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Palindrome Number – Solution & Complexity

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Palindrome Number

1. Handle Impossible Cases First

  • Negative numbers are not palindromes because the minus sign appears only on the left.
  • Positive numbers ending in 0 cannot be palindromes unless the number itself is 0.
  • These checks avoid extra work and prevent misleading reversals.
if x < 0 or (x % 10 == 0 and x != 0):
    return False

2. Reverse Only Half of the Number

  • Reversing the whole number works, but reversing half avoids unnecessary work and possible overflow in some languages.
  • Repeatedly move the last digit of x into reversed_half.
  • Stop when reversed_half is greater than or equal to the remaining x.
reversed_half = 0
while x > reversed_half:
    reversed_half = reversed_half * 10 + x % 10
    x //= 10

3. Compare Even and Odd Lengths

  • For an even number of digits, the remaining left half should equal the reversed right half.
  • For an odd number of digits, the middle digit is irrelevant.
  • Dropping the middle digit with reversed_half // 10 handles odd lengths.
return x == reversed_half or x == reversed_half // 10

4. Final Complete Solution

  • The algorithm uses arithmetic only, so it satisfies the no-string follow-up.
  • It processes about half the digits of the input.
  • Time complexity is O(log10 n), and extra space complexity is O(1).
def is_palindrome_number(x: int) -> bool:
    if x < 0 or (x % 10 == 0 and x != 0):
        return False

    reversed_half = 0
    while x > reversed_half:
        reversed_half = reversed_half * 10 + x % 10
        x //= 10

    return x == reversed_half or x == reversed_half // 10

FAQ

See also:

Practice Palindrome Number online (run code in the browser)Climbing Stairs solution (easy)Factorial solution (easy)Fibonacci Number solution (easy)FizzBuzz solution (easy)All coding interview problems
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