easy

Palidrome Number

We can solve the palindrome check problem using two pointers. We can use the two-pointer approach to compare digits from the left and right to check if they form a palindrome.

The idea is to extract the digits of the number from left to right and right to left, and then compare them. To extract the digits, we can use the modulo operator (%) to get the rightmost digit, and the integer division (Math.floor) to remove the rightmost digit.

Let's implement the solution using the two-pointer approach in JavaScript:

/**
 * @param {number} x
 * @return {boolean}
 */
var isPalindrome = function (x) {
  if (x < 0) {
    return false; // Negative numbers cannot be palindromes
  }
 
  let originalX = x;
  let reversedX = 0;
 
  while (x > 0) {
    const digit = x % 10;
    reversedX = reversedX * 10 + digit;
    x = Math.floor(x / 10);
  }
 
  return originalX === reversedX;
};

You can use the isPalindrome function in the same way as mentioned earlier to check if a given integer x is a palindrome. The function returns true if the integer is a palindrome; otherwise, it returns false.

Let's break down the code inside the while loop and understand how it works to reverse the digits of the given integer x.

while (x > 0) {
  const digit = x % 10;
  reversedX = reversedX * 10 + digit;
  x = Math.floor(x / 10);
}

1. x % 10: This operation extracts the rightmost digit of the number x. In JavaScript, the % operator gives the remainder when x is divided by 10. For example, 123 % 10 would give 3.

2. reversedX * 10 + digit: Here, reversedX is used to build the reversed number. For each iteration of the loop, we multiply reversedX by 10 to shift its digits to the left, and then add the extracted digit to the rightmost position. This effectively places each digit extracted in reverse order. For example, if reversedX is 0 and digit is 3, after this operation, reversedX becomes 3.

3. x = Math.floor(x / 10): This operation removes the rightmost digit from x by performing an integer division by 10. Integer division discards any decimal part, effectively removing the rightmost digit. For example, if x is 123, after this operation, x becomes 12.

The loop continues to run until x becomes 0. During each iteration, it extracts the rightmost digit from x, places it in the reversed position in reversedX, and then removes the rightmost digit from x. After the loop, reversedX will contain the reversed digits of the original x.

For example, let's say x is 123:

1st iteration:

• x: 123
• digit: 3
• reversedX: 3 (since reversedX starts as 0, it becomes 0 * 10 + 3 = 3)
• x: 12 (after performing Math.floor(123 / 10))

2nd iteration:

• x: 12
• digit: 2
• reversedX: 32 (since reversedX was 3, it becomes 3 * 10 + 2 = 32)
• x: 1 (after performing Math.floor(12 / 10))

3rd iteration:

• x: 1
• digit: 1
• reversedX: 321 (since reversedX was 32, it becomes 32 * 10 + 1 = 321)
• x: 0 (after performing Math.floor(1 / 10))

The loop ends since x is now 0. The final value of reversedX becomes 321, which is the reversed digits of the original x (123).

After the loop, we compare reversedX with the original value of x. If they are the same, it means x is a palindrome. Otherwise, it's not.

This implementation has a time complexity of O(log10(x)), where x is the given integer, as we are extracting digits from the number. The space complexity is O(1), as we are not using any additional data structures that grow with the input size.