swarathesh

# Note – this is not a bash script (some of the steps require reboot)
# I named it .sh just so Github does correct syntax highlighting.
#
# This is also available as an AMI in us-east-1 (virginia): ami-cf5028a5
#
# The CUDA part is mostly based on this excellent blog post:
# http://tleyden.github.io/blog/2014/10/25/cuda-6-dot-5-on-aws-gpu-instance-running-ubuntu-14-dot-04/

# Install various packages
sudo apt-get update

swarathesh

from LinkedListpract.LinkedList import LinkedList

linkedlist = LinkedList()

linkedlist.insertstart(1)
linkedlist.insertstart(2)
linkedlist.insertend(12)

linkedlist.traverse()
print("\n")

swarathesh

m = int(input("enter m"))
n = int(input("enter n"))
abso = m-n if m-n>0 else (m-n)*-1
print abso

swarathesh

#Print the count of all the vowels that are available in the string
inp = raw_input("enter the string").lower()
countA = inp.count('a')
countE = inp.count('e')
countI = inp.count('i')
countO = inp.count('o')
countU = inp.count('u')

#print the count

swarathesh

#How to check if a given number is a power of 2 ?
inp = int(raw_input("enter number"))

while inp%2 == 0 :
    inp //=2

answer = 'power of 2 ' if inp == 1 else 'not power of 2 '

#pritn the answer
print answer

swarathesh

#How to check if a given number is a power of 2 ?
n = int(raw_input("enter number"))
count =0
while n != 0 :
    n = n & (n - 1)
    count+=1


print(count)

swarathesh

def binary_search_rec(a, key, low, high):
  if low > high:
    return -1

  mid = low + ((high - low) / 2)
  if a[mid] == key:
    return mid
  elif key < a[mid]:
    return binary_search_rec(a, key, low, mid - 1)
  else:

swarathesh

def find_sum_of_two_2(A, val):
  i = 0
  j = len(A) - 1
  while i < j:
    s = A[i] + A[j]
    if s == val:
      return True

    if s < val:

swarathesh

def sm(arr,val):
    low = 0
    hig = len(arr)-1

    while low < hig:
        sum = arr[low]+arr[hig]
        if sum == val :
            return True
        if sum < val  :
            low+=1

swarathesh

def maxset(A):
  minsum =0
  maxsum = 0
  minset=[]
  maxseT=[]
  for i in A :
      if i >0:
        minsum+=i
        minset.append(i)