1922. Count Good Numbers

Problem Statement

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

Intuition

Approach:

We just have to find pow(n,x) in recursion

So we employ this trick

2^5 = 2^(2*2+1) = (2^2)^2 * 2 here, half is 2^2 now and since 5 is odd extra multiply
Go for 2^2 now 

Links

https://leetcode.com/problems/count-good-numbers/description/

Video Links

Approach 1:

C++

class Solution {
public:
    long mod=1e9+7;
    int countGoodNumbers(long n) {
        long even=(n+1)/2;
        long odd=n/2;

        return (int)(pow(5,even)*pow(4,odd)%mod);
    }
    long pow(long x,long n){
        if(n==0){
            return 1;
        }
        long half=pow(x,n/2);
        if(n%2==0){
            return (half*half)%mod;
        }
        return (half*half*x)%mod;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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