Count Good Numbers Leetcode Solution

Count Good Numbers

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).

See Also: Weekly Contest 248

• 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

Solution

Program C++

class Solution {
public:
    const long long MOD = 1000000007;

    long long mul(long long a, long long x, long long p)
    {
        if (x == 0) return 1;
        long long res = mul(a,x/2,p);
        res = res * res % p;
        if (x%2)
            res = res * a  % p;
        return res;
    }
    int countGoodNumbers(long long n) {
        long long pe = n/2 + n%2, po = (n/2);
        return mul(5,pe,MOD) * mul(4,po,MOD) % MOD;
    }
};

Program Python

class Solution:
    def countGoodNumbers(self, n: int) -> int:
        mod = 10 ** 9 + 7
        return pow(5, (n + 1) // 2, mod) * pow(4, n // 2, mod) % mod

Program Java

Credit: Here

1. For each even index, there are 5 options: 0, 2, 4, 6, 8;
2. For each odd index, there are 4 options: 2, 3, 5, 7;
3. If n is even, the solution is (4 * 5) ^ (n / 2); if odd, (4 * 5) ^ (n / 2) * 5.

public int countGoodNumbers(long n) {
        long good = n % 2 == 0 ? 1 : 5;
        long x = 4 * 5;
        for (long i = n / 2; i > 0; i /= 2, x = x * x % 1_000_000_007) {
            if (i % 2 != 0) {
                good = good * x % 1_000_000_007;
            }
        }
        return (int)good;
    }

    def countGoodNumbers(self, n: int) -> int:
        good, x, i = 5 ** (n % 2), 4 * 5, n // 2
        while i > 0:
            if i % 2 == 1:
                good = good * x % (10 ** 9 + 7)
            x = x * x % (10 ** 9 + 7)
            i //= 2
        return good
Analysis:

Time: O(logn), space: O(1).

Time: O(logn), space: O(1).