2802. Find The K-th Lucky Number 🔒 

Description

We know that 4 and 7 are lucky digits. Also, a number is called lucky if it contains only lucky digits.

You are given an integer k, return the k^th lucky number represented as a string.

Example 1:

Input: k = 4
Output: "47"
Explanation: The first lucky number is 4, the second one is 7, the third one is 44 and the fourth one is 47.

Example 2:

Input: k = 10
Output: "477"
Explanation: Here are lucky numbers sorted in increasing order:
4, 7, 44, 47, 74, 77, 444, 447, 474, 477. So the 10^th lucky number is 477.

Example 3:

Input: k = 1000
Output: "777747447"
Explanation: It can be shown that the 1000^th lucky number is 777747447.

Constraints:

1 <= k <= 10⁹

Solutions

Solution 1: Mathematics

According to the problem description, a lucky number only contains the digits $4$ and $7$, so the number of $n$-digit lucky numbers is $2^n$.

We initialize $n=1$, then loop to check whether $k$ is greater than $2^n$. If it is, we subtract $2^n$ from $k$ and increment $n$, until $k$ is less than or equal to $2^n$. At this point, we just need to find the $k$-th lucky number among the $n$-digit lucky numbers.

If $k$ is less than or equal to $2^{n-1}$, then the first digit of the $k$-th lucky number is $4$, otherwise the first digit is $7$. Then we subtract $2^{n-1}$ from $k$ and continue to determine the second digit, until all digits of the $n$-digit lucky number are determined.

The time complexity is $O(\log k)$, and the space complexity is $O(\log k)$.

Python3

class Solution:
    def kthLuckyNumber(self, k: int) -> str:
        n = 1
        while k > 1 << n:
            k -= 1 << n
            n += 1
        ans = []
        while n:
            n -= 1
            if k <= 1 << n:
                ans.append("4")
            else:
                ans.append("7")
                k -= 1 << n
        return "".join(ans)

Java

class Solution {
    public String kthLuckyNumber(int k) {
        int n = 1;
        while (k > 1 << n) {
            k -= 1 << n;
            ++n;
        }
        StringBuilder ans = new StringBuilder();
        while (n-- > 0) {
            if (k <= 1 << n) {
                ans.append('4');
            } else {
                ans.append('7');
                k -= 1 << n;
            }
        }
        return ans.toString();
    }
}

C++

class Solution {
public:
    string kthLuckyNumber(int k) {
        int n = 1;
        while (k > 1 << n) {
            k -= 1 << n;
            ++n;
        }
        string ans;
        while (n--) {
            if (k <= 1 << n) {
                ans.push_back('4');
            } else {
                ans.push_back('7');
                k -= 1 << n;
            }
        }
        return ans;
    }
};

Go

func kthLuckyNumber(k int) string {
	n := 1
	for k > 1<<n {
		k -= 1 << n
		n++
	}
	ans := []byte{}
	for n > 0 {
		n--
		if k <= 1<<n {
			ans = append(ans, '4')
		} else {
			ans = append(ans, '7')
			k -= 1 << n
		}
	}
	return string(ans)
}

TypeScript

function kthLuckyNumber(k: number): string {
    let n = 1;
    while (k > 1 << n) {
        k -= 1 << n;
        ++n;
    }
    const ans: string[] = [];
    while (n-- > 0) {
        if (k <= 1 << n) {
            ans.push('4');
        } else {
            ans.push('7');
            k -= 1 << n;
        }
    }
    return ans.join('');
}