LeetCode #2443 — MEDIUM

Sum of Number and Its Reverse

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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.

Example 1:

Input: num = 443
Output: true
Explanation: 172 + 271 = 443 so we return true.

Example 2:

Input: num = 63
Output: false
Explanation: 63 cannot be expressed as the sum of a non-negative integer and its reverse so we return false.

Example 3:

Input: num = 181
Output: true
Explanation: 140 + 041 = 181 so we return true. Note that when a number is reversed, there may be leading zeros.

Constraints:

0 <= num <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

The constraints are small enough that we can check every number.
To reverse a number, first convert it to a string. Then, create a new string that is the reverse of the first one. Finally, convert the new string back into a number.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2443: Sum of Number and Its Reverse
class Solution {
    public boolean sumOfNumberAndReverse(int num) {
        for (int x = 0; x <= num; ++x) {
            int k = x;
            int y = 0;
            while (k > 0) {
                y = y * 10 + k % 10;
                k /= 10;
            }
            if (x + y == num) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #2443: Sum of Number and Its Reverse
func sumOfNumberAndReverse(num int) bool {
	for x := 0; x <= num; x++ {
		k, y := x, 0
		for k > 0 {
			y = y*10 + k%10
			k /= 10
		}
		if x+y == num {
			return true
		}
	}
	return false
}

# Accepted solution for LeetCode #2443: Sum of Number and Its Reverse
class Solution:
    def sumOfNumberAndReverse(self, num: int) -> bool:
        return any(k + int(str(k)[::-1]) == num for k in range(num + 1))

// Accepted solution for LeetCode #2443: Sum of Number and Its Reverse
impl Solution {
    pub fn sum_of_number_and_reverse(num: i32) -> bool {
        for i in 0..=num {
            if i + ({
                let mut t = i;
                let mut j = 0;
                while t > 0 {
                    j = j * 10 + (t % 10);
                    t /= 10;
                }
                j
            }) == num
            {
                return true;
            }
        }
        false
    }
}

// Accepted solution for LeetCode #2443: Sum of Number and Its Reverse
function sumOfNumberAndReverse(num: number): boolean {
    for (let i = 0; i <= num; i++) {
        if (i + Number([...(i + '')].reverse().join('')) === num) {
            return true;
        }
    }
    return false;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.