README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0728.Self%20Dividing%20Numbers/README_EN.md	Math

728. Self Dividing Numbers

中文文档

Description

A self-dividing number is a number that is divisible by every digit it contains.

• For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.

A self-dividing number is not allowed to contain the digit zero.

Given two integers left and right, return a list of all the self-dividing numbers in the range [left, right] (both inclusive).

 

Example 1:

Input: left = 1, right = 22
Output: [1,2,3,4,5,6,7,8,9,11,12,15,22]

Example 2:

Input: left = 47, right = 85
Output: [48,55,66,77]

 

Constraints:

• 1 <= left <= right <= 104

Solutions

Solution 1: Simulation

We define a function $\textit{check}(x)$ to determine whether $x$ is a self-dividing number. The implementation idea of the function is as follows:

We use $y$ to record the value of $x$, and then continuously divide $y$ by $10$ until $y$ is $0$. During this process, we check whether the last digit of $y$ is $0$, or whether $x$ cannot be divided by the last digit of $y$. If either of these conditions is met, then $x$ is not a self-dividing number, and we return $\text{false}$. Otherwise, after traversing all the digits, we return $\text{true}$.

Finally, we traverse all the numbers in the interval $[\textit{left}, \textit{right}]$, and for each number, we call $\textit{check}(x)$. If it returns $\text{true}$, we add this number to the answer array.

The time complexity is $O(n \times \log_{10} M)$, where $n$ is the number of elements in the interval $[\textit{left}, \textit{right}]$, and $M = \textit{right}$, which is the maximum value in the interval.

Python3

class Solution:
    def selfDividingNumbers(self, left: int, right: int) -> List[int]:
        def check(x: int) -> bool:
            y = x
            while y:
                if y % 10 == 0 or x % (y % 10):
                    return False
                y //= 10
            return True

        return [x for x in range(left, right + 1) if check(x)]

Java

class Solution {
    public List<Integer> selfDividingNumbers(int left, int right) {
        List<Integer> ans = new ArrayList<>();
        for (int x = left; x <= right; ++x) {
            if (check(x)) {
                ans.add(x);
            }
        }
        return ans;
    }

    private boolean check(int x) {
        for (int y = x; y > 0; y /= 10) {
            if (y % 10 == 0 || x % (y % 10) != 0) {
                return false;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    vector<int> selfDividingNumbers(int left, int right) {
        auto check = [&](int x) -> bool {
            for (int y = x; y; y /= 10) {
                if (y % 10 == 0 || x % (y % 10)) {
                    return false;
                }
            }
            return true;
        };
        vector<int> ans;
        for (int x = left; x <= right; ++x) {
            if (check(x)) {
                ans.push_back(x);
            }
        }
        return ans;
    }
};

Go

func selfDividingNumbers(left int, right int) (ans []int) {
	check := func(x int) bool {
		for y := x; y > 0; y /= 10 {
			if y%10 == 0 || x%(y%10) != 0 {
				return false
			}
		}
		return true
	}
	for x := left; x <= right; x++ {
		if check(x) {
			ans = append(ans, x)
		}
	}
	return
}

TypeScript

function selfDividingNumbers(left: number, right: number): number[] {
    const check = (x: number): boolean => {
        for (let y = x; y; y = Math.floor(y / 10)) {
            if (y % 10 === 0 || x % (y % 10) !== 0) {
                return false;
            }
        }
        return true;
    };
    return Array.from({ length: right - left + 1 }, (_, i) => i + left).filter(check);
}

Rust

impl Solution {
    pub fn self_dividing_numbers(left: i32, right: i32) -> Vec<i32> {
        fn check(x: i32) -> bool {
            let mut y = x;
            while y > 0 {
                if y % 10 == 0 || x % (y % 10) != 0 {
                    return false;
                }
                y /= 10;
            }
            true
        }

        (left..=right).filter(|&x| check(x)).collect()
    }
}