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from random import randint

board = []

for x in range(5):
    board.append(["O"] * 5)

def print_board(board):
    for row in board:
        print " ".join(row)

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# This is written after refering the CLRS book and hints from the site http://en.literateprograms.org/Merge_sort_(Python)

#The merge method takes in the two subarrays and creates a output array
def merge(left, right):
  result = []
	i , j = 0 , 0
	while i < len (left) and j < len (right): # iterate through both arrays and arrange the elements in sorted order
		if left[i] <= right [j]:
			result.append(left[i])
			i+=1

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#First partition the input array into two groups
# Partitioning alogorithm creates smaller array with elements on left and 
# right being smaller and greater than the pivot

def partiton(a, left, right):
  i = left + 1 # Initialiase i to one more assuming pivot is at position left
	pivot = a[left]

	for j in range(left+1, right+1): #Iterate over the remaining array
		if (a[j] < pivot):

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    for (NSString *familyName in [UIFont familyNames]) {
        for (NSString *fontName in [UIFont fontNamesForFamilyName:familyName]) {
            NSLog(@"%@", fontName);
        }
    }

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#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))

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#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))

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#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

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'''
		1
	2		3
  4   5   6    7


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


'''

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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

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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