Problem Description

You need to count how many "good" digit strings can be formed with a given length n.

A digit string is considered good when:

• Digits at even indices (0, 2, 4, ...) must be even numbers (0, 2, 4, 6, 8)
• Digits at odd indices (1, 3, 5, ...) must be prime numbers (2, 3, 5, 7)

Note that indexing starts from 0.

For example:

• "2582" is good because:

  • Index 0: 2 is even ✓
  • Index 1: 5 is prime ✓
  • Index 2: 8 is even ✓
  • Index 3: 2 is prime ✓

• "3245" is not good because:

  • Index 0: 3 is odd (should be even) ✗

Given an integer n, you need to find the total number of good digit strings of length n. Since the result can be very large, return it modulo 10^9 + 7.

The key insight is that for a string of length n:

• There are ⌈n/2⌉ positions at even indices, each can be filled with 5 choices (0, 2, 4, 6, 8)
• There are ⌊n/2⌋ positions at odd indices, each can be filled with 4 choices (2, 3, 5, 7)

Therefore, the total count is: 5^(⌈n/2⌉) × 4^(⌊n/2⌋) mod (10^9 + 7)

Intuition

When we look at this problem, we need to count valid combinations, not generate them. This is a classic combinatorics problem.

Let's think about building a good string position by position. For each position, we have independent choices that don't affect other positions:

• At position 0 (even index): We can choose from 5 even digits (0, 2, 4, 6, 8)
• At position 1 (odd index): We can choose from 4 prime digits (2, 3, 5, 7)
• At position 2 (even index): Again 5 choices
• At position 3 (odd index): Again 4 choices
• And so on...

Since each position's choice is independent, we can use the multiplication principle. If we have a string of length 4, the total number of good strings would be 5 × 4 × 5 × 4.

We can see a pattern emerging:

• For length 1: 5 (only one even-indexed position)
• For length 2: 5 × 4
• For length 3: 5 × 4 × 5 = 5² × 4
• For length 4: 5 × 4 × 5 × 4 = 5² × 4²
• For length 5: 5 × 4 × 5 × 4 × 5 = 5³ × 4²

The pattern shows that for a string of length n:

• Number of even-indexed positions = ⌈n/2⌉ (ceiling of n/2)
• Number of odd-indexed positions = ⌊n/2⌋ (floor of n/2)

Therefore, the total count becomes 5^(number of even positions) × 4^(number of odd positions).

Since the numbers can get very large, we use modular arithmetic and fast exponentiation to efficiently compute the result modulo 10^9 + 7.

Learn more about Recursion and Math patterns.

Solution Approach

The solution uses fast exponentiation to efficiently compute large powers modulo 10^9 + 7.

Based on our analysis, we need to calculate: $ans = 5^{\lceil \frac{n}{2} \rceil} \times 4^{\lfloor \frac{n}{2} \rfloor}$

Let's break down the implementation:

1. Calculate the number of even and odd positions:

   • Even-indexed positions: ⌈n/2⌉ = ⌊(n+1)/2⌋
   • Odd-indexed positions: ⌊n/2⌋

   In the code, this is done using bit operations:

   • (n + 1) >> 1 is equivalent to ⌊(n+1)/2⌋ (right shift by 1 divides by 2)
   • n >> 1 is equivalent to ⌊n/2⌋

2. Use Python's built-in pow function with three arguments:

   • pow(base, exponent, mod) efficiently computes base^exponent mod mod using fast exponentiation
   • This handles large exponents efficiently in O(log n) time

3. Multiply the results:

   • First compute 5^(⌈n/2⌉) mod (10^9 + 7)
   • Then compute 4^(⌊n/2⌋) mod (10^9 + 7)
   • Multiply them together and take modulo again to get the final answer

The implementation:

def countGoodNumbers(self, n: int) -> int:
    mod = 10**9 + 7
    return pow(5, (n + 1) >> 1, mod) * pow(4, n >> 1, mod) % mod

This approach avoids overflow issues and runs in O(log n) time complexity due to the fast exponentiation algorithm, making it efficient even for very large values of n.

Example Walkthrough

Let's walk through the solution for n = 4 to see how we count good digit strings.

Step 1: Identify positions and their constraints For a string of length 4, we have positions [0, 1, 2, 3]:

• Position 0 (even index): must be even digit → choices: {0, 2, 4, 6, 8} = 5 options
• Position 1 (odd index): must be prime digit → choices: {2, 3, 5, 7} = 4 options
• Position 2 (even index): must be even digit → choices: {0, 2, 4, 6, 8} = 5 options
• Position 3 (odd index): must be prime digit → choices: {2, 3, 5, 7} = 4 options

Step 2: Count positions by type

• Even-indexed positions: positions 0 and 2 → count = 2 = ⌈4/2⌉
• Odd-indexed positions: positions 1 and 3 → count = 2 = ⌊4/2⌋

Step 3: Apply the multiplication principle Since each position's choice is independent:

• Total = (choices for even positions)^(count of even positions) × (choices for odd positions)^(count of odd positions)
• Total = 5² × 4² = 25 × 16 = 400

Step 4: Verify with code approach Using our formula with bit operations:

• Even positions: (n + 1) >> 1 = (4 + 1) >> 1 = 5 >> 1 = 2
• Odd positions: n >> 1 = 4 >> 1 = 2
• Result: pow(5, 2, mod) * pow(4, 2, mod) % mod = 25 * 16 % mod = 400

Therefore, there are 400 good digit strings of length 4.

Some examples of valid strings: "0202", "2582", "4375", "8753" Example of invalid string: "1234" (position 0 has '1' which is odd, not even)

Solution Implementation

Time and Space Complexity

The time complexity of this code is O(log n). This is because the built-in pow function with three arguments uses fast exponentiation (also known as exponentiation by squaring) to compute the modular exponentiation. Each call to pow(base, exponent, mod) takes O(log exponent) time. In this code:

• pow(5, (n + 1) >> 1, mod) computes 5^⌈n/2⌉ mod (10^9 + 7) in O(log((n + 1) >> 1)) = O(log n) time
• pow(4, n >> 1, mod) computes 4^⌊n/2⌋ mod (10^9 + 7) in O(log(n >> 1)) = O(log n) time

Since these operations are performed sequentially and then multiplied together, the overall time complexity remains O(log n).

The space complexity is O(1). The algorithm only uses a constant amount of extra space to store:

• The mod variable
• Intermediate results from the pow function calls
• The final result before returning

The fast exponentiation algorithm implemented in Python's built-in pow function uses an iterative approach rather than recursive, so it doesn't consume additional stack space proportional to the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Not Using Modular Arithmetic Properly

Pitfall: A common mistake is computing the powers first and then applying modulo at the end:

# WRONG approach - will cause overflow for large n
result = (5 ** even_positions) * (4 ** odd_positions) % MOD

This fails because 5^even_positions and 4^odd_positions can become astronomically large before the modulo operation, causing memory errors or overflow.

Solution: Use modular exponentiation throughout the calculation:

# CORRECT approach
even_count = pow(5, even_positions, MOD)
odd_count = pow(4, odd_positions, MOD)
result = (even_count * odd_count) % MOD

2. Incorrect Position Count Calculation

Pitfall: Confusing the count of even and odd positions:

# WRONG - swapped the calculations
even_positions = n // 2
odd_positions = (n + 1) // 2

For a string of length n:

• When n=1: only position 0 (even) exists
• When n=2: positions 0 (even) and 1 (odd) exist
• When n=3: positions 0, 2 (even) and 1 (odd) exist

Solution: Remember that even positions always equal or exceed odd positions:

# CORRECT
even_positions = (n + 1) // 2  # Ceiling of n/2
odd_positions = n // 2          # Floor of n/2

3. Missing Final Modulo Operation

Pitfall: Forgetting to apply modulo after multiplication:

# WRONG - missing final modulo
return pow(5, even_positions, MOD) * pow(4, odd_positions, MOD)

Even though each pow() returns a value less than MOD, their product can exceed MOD.

Solution: Always apply modulo after multiplication:

# CORRECT
return (pow(5, even_positions, MOD) * pow(4, odd_positions, MOD)) % MOD

4. Using Custom Power Function Instead of Built-in

Pitfall: Implementing your own power function with modulo:

# INEFFICIENT and error-prone
def power_mod(base, exp, mod):
    result = 1
    for _ in range(exp):
        result = (result * base) % mod
    return result

This has O(n) time complexity and is much slower for large exponents.

Solution: Use Python's built-in pow(base, exp, mod) which implements fast exponentiation with O(log n) complexity:

# EFFICIENT
result = pow(base, exponent, MOD)