A Number After a Double Reversal - Practice Coding Problems

A Number After a Double Reversal - Problem

Math Easy

Reversing an integer means to reverse all its digits.

For example, reversing 2021 gives 1202. Reversing 12300 gives 321 as the leading zeros are not retained.

Given an integer num, reverse num to get reversed1, then reverse reversed1 to get reversed2. Return true if reversed2 equals num. Otherwise return false.

Input & Output

Example 1 — Number Without Trailing Zeros

$ Input: num = 526

› Output: true

💡 Note: First reversal: 526 → 625. Second reversal: 625 → 526. Since 526 == 526, return true.

Example 2 — Number With Trailing Zero

$ Input: num = 1800

› Output: false

💡 Note: First reversal: 1800 → 81 (trailing zeros lost). Second reversal: 81 → 18. Since 18 ≠ 1800, return false.

Example 3 — Single Digit

$ Input: num = 0

› Output: true

💡 Note: First reversal: 0 → 0. Second reversal: 0 → 0. Since 0 == 0, return true.

Constraints

0 ≤ num ≤ 10⁶

Visualization

Tap to expand

Understanding the Visualization

Input

Given integer num

Double Reverse

Reverse twice and check equality

Pattern

Numbers with trailing zeros always fail

Key Takeaway

🎯 Key Insight: Numbers with trailing zeros always fail double reversal because leading zeros are dropped

Asked in

G Google 15 a Amazon 8

The key insight is that a number survives double reversal if and only if it has no trailing zeros. Numbers ending in zero lose those zeros during the first reversal. Best approach is trailing zero detection with Time: O(1), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Double Reversal Simulation	O(log n)	O(1)	Actually reverse the number twice and check if it equals the original
Trailing Zero Detection	O(1)	O(1)	Check if the number ends with zero - those numbers always fail double reversal

Double Reversal Simulation — Algorithm Steps

Convert number to string
Reverse digits and convert back to integer
Reverse again and compare with original

Visualization

Tap to expand

Step-by-Step Walkthrough

Original Number

Start with the input number

First Reversal

Reverse digits to get reversed1

Second Reversal

Reverse reversed1 to get reversed2

Compare

Check if reversed2 equals original

Code -

solution.c — C

#include <stdio.h>

int reverseNumber(int n) {
    int result = 0;
    while (n > 0) {
        result = result * 10 + n % 10;
        n /= 10;
    }
    return result;
}

int solution(int num) {
    return reverseNumber(reverseNumber(num)) == num;
}

int main() {
    int num;
    scanf("%d", &num);
    int result = solution(num);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

We process each digit twice, where the number of digits is log₁₀(n)

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables for the reversal process

✓ Linear Space

28.9K Views

Medium Frequency

~5 min Avg. Time

824 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Double Reversal Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler