LeetCode #7: Reverse Integer

The Problem

Given a signed 32-bit integer x, return x with its digits reversed.

If the reversed integer overflows (outside [-2^31, 2^31 - 1]), return 0.

Examples

Step-by-Step: How to Reverse

We can reverse the digits by repeatedly extracting the last digit:

1. Get the last digit: x % 10
2. Add it to the result: reversed = reversed * 10 + digit
3. Remove the last digit: x = Math.trunc(x / 10)
4. Repeat until x === 0
5. Preserve the sign at the end

✅ Solution (JavaScript)

var reverse = function(x) {
  const INT_MAX = 2 ** 31 - 1; // 2147483647
  const INT_MIN = -(2 ** 31);  // -2147483648

  let reversed = 0;
  const isNegative = x < 0;
  x = Math.abs(x);

  while (x !== 0) {
    const digit = x % 10;
    x = Math.trunc(x / 10);

    // Check for overflow before multiplying
    if (reversed > Math.floor(INT_MAX / 10)) {
      return 0;
    }

    reversed = reversed * 10 + digit;
  }

  // Restore sign
  if (isNegative) reversed = -reversed;

  // Final overflow check
  if (reversed < INT_MIN || reversed > INT_MAX) {
    return 0;
  }

  return reversed;
};

⚠️ Why Check Overflow Before Multiplying?

We can't check reversed * 10 + digit > INT_MAX because that operation itself might overflow!

Instead, we check:

if (reversed > Math.floor(INT_MAX / 10)) {
  return 0;
}

This is safe because Math.floor(INT_MAX / 10) = 214748364

Try It: Digit Reversal Animation

Watch how each digit is extracted and built into the reversed number.