The Next Palindrome number

Question

I am beginner in programming, So can you please tell me what's wrong with my code?

I want to print next palindrome number if the number entered by the user (n) is not palindrome

n = int(input("Enter any number :- "))

reverse = 0
temp = n

while (n!=0):
    reverse = reverse * 10
    reverse = reverse + n%10
    n=n//10 
if(temp==reverse):
    print ("Already palindrome:: ")

if(temp != reverse):
     new_temp = temp
     new_reverse = 0
     for i in range(new_temp,new_temp+10):
        while(temp != 0):
            new_reverse = new_reverse * 10
            new_reverse = new_reverse + temp%10
            temp = temp//10
         if(new_temp==new_reverse):
             print ("Next pallindrome is :- ",new_temp)
             break
         if(new_temp != new_reverse):
             temp = new_temp+1

Answer 1

if a definite range is given:-

funcion to check if the number is a palindrome

def palin(x):                 
    s=str(x)
    if s==s[::-1]:
        return True
    else:
        return False

n=int(input("Enter the number"))

putting up range from the next number till 15 digits

for i in range(n+1,int(10e14)):
    if palin(i) is True:
        print(i)
        break

Answer 2

I have written this for finding next pallindrome number given a pallindrome number..

#given a pallindrome number ..find next pallindrome number
input=999
inputstr=str(input)

inputstr=inputstr
#append 0 in beginning and end of string ..in case like 99 or 9999
inputstr='0'+inputstr+'0'
length=len(inputstr)
halflength=length/2;
#if even length
    if(length%2==0):
    #take left part and reverse it(which is equal as the right part )
    temp=inputstr[:length/2]
    temp=temp[::-1]
    #take right part of the string ,move towards lsb from msb..If msb is 9 turn it to zero and move ahead
    for j,i in enumerate(temp):
        #if number is not 9 then increment it and end loop
        if(i!="9"):

            substi=int(i)+1
            temp=temp[:j]+str(substi)+temp[j+1:]
            break;
        else:

            temp=temp[:j]+"0"+temp[j+1:]
    #now you have right hand side...mirror it and append left and right part
    output=temp[::-1]+temp
#if the length is odd
    if(length%2!=0 ):
    #take the left part with the mid number(if length is 5 take 3 digits
        temp=inputstr[:halflength+1]
    #reverse it
        temp=temp[::-1]
    #apply same algoritm as in above
    #if 9 then make it 0 and move on
    #else increment number and break the loop
        for j,i in enumerate(temp):

            if(i!="9"):

                substi=int(i)+1
                temp=temp[:j]+str(substi)+temp[j+1:]
                break;
            else:

                temp=temp[:j]+"0"+temp[j+1:]
    #now the msb is the middle element so skip it and copy the rest
        temp2=temp[1:]
    #this is the right part mirror it to get left part then left+middle+right isoutput
        temp2=temp2[::-1]
        output=temp2+temp
    print(output)

# similarly for this problem take the left part of given number ...reverse it..store it in temp

  inputstr=str(number)
    if(inputstr==inputstr[::-1])
        print("Pallindrome")
    else:
        temp=inputstr[:length/2]
        temp=temp[::-1]
        for j,i in enumerate(temp):

                if(i!="9"):

                    substi=int(i)+1
                    temp=temp[:j]+str(substi)+temp[j+1:]
                    break;
                else:

                    temp=temp[:j]+"0"+temp[j+1:]
    now depending on length of your number odd or even generate the output..as in the code
    if even then output=temp[::-1]+temp
    if odd then  temp2=temp1[1:]
                 output=temp2[::-1]+temp

    I am not sure about this solution..but hope it helps

Answer 3

**

def nearest_palindrome(number):
    n = len(str(number))//2
    if(len(str(number)) % 2 == 0):
        #number like 1221
        number_1 = int((str(number))[:n]) #12
        number_2 = int((str(number))[n:]) #21
        if(number_1 < number_2):
            number_1 += 1
            number_2 = int(str(number_1)[::-1])
        else:
            number_2 = int(str(number_1)[::-1])
        # if last half part is zero then just reverse the first number
        if number_2 == 0:
            number_2 = str(number_1)[::-1]
        #combining the both parts
        ans = int(str(number_1) + str(number_2))
        return ans
    else:
        #numer like 12510   n=2
        nu = int((str(number))[:n+1]) #add in this number
        number_1 = int((str(number))[:n])  # 12
        number_2 = int((str(number))[n+1:])  # 21
        if (number_1 < number_2):
            nu += 1
            number_2 = int((str(nu))[::-1][1:])
        else:
            number_2 = int((str(nu))[::-1][1:])
        #if last half part is zero then just reverse the first number
        if number_2 == 0:
            number_2 = str(nu)[::-1]
            number_2 = number_2[1:]
        #combinning both parts
        ans = int(str(nu) + str(number_2))
        return ans
number=12331
print(nearest_palindrome(number))

**

Answer 4

To check if a number is a palindrome, you don't need to convert it to a number. In fact, its a lot simpler if you just check the string equivalent of your number.

>>> i = '212'
>>> i == i[::-1]
True
>>> i = '210'
>>> i == i[::-1]
False

Use this to your advantage, and create a function:

def is_palindrome(foo):
   return str(foo) == str(foo)[::-1]

Next, to find the next palindrome, simply increment the number till your palindrome check is true.

Combine all that, and you have:

def is_palindrome(n):
    return str(n) == str(n)[::-1]

n = raw_input('Enter a number: ')
if is_palindrome(n):
   print('Congratulations! {0} is a palindrome.'.format(n))
else:
   n1 = n
   while not is_palindrome(n1):
       n1 = int(n1)+1
   print('You entered {0}, but the next palindrome is {1}'.format(n, n1))

Here is how it works:

$ python t.py
Enter a number: 123
You entered 123, but the next palindrome is 131
$ python t.py
Enter a number: 121
Congratulations! 121 is a palindrome.