LeetCode 1922: Count Good Numbers Solution
Master LeetCode problem 1922 (Count Good Numbers), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation
To solve this problem efficiently, we can use exponentiation by squaring to calculate the powers of 5 and 4 modulo 10^9 + 7. We can then use these powers to compute the total number of good digit strings of length n based on whether n is even or odd.
- If n is even, the total number of good digit strings of length n is (pow(5, n/2) * pow(4, n/2)) % (10^9 + 7).
- If n is odd, we need to consider an extra factor of 5 for the last digit, so the total number of good digit strings of length n is (pow(5, n/2) * pow(4, n/2) * 5) % (10^9 + 7).
Time complexity: O(log n)
Space complexity: O(1)
Solution Code
class Solution {
public int countGoodNumbers(long n) {
long mod = 1000000007;
long res = 1;
long power5 = 5;
long power4 = 4;
while (n > 0) {
if ((n & 1) == 1) {
res = (res * power5) % mod;
n--;
} else {
power4 = (power4 * power4) % mod;
n /= 2;
}
}
return (int) res;
}
}Try It Yourself
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