Problem Description

You have a binary string s (containing only '0's and '1's) and a positive integer k.

You can perform a flip operation on the string as many times as you want. In each operation:

• Select any substring of length exactly k from s
• Flip all characters in that substring (change all '1's to '0's and all '0's to '1's)

Your task is to count how many distinct strings you can create by applying this operation any number of times (including zero times, which gives you the original string).

For example, if you have string "101" and k = 2:

• You could flip the first two characters "10" to get "011"
• You could flip the last two characters "01" to get "110"
• You could apply multiple flips in sequence to get other strings

The answer should be returned modulo 10^9 + 7 since the count can be very large.

Key observations:

• A substring must be exactly k characters long and must be contiguous
• You can apply the operation zero or more times
• The same substring can be flipped multiple times
• We want the total count of unique strings that can be obtained

The mathematical insight is that if the string has length n, there are n - k + 1 possible positions where you can place a substring of length k. Each of these positions can either be flipped or not flipped in the final result, giving us 2^(n - k + 1) total distinct strings.

Intuition

Let's think about what happens when we flip substrings. If we have a string of length n and we can flip substrings of length k, we have n - k + 1 possible positions where we can start our flip operation.

The key insight is that each position can be independently decided - we either flip it or we don't. This is because:

1. Flipping twice cancels out: If we flip the same substring twice, it returns to its original state. So for any position, the only thing that matters is whether we flip it an odd or even number of times.

2. Order doesn't matter: If we flip position i and then position j, we get the same result as flipping position j and then position i. The final string only depends on which positions were flipped an odd number of times.

3. Overlapping flips combine uniquely: When substrings overlap, the overlapping parts get flipped by both operations. But since we can control each position independently (flip or not flip), different combinations still produce distinct results.

Think of it like this: we have n - k + 1 switches, and each switch can be either ON (flip that position) or OFF (don't flip). Each unique combination of switch settings produces a unique string.

Since each of the n - k + 1 positions has 2 choices (flip or not flip), and all choices are independent, the total number of distinct strings we can create is 2^(n - k + 1).

This explains why the solution is simply computing 2^(n - k + 1) mod (10^9 + 7).

Learn more about Math patterns.

Solution Approach

The implementation is straightforward once we understand that we need to calculate 2^(n - k + 1) where n is the length of the string.

Step-by-step implementation:

1. Calculate the exponent: First, we need to find how many positions we can place a substring of length k in string s. This is len(s) - k + 1.

2. Compute the power: We need to calculate 2^(len(s) - k + 1). Python's built-in pow() function efficiently handles this calculation.

3. Apply modulo: Since the result can be very large, we take modulo 10^9 + 7 to keep the number within bounds.

The complete formula:

result = pow(2, len(s) - k + 1) % (10**9 + 7)

Why this works:

• Each of the n - k + 1 substring positions represents an independent choice (flip or not flip)
• With 2 choices per position and all positions being independent, we get 2^(n - k + 1) total combinations
• Each combination produces a unique string

Time Complexity: O(log(n)) for the power calculation using Python's built-in pow() function which uses fast exponentiation.

Space Complexity: O(1) as we only use a constant amount of extra space.

The beauty of this solution is that we don't actually need to generate or track the strings themselves - the mathematical property tells us exactly how many distinct strings are possible.

Example Walkthrough

Let's walk through a concrete example with string s = "10110" and k = 3.

Step 1: Identify possible flip positions

• String length n = 5
• Substring length k = 3
• Number of positions where we can place a substring of length 3: n - k + 1 = 5 - 3 + 1 = 3

These positions are:

• Position 0: substring "101" (indices 0-2)
• Position 1: substring "011" (indices 1-3)
• Position 2: substring "110" (indices 2-4)

Step 2: Understand the choices For each position, we have 2 choices:

• Don't flip (0)
• Flip (1)

This gives us 2^3 = 8 total combinations:

Step 3: Verify distinctness Let's trace through one example - flipping positions 0 and 1:

• Original: "10110"
• Flip position 0 (substring "101"): "01010"
• Flip position 1 (substring "101"): "00100"

Notice how overlapping flips affect shared characters:

• Character at index 1 gets flipped by position 0 (changes 0→1)
• Character at index 1 gets flipped again by position 1 (changes 1→0)
• Character at index 2 gets flipped by both operations (1→0→1→0)

Step 4: Calculate the answer Using our formula: 2^(n - k + 1) = 2^3 = 8

All 8 strings we generated are distinct, confirming our mathematical approach. The answer would be 8 % (10^9 + 7) = 8.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. This is because:

• len(s) takes O(n) time to compute the length of the string
• The arithmetic operations len(s) - k + 1 take O(1) time
• The pow(2, len(s) - k + 1) operation takes O(len(s) - k + 1) time in the worst case, which is O(n) since the exponent is bounded by n
• The modulo operation % (10**9 + 7) takes O(1) time for the final result

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

• The intermediate calculation results
• The final result value

No additional data structures that scale with input size are created.

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

Common Pitfalls

1. Incorrect Modulo Application

Pitfall: Using pow(2, len(s) - k + 1) % MOD instead of pow(2, len(s) - k + 1, MOD).

When dealing with large exponents, computing 2^(n-k+1) first and then taking modulo can cause integer overflow issues, even in Python which handles big integers. The three-argument form of pow() computes the result using modular exponentiation, which is much more efficient.

Wrong approach:

result = pow(2, len(s) - k + 1) % (10**9 + 7)  # Can be inefficient for large exponents

Correct approach:

result = pow(2, len(s) - k + 1, 10**9 + 7)  # Efficient modular exponentiation

2. Edge Case: When k > len(s)

Pitfall: Not handling the case where k is greater than the string length.

If k > len(s), we cannot select any substring of length k, so only the original string is possible. The formula 2^(n-k+1) would give a negative or zero exponent, which might not represent the correct answer.

Solution:

def countDistinctStrings(self, s: str, k: int) -> int:
    MOD = 10**9 + 7
    n = len(s)
  
    # Handle edge case where k is larger than string length
    if k > n:
        return 1  # Only the original string is possible
  
    return pow(2, n - k + 1, MOD)

3. Misunderstanding the Problem Statement

Pitfall: Confusing the flip operation with replacement operation.

The code comment mentions "replacing it with '0' or '1'" which is incorrect. The problem is about flipping substrings (toggling bits), not arbitrary replacement. While the mathematical result might be the same due to the independence of positions, the conceptual understanding matters for variations of this problem.

Correct understanding:

• Flip operation: '0' → '1' and '1' → '0'
• Each substring of length k can be in either flipped or unflipped state
• Total states = 2^(number of possible substring positions)

4. Off-by-One Error in Counting Positions

Pitfall: Incorrectly calculating the number of possible substring positions.

Some might think there are n - k positions instead of n - k + 1.

Example verification:

• String "1010" with k=2
• Possible positions: [0:2], [1:3], [2:4]
• That's 3 positions = 4 - 2 + 1 ✓

Correct formula: positions = len(s) - k + 1