that correct? lemme check P.S. the answer is 5,3,2 in the doc and it seems my disappointment is immeasurable, and my day is ruined.

Is it possible .. to get the the list or all M primes and thn apply Target sum and count minimum counts

This qn is similar to coin change

i don't get the first question, what does it mean? help?

can we do sieve and binary search on search to get lower bound till n=0 if anyway in mid way lower bound out of sieve return -1

7:33 LOL, sieve of.....what !!? 😂

Store all possible M primes in an array in the sorted order then start subtracting given N from last element of the array e.g. if N=19 is given and M is 3 then arr[]{2,3,5} 19%(last element in arr)= 4 now the main twist comes if I go with second last element then it will give me 1 as remainder which isn't a prime number we cannot add this number in our solution array. so if the remainder is 1 we have to switch to 3rd last element which is 2 : 19/5=3 + (19%5/2)=2= 3+2=5. I've not checked all testcases correct me if I'm wrong.

yaar itna time kaha dete hain wo log bas 5-10 min and btw ye question 10 min ka hai baki aap jaisa interviewer milega to badhiya kismat

why he just started with tabulation i.e most optimal approach rather giving bruteforce approach first, at 22:02

At 7:34 which thing he mentioned he can use to find first n primes?

Unbound Knapsack ... Dekhte hi bola 😂😂😂

I also solver the question but take an hour 🙁

7:35 whaaaat

Can anyone have links to the questions

Why can't we use greedy approach here? Like for N=11 and M=3, I could have started from the biggest prime no. that's gonna be 5 and subtract that value from N and update value of N until I don't get value of N less than the current prime no or value of N==current prime no+1. In that case, I will move towards the smaller prime no and try to do the same thing with that. If N!=0 then we can't make N with the help of first M prime no.

2+3+5 is 10 right. Why he told its 11?

Striver se sikh ke striver ko interview dene ka maza hi alag h😂

his voice was shaking

test

Step1: Make an array with first M prime numbers Step2: Use unbounded knapsack code: import java.util.*; class Solution { public static int minPrimes(int N, int M) { int[] arr = findPrimes(M); int ans = unbounded_Knapsack(arr, N); return ans; } public static int unbounded_Knapsack(int[] nums, int N) { int len = nums.length; int[][] t = new int[len + 1][N + 1]; for (int i = 0; i

All fake and bogus

How can I get a chance to give a live mock with striver?? @striver @take U forward