Sum of Number and Its Reverse

Sum of Number and Its Reverse - Problem

Math Enumeration Medium

Given a non-negative integer num, return true if num can be expressed as the sum of any non-negative integer and its reverse, or false otherwise.

The reverse of an integer is obtained by reversing the order of its digits. For example, the reverse of 123 is 321.

Example: If num = 443, we can express it as 181 + 181 (where 181 reversed is 181), so return true.

Input & Output

Example 1 — Palindromic Sum

$ Input: num = 443

› Output: true

💡 Note: 443 can be expressed as 370 + 73, where 73 is the reverse of 370.

Example 2 — Simple Case

$ Input: num = 63

› Output: true

💡 Note: 63 can be expressed as some number plus its reverse. For example, we need to find x where x + reverse(x) = 63.

Example 3 — Impossible Case

$ Input: num = 4

› Output: true

💡 Note: 4 can be expressed as 2 + 2, where the reverse of 2 is 2.

Constraints

0 ≤ num ≤ 10⁶

Visualization

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Asked in

M Meta 25 G Google 20

The key insight is to iterate through all possible numbers from 0 to num and check if any number plus its reverse equals the target. Best approach uses mathematical reverse calculation to avoid string operations. Time: O(n×d), Space: O(1)

Common Approaches

✓ Brute Force Enumeration

⏱️ Time: O(n × d) Space: O(d)

Iterate through all possible numbers from 0 to num. For each number, calculate its reverse by converting to string, reversing, and converting back. Check if the sum equals our target num.

Mathematical Reverse Calculation

⏱️ Time: O(n × d) Space: O(1)

Instead of converting to string and back, calculate the reverse using mathematical operations with modulo and division. This avoids string allocation overhead while maintaining the same logic.

Brute Force Enumeration — Algorithm Steps

Iterate from 0 to num
For each number, calculate its reverse
Check if number + reverse equals num
Return true if found, false otherwise

Visualization

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Step-by-Step Walkthrough

Start Loop

Check numbers 0, 1, 2, ... up to num

Calculate Reverse

For each number, find its reverse

Check Sum

Test if number + reverse = target

Code -

solution.c — C

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

int reverseNumber(int n) {
    char str[20];
    sprintf(str, "%d", n);
    int len = strlen(str);
    for (int i = 0; i < len / 2; i++) {
        char temp = str[i];
        str[i] = str[len - 1 - i];
        str[len - 1 - i] = temp;
    }
    return atoi(str);
}

int solution(int num) {
    for (int i = 0; i <= num; i++) {
        int reversedI = reverseNumber(i);
        if (i + reversedI == num) {
            return 1;
        }
    }
    return 0;
}

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(n × d)

Loop runs n times, each reverse operation takes O(d) where d is number of digits

✓ Linear Growth

Space Complexity

O(d)

String conversion requires space for d digits

✓ Linear Space

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Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

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Code -

Time & Space Complexity

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