Palindrome number

Leetcode 9 problem.

Palindrome number is a natural number that has reflectional symmetry across a vertical axis. It remains the same when its digits are reversed.

Go

Time complexity: $O(log_{10}(n))$ - each iteration divides a parameter by 10
Auxiliary space: $O(1)$ - constant amount of space

func isPalindrome(x int) bool {
    if x < 0 || (x % 10 == 0 && x != 0) {
            return false;
    }
    
    var reversedNumber int = 0
    for x > reversedNumber {
            remainder := x % 10
            reversedNumber = reversedNumber * 10 + remainder
            x = x / 10
    }
    
    // odd and even cases
    return x == reversedNumber || x == reversedNumber / 10;
}

Typescript

Time complexity: $O(n)$ - n is a length of a string representation of a number parameter
Auxiliary space: $O(1)$ - constant amount of space

function isPalindrome(x: number): boolean {
  const reversedNumber: Number = Number(
    x.toString().split("").reverse().join(""),
  );

  return x === reversedNumber;
}

C++

Number solution:

Time complexity: $O(log_{10}(n))$ - each iteration divides a parameter by 10
Auxiliary space: $O(1)$ - constant amount of space

bool isPalindrome(int x) {

    // can not be 0 or less than 10
    if(x < 0 || (x % 10 == 0 && x != 0)) {
        return false;
    }
    
    long int initialNumber = x, reversedNumber = 0, remainder = 0;

    // can not be less than 10 (result of x / 10)
    while(x != 0)
    {
        remainder = x % 10;
        
        // reverses order of digits
        // takes reversed accumulated last digits * 10 + current last digit
        reversedNumber = reversedNumber * 10 + remainder;
        
        // decreases the number digits by division by 10
        x = x / 10;
    }
    
    return initialNumber == reversedNumber;
};

String solution:

Time complexity: $O(n/2)$ - n is a length of a string representation of a number parameter
Auxiliary space: $O(1)$ - constant amount of space

#include <string>

bool isPalindrome(int x) {
    string s = to_string(x);

    bool palindrome = true;
    int length = s.length();
    int stringMiddleChar = length/2;

    // iterate to the middle of the string
    for(int i = 0; i < stringMiddleChar; ++i) {
    
        // compare the chars on the left/right ends
        if(s[i] != s[length-1-i]) {
            palindrome = false;
            break;
        }
    }
    
    return palindrome;
}

thoughtbook

Palindrome number

Go

Typescript

C++

Number solution:

String solution: