SSDC - HIVE OF DEVELOPERS

 Makarov's Ambition - Mindstormer April 20  Question Brief   :            Two numbers s (number of states) and d (number of districts) followed by a series of 'n' numbers (0 and 1) will be given to you such that s*d=n. All you have to do is check that is there any way to divide the series into n parts such that in majority of states, the guild is victorious. The guild wins if more than half of the districts are won by the guild. Also the division will be consecutive. For example: if s=3, d=2 and series is a1, a2, a3, a4, a5, a6. so division can be made in only these two ways : (a1, a2), (a3, a4), (a5, a6) (a2, a3), (a4, a5), (a6,a1) Prequisite   :   Basic Programming Approach   :         We have to take two arrays, one to store the series and other stores the sum of series till that index. There can be only that number of ways to divide the states as many districts a state can have. All you have to do is check for the majority, by checking the

Happy Group Question Breif A group of blocks is said to be connected if we can reach from any given block to any other block in the same group, and this group is known as happy group . Given N blocks (numbered from 1 to N) and two lists of size M (u and v) denoting block u[i] is connected to block v[i] and vice versa . Can you count the number of happy groups.  Note: A block is always connected to himself Link to question: https://www.hackerrank.com/contests/mindstormer-apr20/challenges/happy-group Approach: Graph-  BFS we keep arrays u and v as given in the question. We create a  2D array-edges(ed) which keep record of whichblocks are connected with each other. We maintain  visited array(vis) that keep record of which block is visited and which is left. Further explanation is in the comments with the code. The main thing used is BFS and graph to solve this problem

PRIME DISTANCE - MINDSTORMER QUESTION BRIEF : In the question A number X(secret no) was given to you, and you had to recursively do following things. Represent X in the form of  2 n -z  Find Nearest Prime no(y) to z Assign x=y Repeat the process, untill x becomes less than or equal to 2, because there is no prime no less than 2. Prerequisite  : Preprocessing, Recursion,Brute Force Approach : According to constraint, x may be upto 500000 and the no of queries asked for x might be upto 100000. So for perfect solution , all the three points must be done not more than O(1) for a step and for any starting value of x , it will converge to 2 very quickly. Preprocessing : For number ranging from 1 to 1000000, we will store these three calculations already  Prime no just less than it Bool (True or False) if current no is a prime Just next big Power of 2 Do Recursion and do each iteration in O(1) only. Algorithm : Intialize a list or "

Satisfy the condition: x*p^x ≡ q (mod r) Problem Brief   : Tanu is a pre final year cs student. She has been given an integer I. And has been asked to compute how many integers x ( 1 <= x <= I) satisfy the below condition:- x*p x   ≡  q (mod r) Prerequisite   :    Basic Mathematics Approach   : Brute force solution won't work as 'I' is in range of 10^10 .  So we need to find out some features of that given equation. We have p r-1 ≡1(mod r) where r is a prime number by  Fermat's Little Theorem , it is obvious that p z  mod r falls into a loop and the looping section is r−1. Also, z mod r has a looping section r .We can try to list a chart to show what  x*p x  is with some specific i,j. (In the chart shown below, x is equal to i*(r-1) + j. Proof for the chart shown above: For a certain i,j  we can see that  x *p^x mod r= ((i*(r−1)+j)mod r)* x ^ (i*(r−1)+j) mod (r−1) =( j−i)*x^j mod r.  W e can also prove that  x*p^x mod r has a looping secti

Wonder String Question Breif Given a string containing just the characters 'a', 'b', 'c', 'd', 'e' and 'f', determine if the input string is wonder or not. Pairs are known as (opening element,closing element) : (a,b), (c,d), (e,f) An input string is wonder if: First(opening element) element of pair must be closed by the second(closing element) element of the pair Opening element must be closed by respective closing element in the correct order Note that an empty string is also considered wonder string: Link To Question: https://www.hackerrank.com/contests/mindstormer-apr20/challenges/wonder-string-1 Approach: Hint:  Use Stack If an opening element occurs we push it into stack and when the closing element occurs, we check if the element at the top of stack is corresponding opening element . If its some other element then we return false. Else we pop out  the top most element. If at the end of s

Product Of All Perfect Square Term Upto n Question Brief   :           A Number 'n' is given to you , you are said to find the product of all the perfect square term less than or equal to n. Prequisite   :   Basic Mathematics, Preprocessing Approach   :      Perfect Square numbers upto any number n are only these:          (1 2 ,2 2 ,........,(√n) 2 )      In Question it was asked to find the product of all perfect square numbers.i.e. ans =  (1 2  x 2 2   x....... x (√n) 2 ) it can be easily transformed to: ans =  (1   x 2   x....... x √n ) 2 ans= (√n!) 2 So, in each query , for given value of any n, we had to print the square of factorial of square root of n. This can be easily answered in O(1) if we already preprocess the number 1 to sqrt(N) and store their factorial already. SETTER CODE  :