2443. Sum of Number and Its Reverse

Description

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

Solutions

Solution 1: Brute Force Enumeration

Enumerate $k$ in the range $[0,.., num]$, and check whether $k + reverse(k)$ equals $num$.

The time complexity is $O(n \times \log n)$, where $n$ is the size of $num$.

Python Code
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class Solution:
    def sumOfNumberAndReverse(self, num: int) -> bool:
        return any(k + int(str(k)[::-1]) == num for k in range(num + 1))

Java Code
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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;
    }
}

C++ Code
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class Solution {
public:
    bool 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;
    }
};

Go Code
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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
}

TypeScript Code
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function sumOfNumberAndReverse(num: number): boolean {
    for (let i = 0; i <= num; i++) {
        if (i + Number([...(i + '')].reverse().join('')) === num) {
            return true;
        }
    }
    return false;
}

Rust Code
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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
    }
}

C Code
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bool sumOfNumberAndReverse(int num) {
    for (int i = 0; i <= num; i++) {
        int t = i;
        int j = 0;
        while (t > 0) {
            j = j * 10 + t % 10;
            t /= 10;
        }
        if (i + j == num) {
            return 1;
        }
    }
    return 0;
}