ALGO NOTEBOOK

Euler's Totient Function Euler's Totient function returns the number of positive integers less than or equal to n which are co-prime with n. Two numbers are co-prime when they have no common divisors or their gcd is 1. For instance,  phi(8)=4  , since there's four integers 1,3,5,7 that are co-prime with 8. Prime numbers are co-prime with any number below it. if P is a prime number then  phi(P)=P-1  . For instance  phi(7)=6  , since 1,2,3,4,5,6 all are co-prime with 7. Phi function is also multiplicative. That means if n=a*b , then, Phi(n) = Phi(a * b) = Phi(a) * Phi(b) if a and b are not relatively co-prime then Phi(n) = Phi(a * b) = Phi(a) * Phi(b) * d/phi(d) , where d=gcd(a,b) P1 and P2 are prime numbers and n can be represented in the form of n=P1*P2, then Phi(n) = Phi(P1 * P2) = Phi(P1) * Phi(P2) = (P1-1) * (P2-1)  for instance , 21 = 3 * 7 Phi(21)=Phi(3) * Phi(7) =(3-1) * (7-1) ...

The  problem  statement actually refers to finding out number of pairs of integers that have LCM n ( which is the given value). There can be 2000 test cases and value of n can be at most 10 14 . let's assume that the prime factorization of n can be represented as n = p1 q1 .p2 q2 .......px qx , where p1,p2....px are prime numbers and q1,q2...qx are corresponding powers. lets assume that any two integers are i and j such that LCM(i,j)=n. since n is their LCM, the prime factorization of i and j must have the following qualities. 1. if i = px qx then j must be either j = px qx-1 or j = px qx-2 or ..... j = px 0 2. if j = px qx then i must be either i = px qx-1 or i = px qx-2 or ..... i = px 0 3. i = px qx and j = px qx for the 1st scenario we can see there are qx combinations for the 2nd scenario we can see there are qx combinations for the 3rd scenario we can see there are 1 combinations so, there can be a total of 2*qx + 1 combinations for each of the pr...

Cycle in a 2D graph with DFS Problem Link :  Fox And Two Dots Solution Idea : This problem can be solved using DFS. You can Check my  Solution  . Coloring a graph  Problem Link :  Coloring a Tree Solution Idea : Using DFS we can color a node and its subtree. We are coloring only subtrees. so we can't visit upwards of a tree while coloring. That's why we are using directed graph. if a tree is already colored in it's required color, we don't need to color it anymore. You can Check my  Solution . Also solved  Customized Chess Board  ( brute force approach )

Problem link :  MARBLES In order to solve this problem we need to know basic permutation and combination. We can address this problem with the tag "Selection With Repetition" . If we have n type of object or n objects and we have to choose k objects from these where each object can be choosen multiple times, then number of ways we can do it is  (n+k-1) C (k). Let's come to our problem. we have K type of objects and we have to choose N objects from them. So, now we can convert the given formula into (K+N-1) C (N). Suppose , we have 30 objects to choose and we have 7 types of objects. This formula will give us (36)C(30) = 1947792. Which is much much greater that the required answer. This happens because we have calculated all the possibilities without filtering those possibilities where we haven't taken at least one from each types (which is a key condition) . Let's assume another example - we have two letters A , B and we have to choose 3 letters each ti...