Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1922.html

1922. Count Good Numbers

Level

Medium

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 10^9 + 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 <= 10^15

Solution

A good digit string has five possible digits at each even index and four possible digits at each odd index. For each pair of adjacent indices, there are 20 possible values. Let n = 2 * m + k where m is an integer and k is either 0 or 1, and the number of good digits with length n is m^20*5^k.

  • class Solution {
    static final int MODULO = 1000000007;

    public int countGoodNumbers(long n) {
        long halfN = n / 2;
        int remainder = (int) (n % 2);
        long count = myPow(5 * 4, halfN);
        if (remainder == 1)
            count *= 5;
        count %= MODULO;
        return (int) count;
    }

    public long myPow(long x, long n) {
        long power = 1;
        while (n > 0) {
            if (n % 2 == 1)
                power = power * x % MODULO;
            x = x * x % MODULO;
            n /= 2;
        }
        return power;
    }
}

############

class Solution {
    private int mod = 1000000007;

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

    private long myPow(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;
    }
}

  • // OJ: https://leetcode.com/problems/count-good-numbers/
// Time: O(logN)
// Space: O(1)
class Solution {
    long mod = 1e9 + 7;
    long modpow(long base, long exp) {
        long ans = 1;
        while (exp) {
            if (exp & 1) ans = ans * base % mod;
            base = base * base % mod;
            exp >>= 1;
        }
        return ans;
    }
public:
    int countGoodNumbers(long long n) {
        return modpow(5, (n + 1) / 2) * modpow(4, n / 2) % mod;
    }
};

  • class Solution:
    def countGoodNumbers(self, n: int) -> int:
        mod = 10**9 + 7

        def myPow(x, n):
            res = 1
            while n:
                if (n & 1) == 1:
                    res = res * x % mod
                x = x * x % mod
                n >>= 1
            return res

        return myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % mod

############

# 1922. Count Good Numbers
# https://leetcode.com/problems/count-good-numbers

class Solution:
    def countGoodNumbers(self, n: int) -> int:
        M = 10 ** 9 + 7
        
        def power(base, exp):
            if exp == 0: return 1
            
            v = power(base, exp // 2)
            
            if exp % 2 == 0:
                return (v * v) % M
            else:
                return (v * v * base) % M
        
        even = int(n // 2 + int(n % 2 != 0))
        odd = int(n // 2)
         
        e = power(5, even)
        o = power(4, odd)
        
        return (e * o) % M


  • 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
}

All Problems

All Solutions