How do you solve Count Good Numbers on LeetCode?

Count Good Numbers (LeetCode #1922) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Good numbers have a specific structure based on the index of digits.

What is the brute force solution for Count Good Numbers?

The brute force approach for Count Good Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible digit strings of length n and checks if each string meets the criteria of being a good number. This is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Count Good Numbers?

Count Good Numbers uses the following concepts: Math, Recursion. The recommended patterns to study are: Combinatorial Counting, Fast Exponentiation.

What are the interview tips for Count Good Numbers?

Always consider the constraints of the problem before deciding on an approach. Explain your thought process clearly; interviewers appreciate understanding your reasoning. Practice writing clean and efficient code, as clarity can be just as important as correctness.

What are common mistakes when solving Count Good Numbers?

Students often try to brute force without considering the size of n, leading to timeouts. Not using modular arithmetic can lead to overflow errors.

#1922

Count Good Numbers

Medium

MathRecursionCombinatorial CountingFast Exponentiation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(1)

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Intuition

Time O(log n)Space O(1)

Instead of generating all strings, we can use combinatorial counting. For even indices, we have 5 choices (0, 2, 4, 6, 8) and for odd indices, we have 4 choices (2, 3, 5, 7). We can calculate the total number of good strings using powers of these choices.

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Algorithm

3 steps

1Step 1: Determine the number of even and odd indices based on n.
2Step 2: Calculate the number of good strings as (5^(even_count) * 4^(odd_count)) % (10^9 + 7).
3Step 3: Use fast exponentiation to compute powers efficiently.

solution.py14 lines

1def countGoodNumbers(n):
2    MOD = 10**9 + 7
3    even_count = (n + 1) // 2
4    odd_count = n // 2
5    def power(x, y, p):
6        res = 1
7        x = x % p
8        while y > 0:
9            if (y & 1):
10                res = (res * x) % p
11            y //= 2
12            x = (x * x) % p
13        return res
14    return (power(5, even_count, MOD) * power(4, odd_count, MOD)) % MOD

ℹ

Complexity note: The time complexity is O(log n) due to the fast exponentiation method, which reduces the number of multiplications needed to compute powers.

1Good numbers have a specific structure based on the index of digits.
2Using combinatorial counting allows us to avoid generating all possible strings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.