HACKERRANK: Power Of Large Numbers

Link:- https://www.hackerrank.com/challenges/power-of-large-numbers

Summary:
Given A and B.
Find (A^B)%(10^9+7)

Constraints:
1 <= T <= 10 
1 <= A,B <= 10100000 
A % (109 + 7) != 0

For Example:
Sample Input:
4
3 2
4 5
7 4
34534985349875439875439875349875  93475349759384754395743975349573495

Sample Output:
9
1024
2401
735851262

Logic:
(A^B)%MOD = (A%MOD)^B
Since A is very very large number can't be stored in int or long so using modulo technique bring A in the range of int or long as:
Let 'a' be an array contains digit of A so to construct A%MOD. Let A = 0
for i 0 to len(a):
    A = (A*10 + a[i])%MOD
This gives A%MOD

Now B is also very very large but we can't apply same technique as we done in A as
(A%MOD)^B%MOD != (A%MOD)^(B%MOD)%MOD
So we use Fermet's Little Theorem which says:

where p is prime number.
divide by a both side


Since MOD = 10^9+7 is a prime number

we need to find A^B%MOD
so if  B is divisible by (p-1) or (MOD-1) then B = (MOD-1)*x
so (A^((MOD-1)*x))%MOD = (A^(MOD-1)%MOD)^x%MOD = 1^x%MOD = 1

and if B is not divisible by (MOD-1) then B = (MOD-1)*x + y
so
(A^((MOD-1)*x+y))%MOD = ((A^(MOD-1)%MOD)^x)%MOD*(a^y)%MOD =  1^x*(a^y)%MOD
and y = B%(MOD-1)
so y can be computed as:
y = 0
b is an array contains digit of number B.
for i 0 to len(b)
    y = (y*10 + b[i])%(MOD-1)
if y==0
answer = 1
else
answer = A^y%MOD

To View My Solution:
https://github.com/shivam04/codes/blob/master/hackerrank/math/number%20theory/medium/power-of-large-number.py

CODEFORCES: Sereja and Suffixes(368B)

Problem Link: http://codeforces.com/problemset/problem/368/B

Summary:
Given 'n' number of elements in array.

Given 'm' i.e 'm' different queries each query contains an integer 'l'.

Find the number of distinct elements from 'l' to 'n'  in array.

For Example:

input

10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10

output

6
6
6
6
6
5
4
3
2
1

Topic:
Dynamic Programming
Logic:

Let dp[i] be the answer i.e number of distinct elements from i to n.

and dp[i] = dp[i+1] +1 when element at a[i] has not been visited yet or visit[a[i]] = 0 otherwise dp[i] = dp[i+1] i.e visit[a[i]] = 1

For any 'l' print dp[l]  is the answer.

To View my solution:

https://github.com/shivam04/codes/blob/master/codeforces/seerja_and_suffixes.py

HACKER RANK: Journey to the Moon

Problem Link: https://www.hackerrank.com/challenges/journey-to-the-moon

Summary:

There are  trained astronauts numbered from  to . But those in charge of the mission did not receive information about the citizenship of each astronaut. The only information they have is that some particular pairs of astronauts belong to the same country.

Your task is to compute in how many ways they can pick a pair of astronauts belonging to different countries. Assume that you are provided enough pairs to let you identify the groups of astronauts even though you might not know their country directly. For instance, if  are astronauts from the same country; it is sufficient to mention that  and  are pairs of astronauts from the same country without providing information about a third pair .

Print An integer that denotes the number of permissible ways to choose a pair of astronauts.

For Example:

Sample Input:

4 2
0 1
2 3

Sample Output:

4

Topic:

Graph Theory, BFS(Breadth First Search)

Logic:

I have create an adjacency list.

Then Using BFS I have count total number of sub graphs in a graph such that there is no edges between sub graphs and also count total number of nodes in each sub graph.

Then compute number of ways 2 astronauts from different country can be sent to Moon.

For example:

Like From America There are 3 trained astronauts.

From India There are 4 trained astronauts.

From Japan There are 2 trained astronauts.

From China, Australia and Russia There are Only One astronaut.

So number_of_ways = 3*(4+2+1+1+1) + 4*(2+1+1+1) + 2*(1+1+1) + 2 + 1 = 56

To View My Solution:

https://github.com/shivam04/codes/blob/master/hackerrank/algo/graph/medium/journey-to-the-moon.py

HACKER RANK: Play Game

Problem Link: https://www.hackerrank.com/challenges/play-game

Summary:
Given N number of elements in stack.

first line will contain a number N i.e. number of elements in the stack and next line will contain N numbers i.e. numbers etched on bricks from top to bottom.

You and your friend decide to play a game using a stack consisting of N bricks. In this game, you can alternatively remove 1, 2 or 3 bricks from the top, and the numbers etched on the removed bricks are added to your score. You have to play so that you obtain the maximum possible score. It is given that your friend will also play optimally and you make the first move.

Print a single line containing your maximum score.

For Example:

Sample Input:

2

5

999 1 1 1 0

5

0 1 1 1 999

Sample Output:

1001

999

Topic:
Dynamic Programming Approach: https://en.wikipedia.org/wiki/Dynamic_programming

Logic:

I took dp[2][n]

dp[0]->My Score

dp[1]->Minimum Score other player leaves for me as he also play optimally

When n = 1,2,3 then you took all elements in stack

dp[0][n-1]=a[n-1]

dp[0][n-2]=a[n-1]+a[n-2]

dp[0][n-3]=a[n-1]+a[n-2]+a[n-3]

so for i n-4 to 0:

    dp[1][i] = min(dp[0][i+1],min(dp[0][i+2],dp[0][i+3]))

    dp[0][i] = max(a[i]+dp[1][i+1],max(a[i]+a[i+1]+dp[1][i+2],a[i]+a[i+1]+a[i+2]+dp[1][i+3]))

ans = dp[0][0]

print ans

In this code dp[1][i] is minimum left by other player for you when he plays optimally i.e chooses maximum sum of elements for itself.

and dp[0][i] is maximum of taking one element, two element or three element plus minimum left by other player when he chooses maximum sum of elements for itself.

Time Complexity: O(n)

To View My Solution:

https://github.com/shivam04/hackerrank/blob/master/play-game.cpp

HACKER RANK: Red John is Back

Problem Link: https://www.hackerrank.com/challenges/red-john-is-back

Summary:
Given N find number of ways to to arrange 4X1 and 1X4 bricks in 4XN block. Say there are M ways so print P number of prime up to M(i.e. <=M).

For Example:

Sample Input:
4
1
3

Sample Output:
0
3

For N=1
There is only one way so P = 0 as there is no prime number less than M=1

For N=7
there is M=5 ways so answer P=3 as 2,3,5 is less than than and equal to 5.

Topic:
Dynamic Programming Approach: https://en.wikipedia.org/wiki/Dynamic_programming

Logic:
What I have done is take number N as a block of weight N kg and try to adjust weight 1kg and 4kg weight in different ways.
So number of ways can be recursively computed as:
F(n) = F(n-1) + F(n-4)
if we place weight of 1kg then F(n-1) need to be computed.
if we place weight of 4kg then F(n-4) need to be computed.

Then taken sum of all computes M = F(n)

Now using Sieve Of Eratosthenes(http://www.geeksforgeeks.org/sieve-of-eratosthenes/) we compute P number of prime numbers up to M.

To View My Solution:
https://github.com/shivam04/hackerrank/blob/master/red-john-is-back.cpp

HACKER RANK: Common Child

Problem Link: https://www.hackerrank.com/challenges/common-child

Summary:
Given 2 strings of equal length what's the longest string that can be constructed such that it is a child of both.

For example:

Sample Input 1:
HARRY
SALLY

Sample Output 1:
2

Longest possible string is "AY"

Sample Input 2:
ABCDEF
FBDAMN

Sample Output 2:
2

Longest possible string is "BD"

Topic:
Dynamic Programming Approach: https://en.wikipedia.org/wiki/Dynamic_programming
Longest String Subsequence: https://en.wikipedia.org/wiki/Longest_common_substring_problem
                                                http://www.geeksforgeeks.org/longest-common-substring/

My Logic:

Let string str1 = HARRY and str2 = SALLY
So this is how i solved this question using DP(Dynamic Programming).

To view My Solution:
https://github.com/shivam04/hackerrank/blob/master/common-child.java

HACKER RANK: Sherlock and Anagrams

Problem Link: https://www.hackerrank.com/challenges/sherlock-and-anagrams

Given a string S, find the number of "unordered anagrammatic pairs" of substrings.

Input Format

First line contains T, the number of testcases. Each testcase consists of string S in one line.

Constraints 
1≤T≤10 
2≤length(S)≤100 
String S contains only the lowercase letters of the English alphabet.

Output Format

For each testcase, print the required answer in one line.

Logic:

Let's say S[i,j] denotes the substring Si,Si+1,⋯,Sj.

Match Subset Si  With Si+1, ....... Sj For anagram.

To View My Solution:

https://github.com/shivam04/hackerrank/blob/master/sherlock-and-anagrams.java