Recursion - Medium Level - Question 1

Recursion - Medium Level - Question 1

Leetcode 50 Pow(x, n)

Implement pow(x, n), which calculates x raised to the power n (i.e., x^n).

Constraints:

-100.0 < x < 100.0

-2^31 <= n <= 2^31-1

-10^4 <= x^n <= 10^4

Analysis:

If n == 0, then it always return 1;

If n == 1, it returns x.

For other n, consider the positive n first (since the negative value can be calculated with the positive ones).

If n is even, then pow(x, n) can be rewritten as pow(x, n/2) * pow(x, n/2); if odd, pow(x, n/2) * pow(x, n/2) * x. One of the key steps is that we just need to calculate pow(x, n/2) once, for the rest calls (or the second time to this question), just re-use the calculated results.

One corner case is pow(x, INT_MIN), which will cause overflow if simply timing with (-1). Some details need to take.

See the code below:

class Solution {
public:
    double myPow(double x, int n) {
        // the nagative case (handled the corner case)
        if(n<0) return 1.0/(x*myPow(x, -(n+1)));
        // base case
        if(n==0) return 1;
        // only calculate once
        double t = myPow(x, n/2);
        if(n&1) return t*t*x;
        return t*t;
    }
};

Question 2

Leetcode 1922 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).

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.

Constraints:

1 <= n <= 10^15

Analysis:

One of the keys is how to keep the question still the "same question".

It is clear that for this question, the total number is "5 * 4 * 5 * 4 * 5 * 4 ...", which are apparently repetitive operations. So we do not need to calculate it one by one, which is NOT the most efficient method.

We can divide it into two parts: depending on its length, 

 if the first part is even, then the second part can always start with 5, or the same as the original question;

if the first part is odd, then the second part will start with 4, which is NOT the same as the original question. But we make it to the same: since we know it will start with 4, we can re-write it as 4 * (the remaining part), where the remaining part starts with 5 again.

Another key is to just calculate one for each length.

See the code below:

class Solution {
public:
    int countGoodNumbers(long long n) {
        int mod = 1e9 + 7;
        if(n==0) return 1;
        if(n==1) return 5;
        if(n==2) return 20;
        long d = n/2, t = 1;
        if(n&1) {
            t = countGoodNumbers(d);
            // n = 3, 7, ...
            if(d&1) return t*4*t%mod;
            // n = 5, 9, ...
            return t*5*t%mod;
        }
        if(d&1) {
            // n = 6, 10, ...
            t = countGoodNumbers(d-1);
            return t*5%mod*4*t%mod;
        }
        // n = 4, 8, ...
        t = countGoodNumbers(d);
        return t*t%mod;
    }
};

A different way to divide the n:

We can always to take the power of 2 as the first part, then it will always to be even (even after division by half). 

See the code below:

class Solution {
public:
    int countGoodNumbers(long long n) {
        if(n==0) return 1;
        if(n==1) return 5;
        if(n==2) return 20;
        // this line (or base case) is needed
        if(n==3) return 100;
        long mod = 1e9 + 7, s = pow(2, floor(log2(n)));
        long t = countGoodNumbers(s/2)%mod;
        long r = countGoodNumbers(n - s)%mod;
        return (t*t%mod)*r%mod;
    }
};

Please be noted that the above code may re-calculate some intermediate states, but it is till fast enough to coverge.

Upper Layer