How do you solve Sum of Number and Its Reverse on LeetCode?

Sum of Number and Its Reverse (LeetCode #2443) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The sum of a number and its reverse can be efficiently checked by limiting the range of candidates.

What is the brute force solution for Sum of Number and Its Reverse?

The brute force approach for Sum of Number and Its Reverse is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible non-negative integer to see if it can sum with its reverse to equal the given number. This is straightforward but inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Number and Its Reverse?

Sum of Number and Its Reverse uses the following concepts: Math, Enumeration. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Sum of Number and Its Reverse?

Always think about edge cases, such as 0 or small numbers. Discuss your thought process and approach before coding. Optimize your solution after implementing the brute force to show your understanding of efficiency.

What are common mistakes when solving Sum of Number and Its Reverse?

Not considering leading zeros in the reverse. Looping through all numbers instead of limiting to half of num.

#2443

Sum of Number and Its Reverse

Medium

MathEnumerationTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Instead of checking every number, we can iterate through potential candidates for the reverse sum. This reduces unnecessary calculations and focuses on valid pairs.

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Algorithm

4 steps

1Step 1: Loop through all integers from 0 to num/2 (since both numbers must be non-negative).
2Step 2: For each integer, calculate its reverse.
3Step 3: Check if the sum of the integer and its reverse equals num. If yes, return true.
4Step 4: If no such pair is found after the loop, return false.

solution.py9 lines

1# Full working Python code
2
3def isSumOfNumberAndReverse(num):
4    def reverse(n):
5        return int(str(n)[::-1])
6    for i in range(num // 2 + 1):
7        if i + reverse(i) == num:
8            return True
9    return False

ℹ

Complexity note: The time complexity is O(n) because we only loop through half of the numbers up to num, and reversing a number takes constant time. The space complexity is O(1) since we use a constant amount of space.

1The sum of a number and its reverse can be efficiently checked by limiting the range of candidates.
2Reversing a number can be done easily with string manipulation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.