2443. Sum of Number and Its Reverse

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Problem Description

In the context of this LeetCode problem, one is tasked with determining whether a non-negative integer num can be represented as a sum of any non-negative integer and its reverse. The reverse of an integer is the number obtained by reversing the order of its digits. For instance, the reverse of 123 is 321. The problem requires the function to return true if such a representation is possible for the given num, or false otherwise.


The straightforward approach to solve this problem relies on the simple brute force method. We consider all non-negative integers from 0 up to num inclusive, because the sum of a number and its reverse cannot be greater than num itself. For each of these integers, named k, we calculate its reverse by converting k to a string, reversing the string, and converting it back to an integer. Then, we check whether the original number k plus its reverse equals num. If we find any such k that satisfies this condition, the function returns true. If the loop completes without finding any valid k, the function returns false. This approach ensures that all possibilities are checked.

This process is efficiently executed in the solution code using a generator expression within the any function, which iteratively checks each number until a match is found, and returns true as soon as a satisfying number is encountered.

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Solution Approach

For this problem, the implementation is fairly straightforward and does not involve complex data structures or advanced algorithms. It's a direct translation of the brute force approach into Python code.

The solution defines a method sumOfNumberAndReverse within the Solution class. This method takes one parameter, num, which is the non-negative integer that we want to examine.

The core of the implementation is the expression:

1any(k + int(str(k)[::-1]) == num for k in range(num + 1))

Here's a step-by-step walk-through of what's happening:

  1. range(num + 1): We create an iterable sequence of numbers starting from 0 up to and including num. This is because the largest number that, when added to its reverse, could potentially equal num is num itself.

  2. str(k)[::-1]: For each number k in the range, we convert k into a string using str(k) then reverse the string by applying the slicing operation [::-1]. This slice notation is a Python idiom for reversing sequences.

  3. int(...): The reversed string is converted back to an int because we need to perform arithmetic with it.

  4. k + int(str(k)[::-1]) == num: We add the integer k to its reverse and check if the sum is equal to num.

  5. any(...): This is a built-in Python function that takes an iterable and returns True if at least one element in the iterable is True. In this case, it iterates over the generator expression, which yields True or False for each k in the sequence based on whether k plus its reverse equals num.

  6. return ...: Finally, the method returns the result of the any function. If any value of k found satisfies the condition, True is returned; otherwise, False is returned.

This solution is elegant and concise thanks to Python's high-level abstractions but comes with an O(n) time complexity, as it might need to check all integers from 0 to num. The space complexity, on the other hand, is O(1) since there are no additional data structures consuming memory based on the input size; the integers are generated one by one.

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Example Walkthrough

Let's consider a small example using the number num = 121 to illustrate the solution approach. We want to determine if there exists a non-negative integer k such that when k is added to its reverse, the sum equals 121.

  1. First, we generate a sequence of numbers from 0 to 121 inclusive, because these are all the potential candidates for k that, when combined with their reverse, could equal num.

  2. We then iterate through these numbers one by one. For each iteration, let's denote our current number as k.

  3. We reverse the digits of k. If k was 12, the reversed version would be 21 which is obtained by converting k to a string ("12"), reversing the string ("21"), and converting it back to an integer (21).

  4. We add the original k to its reversed version. For k = 12, this would be 12 + 21 which equals 33.

  5. We check if this sum equals num (121 in this case). Since 33 is not equal to 121, we continue this process with the next number in the sequence.

  6. If at any point the sum of k and its reverse equals 121, we will return True. For example, if k were 112, its reverse would be 211, and adding those together yields 112 + 211 = 323 which is not 121. So we move on.

  7. If we have gone through all numbers up to 121 and have not found a sum that equals 121, we will return False.

  8. Luckily, when k = 29, we find that its reverse is 92, and adding them together yields 29 + 92 = 121. Since this satisfies our condition, the any function will immediately return True, indicating that num = 121 can indeed be expressed as the sum of a number and its reverse.

Solution Implementation

1class Solution:
2    def sum_of_number_and_reverse(self, num: int) -> bool:
3        # Iterate over all numbers from 0 to num, inclusive
4        for integer in range(num + 1):
5            # Calculate the reverse of the current number by converting it to a string, 
6            # reversing it, and then casting it back to an integer
7            reverse_integer = int(str(integer)[::-1])
9            # Check if the sum of the current number and its reverse equals the input number
10            if integer + reverse_integer == num:
11                # If a match is found, return True
12                return True
14        # If no match is found in the iteration, return False
15        return False
1class Solution {
2    /**
3     * Checks whether a given number can be expressed as 
4     * the sum of a number and its reverse.
5     *
6     * @param num The number to check.
7     * @return true if the number can be expressed as the sum 
8     *         of a number and its reverse, otherwise false.
9     */
10    public boolean sumOfNumberAndReverse(int num) {
11        // Loop from 0 to the given number (inclusive)
12        for (int originalNumber = 0; originalNumber <= num; ++originalNumber) {
13            int reversedNumber = 0;
14            int temp = originalNumber;
16            // Reverse the current number
17            while (temp > 0) {
18                int lastDigit = temp % 10;
19                reversedNumber = reversedNumber * 10 + lastDigit;
20                temp /= 10;
21            }
23            // Check if the sum of the original and reversed number is equal to the input number
24            if (originalNumber + reversedNumber == num) {
25                return true; // Found a pair that satisfies the condition
26            }
27        }
29        // If no such pair is found in the loop, return false
30        return false;
31    }
1class Solution {
3    // Checks if a number can be expressed as the sum of a number and its reverse
4    bool sumOfNumberAndReverse(int num) {
5        // Loop through all numbers starting from 0 up to the given number
6        for (int original_number = 0; original_number <= num; ++original_number) {
7            int remaining = original_number;
8            int reversed_number = 0;
9            // Reverse the original_number
10            while (remaining > 0) {
11                reversed_number = reversed_number * 10 + remaining % 10; // Append the last digit of remaining to reversed_number
12                remaining /= 10; // Remove the last digit from remaining
13            }
14            // Check if the sum of the original number and its reverse equals the given number
15            if (original_number + reversed_number == num) {
16                return true; // If the condition is met, return true
17            }
18        }
19        return false; // If no such pair is found, return false
20    }
1// Checks if a given number is equal to the sum of another number and its reverse.
2// num: Number to check for this property.
3// Returns a boolean value indicating whether such a pair exists.
5function sumOfNumberAndReverse(num: number): boolean {
6    // Iterate over all numbers from 0 to the input number
7    for (let i = 0; i <= num; i++) {
8        // Calculate the reverse of the current number 'i'
9        const reversedNumber = Number([...(i.toString())].reverse().join(''));
11        // Check if the current number plus its reverse equals the input number
12        if (i + reversedNumber === num) {
13            // Return true if the condition holds for the current number
14            return true;
15        }
16    }
17    // After checking all numbers, return false if no suitable pair was found
18    return false;

Time and Space Complexity

Time Complexity

The given function sumOfNumberAndReverse performs a linear search from 0 to num inclusive. For each value k in this range, the function calculates the reverse of the number by converting it to a string, reverse the string (this is done using str(k)[::-1]), and then converting it back to an integer. After this, it checks if the sum of the number k and its reverse equals the input number num.

Therefore, we can say the time complexity of the function is O(n * m), where n is the value of num and m represents the time taken to reverse the number. Since the number of digits d in the number k can be represented as O(log(k)), the reverse operation is O(d) which simplifies to O(log(n)) for the worst case when k is close to num. Thus, the overall time complexity is O(n * log(n)).

Space Complexity

The space complexity of the function is O(1). The reason for this constant space complexity is because aside from a few variables (k and the reversed number), the function does not use any additional space that scales with the input size num. The inputs and variables are of a fixed size, so it doesn't matter how large num is, the space used by the function remains constant.

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