mailpraveens

def findDivergePointForOfTwoLinkedLists(list1, list2):
	# Assuming the list class has a method length to give the length of the list
	lengthList1 = list1.length()
	lengthList2 = list2.length()
	differnce = lengthList2 - lengthList1
	if(lengthList2>=lengthList1):
		while(differnce!=0):
			list2 = list2.next
			differnce -= 1

mailpraveens

#method to check for anagrams within a set
from operator import itemgetter

def checkForAnagrams(input):
	temp = dict()
	#Sort the input values individually
	for k,v in input.items():
		temp[k]=''.join(sorted(v))
	#Now sort the entire list based on the value of the key
	temp2 = sorted(temp.items(),key = itemgetter(1))

mailpraveens

#max heap construction
def heapify(items,idx):
	left = leftChild(idx)
	right = rightChild(idx)
	largest = -1
	if(left < len(items) and items[left] > items[idx]):
		largest = left
	else:
		largest = idx

mailpraveens

'''
		1
	2		3
  4   5   6    7


 should print 1,2,4,5,6,7,3,1


'''

mailpraveens

def findLCA(node,value1,value2):
	if(node == None):
		return None
	#move to right if values are greater
	if(node.value < value1 and node.value < value2):
		return findLCA(node.right,value1, value2)
	#move to left if values are smaller
	if(node.value > value1 and node.value > value2):
		return findLCA(node.left,value1, value2)
	# Whereever it diverges, return the node as its the LCA

mailpraveens

#http://www.quora.com/Whats-a-math-trick-that-is-not-very-well-known/answers/4234574
import sys
def verifySquaringTheorem(n):
	for i in range(25,n):
		nSquaredSupposed = (( i - 25 ) * 100) + pow(i-50,2)
		actualSquare = pow(i,2);
		if nSquaredSupposed == actualSquare:
			print ("Pass for n = " + str(i) + " with value " + str(nSquaredSupposed))
		else:
			print("Fail for n = " + str(i) + "with value " + str(nSquaredSupposed) + "and actualSquare value = " + str(actualSquare))