A solution to the 4th Euler Challenge

Euler4

"""A palindromic number reads the same both ways. Find the
largest palindrome made from the product of two 3-digit numbers."""

#Define a 'surround' function that will be used to construct palindromes
def surround(x,y):
    x=str(x)
    y=str(y)
    return str((y+x+y))

#start with an initial array of all palidromes of length 1 or 2
initialArray=['0','1','2','3','4','5','6','7','8','9','00','11','22','33','44','55','66','77','88','99']

#Now build an array of palindromes less than 1000000
k=1
while k<3:    
    array=[]
    j=0
    while j<10:
        for i in initialArray:
            array.append(surround(i,str(j)))
        j+=1
        
    for x in array:
        initialArray.append(x)
    k+=1     
#Get rid of strings with leading zeroes
def removeLeadingZeroes(x):
    ans=[]
    for y in x:
        if y[0]!='0' or y=='0':
            ans.append(y)
    return ans

array=array+initialArray
array=removeLeadingZeroes(array)

#convert to an array of integers, remove any duplicates, and sort into descending order
def convertToInts(x):
    ans=[]
    for y in x:
        z=int(y)
        ans.append(z)
    return ans
array=list(set(array))
array=convertToInts(array)
array=sorted(array)
array.reverse()

def factor(x):
    z=[]
    i=1
    while (i<=x):
        if x%i==0:
            z.append(i)
        i+=1    
    return z

#Define a test of the 3 digit product condition
def threeDigitTest(x):
    ans=False
    y=factor(x)
    for z in y:
        if (len(str(int(x/z)))==3 and len(str(z))==3):
            ans=True
    return ans

#now search through the array for first element that passes test.
i=0
found=False
while found==False:
    if threeDigitTest(array[i]):
        ans=array[i]
        found=True
    else:
        i+=1
print(ans)
            
#The answer is 906609