Reverse Integer

Reverse Integer - Problem

Math Medium

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2³¹, 2³¹ - 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Input & Output

Example 1 — Basic Positive Number

$ Input: x = 123

› Output: 321

💡 Note: Reverse the digits: 123 becomes 321. The result fits in 32-bit range so return 321.

Example 2 — Negative Number

$ Input: x = -123

› Output: -321

💡 Note: Reverse the digits while preserving the negative sign: -123 becomes -321.

Example 3 — Overflow Case

$ Input: x = 1534236469

› Output: 0

💡 Note: Reversing gives 9646324351, which exceeds 2³¹-1 = 2147483647, so return 0.

Constraints

-2³¹ ≤ x ≤ 2³¹ - 1

Visualization

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Asked in

🍎 Apple 15 a Amazon 12 B Bloomberg 8 M Microsoft 6

The key insight is to extract digits using modulo and check for overflow before multiplying. The mathematical approach is optimal: build the reversed number digit by digit while checking 32-bit bounds. Time: O(log n), Space: O(1)

Common Approaches

✓ Mathematical Approach

⏱️ Time: O(log n) Space: O(1)

Extract digits one by one using modulo 10, and build the reversed number by multiplying previous result by 10 and adding current digit. Check for overflow before each multiplication.

String Conversion

⏱️ Time: O(log n) Space: O(log n)

Convert the integer to string, reverse the string, then convert back to integer. Handle negative signs and check for overflow.

Mathematical Approach — Algorithm Steps

Extract last digit using x % 10
Build result = result * 10 + digit
Remove last digit using x = x / 10
Check for overflow before multiplication
Return result when x becomes 0

Visualization

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Step-by-Step Walkthrough

Extract Digit

Use x % 10 to get last digit

Build Result

result = result * 10 + digit

Remove Digit

x = x / 10, repeat until x = 0

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int solution(int x) {
    if (x == 0) return 0;
    
    int result = 0;
    int negative = x < 0;
    
    if (x == INT_MIN) return 0;
    
    x = abs(x);
    
    while (x != 0) {
        int digit = x % 10;
        
        // Check for overflow before multiplying
        if (result > INT_MAX / 10) {
            return 0;
        }
        if (result == INT_MAX / 10 && digit > 7) {
            return 0;
        }
        
        result = result * 10 + digit;
        x /= 10;
    }
    
    if (negative) {
        result = -result;
    }
    
    return result;
}

int main() {
    int x;
    scanf("%d", &x);
    printf("%d\n", solution(x));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Loop runs for each digit, which is O(log n) where n is the input value

⚡ Linearithmic

Space Complexity

O(1)

Only using a few integer variables, no extra data structures

✓ Linear Space

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Input & Output

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Mathematical Approach — Algorithm Steps

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Code -

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