Skip to main content

PRIME DISTANCE : MINDSTORMER - APR20

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 2n-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 
  1. Prime no just less than it
  2. Bool (True or False) if current no is a prime
  3. Just next big Power of 2
Do Recursion and do each iteration in O(1) only.
  • Algorithm :
  1. Intialize a list or "array of tuple" of [1000000][3] with default value [0,False,0]
  2. Run Seive Algorithm and correct the bool specified for list, change False to True for prime positions
  3. Make an array or list of size 20 since (2^20 > 1000000) and store the power of 2 in it.Also change the 3rd value of index(2^i) in list initialized in step 1 with index of the array(i)
  4. Now start with list initialized in step 1 , and do following change, and update the  following thing: Previous prime no, next index of power 2 (i.e. 1st and 3rd value)
  5. Do Recursion as you want either with DP or Loop
Setter Code :
    Uncomment all the print command , What's happening inside the code will be printed.

Labels:

Comments

Post a Comment

Popular posts from this blog

Makarov's Ambition - Mindstormer April 20

 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
Post a Comment
Read more

Brilliant Mind - Mindstormer April 20 (Product Of All Perfect Square Term Upto A Given No 'n')

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  :
Post a Comment
Read more

Happy Group - Mindstormer April 20

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
Post a Comment
Read more