1922. Count Good Numbers 

Description

A digit string is good if the digits (0-indexed) at even indices are even and the digits at odd indices are prime (2, 3, 5, or 7).

For example, "2582" is good because the digits (2 and 8) at even positions are even and the digits (5 and 2) at odd positions are prime. However, "3245" is not good because 3 is at an even index but is not even.

Given an integer n, return the total number of good digit strings of length n. Since the answer may be large, return it modulo 10⁹ + 7.

A digit string is a string consisting of digits 0 through 9 that may contain leading zeros.

Example 1:

Input: n = 1
Output: 5
Explanation: The good numbers of length 1 are "0", "2", "4", "6", "8".

Example 2:

Input: n = 4
Output: 400

Example 3:

Input: n = 50
Output: 564908303

Constraints:

1 <= n <= 10¹⁵

Solutions

Solution 1: Fast Exponentiation

For a "good number" of length $n$, the even-indexed positions have $\lceil \frac{n}{2} \rceil = \lfloor \frac{n + 1}{2} \rfloor$ digits, and these positions can be filled with $5$ different digits ($0, 2, 4, 6, 8$). The odd-indexed positions have $\lfloor \frac{n}{2} \rfloor$ digits, and these positions can be filled with $4$ different digits ($2, 3, 5, 7$). Therefore, the total number of "good numbers" of length $n$ is:

$$ ans = 5^{\lceil \frac{n}{2} \rceil} \times 4^{\lfloor \frac{n}{2} \rfloor} $$

We can use fast exponentiation to compute $5^{\lceil \frac{n}{2} \rceil}$ and $4^{\lfloor \frac{n}{2} \rfloor}$. The time complexity is $O(\log n)$, and the space complexity is $O(1)$.

Python3

class Solution:
    def countGoodNumbers(self, n: int) -> int:
        mod = 10**9 + 7
        return pow(5, (n + 1) >> 1, mod) * pow(4, n >> 1, mod) % mod

Java

class Solution {
    private final int mod = (int) 1e9 + 7;

    public int countGoodNumbers(long n) {
        return (int) (qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod);
    }

    private long qpow(long x, long n) {
        long res = 1;
        while (n != 0) {
            if ((n & 1) == 1) {
                res = res * x % mod;
            }
            x = x * x % mod;
            n >>= 1;
        }
        return res;
    }
}

C++

class Solution {
public:
    int countGoodNumbers(long long n) {
        const int mod = 1e9 + 7;
        auto qpow = [](long long x, long long n) -> long long {
            long long res = 1;
            while (n) {
                if ((n & 1) == 1) {
                    res = res * x % mod;
                }
                x = x * x % mod;
                n >>= 1;
            }
            return res;
        };
        return qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod;
    }
};

Go

const mod int64 = 1e9 + 7

func countGoodNumbers(n int64) int {
	return int(myPow(5, (n+1)>>1) * myPow(4, n>>1) % mod)
}

func myPow(x, n int64) int64 {
	var res int64 = 1
	for n != 0 {
		if (n & 1) == 1 {
			res = res * x % mod
		}
		x = x * x % mod
		n >>= 1
	}
	return res
}

TypeScript

function countGoodNumbers(n: number): number {
    const mod = 1000000007n;
    const qpow = (x: bigint, n: bigint): bigint => {
        let res = 1n;
        while (n > 0n) {
            if (n & 1n) {
                res = (res * x) % mod;
            }
            x = (x * x) % mod;
            n >>= 1n;
        }
        return res;
    };
    const a = qpow(5n, BigInt(n + 1) / 2n);
    const b = qpow(4n, BigInt(n) / 2n);
    return Number((a * b) % mod);
}