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Java Loops | HackerRank solutions -codewithyasar

  Objective In this challenge, we're going to use loops to help us do some simple math. Task Given an integer,  , print its first   multiples. Each multiple   (where  ) should be printed on a new line in the form:  N x i = result . Input Format A single integer,  . Constraints Output Format Print   lines of output; each line   (where  ) contains the   of   in the form: N x i = result . Sample Input 2 Sample Output 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10 2 x 6 = 12 2 x 7 = 14 2 x 8 = 16 2 x 9 = 18 2 x 10 = 20 Explanation:           Here, we just need to use for loops to achieve the result Solution : import java.io.* ; import java.math.* ; import java.security.* ; import java.text.* ; import java.util.* ; import java.util.concurrent.* ; import java.util.regex.* ; public class Solution { public static void main ( String [] args ) t...
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Mountain Peak Sequence

Mountain Peak Sequence

Consider the first three natural numbers 1, 2, 3. These can be arranged in the following ways: 2, 3, 1 and 1, 3, 2. Inboth of these
arrangements, the numbers increase to a certain point and then decrease. A sequence with this property is called a “mountain
peak sequence”.
Given an integer N, write a program to find the remainder of mountain peak arrangements that can be obtained by rearranging the
numbers 1, 2, …, N.

Input Format : One line containing the integer N

Input Constraints : Mod = 10^9 +7 3 ≤ N ≤ 10^9

Output Format : An integer m, giving the remainder of the number of mountain peak arrangements that could be obtained from 1, 2, …, N is divide by Mod

Sample Input :

3

Sample Output :

2

Explanation:

    If mountain like sequence are formed using the number then count it.


Understand the logic:

  • For input 3,6 combinations can be formed,out of these 2 is a mountain sequence.
  • For input 4,24 combinations can be formed,out of these,6  Mountain sequence is formed.
  • By analyze,we form pattern,Like 2**N-1 -2.
  • So that we need to find exponentiation in optimal way and followed by constraints.


Code logic:


  1. Function for exponentiation power.
  2. If N==0 return 1.
  3. else if N==1 return a.
  4. else:
  5. R=power(a,N/2)(Note:Recursive call)
  6. if even return R*R
  7. else return R*a*R(here the a is the base number)
  8. a=2
  9. N input
  10. call the function(a,N-1)-2
  11. Also Include the constraint modulo.
Program:   
def power(a,N,mod):
if N==0:
return 1
elif N==1:
return a
else:
R=power(a,int(N/2),mod)
if N%2==0:
return (R*R)%mod
else:
return (R*a*R)%mod
n=int(input())
mod=1000000007
print(power(2,n-1,mod)-2)
view raw MountainPeakSequence.py hosted with ❤ by GitHub
  
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