Problem Description
Given a binary string s and a positive integer k, you are allowed to repeatedly choose any contiguous substring of length k and flip all its characters (i.e. change 0 to 1 and 1 to 0). Determine the number of distinct binary strings that can be obtained through any sequence of such operations. Return the answer modulo 10^9+7.
Key Insights
- Flipping a substring is an involution (flipping twice is equivalent to no change).
- Every operation can be treated as a binary decision (flip or not flip) on each possible starting index.
- The effect of each flip spans k consecutive bits and these effects are applied modulo 2.
- Mapping the sequence of flip decisions (on all valid starting indices) to the final binary string is a linear transformation over GF(2).
- It turns out that the transformation is injective, and the number of independent operations equals the number of valid starting positions, i.e. (n – k + 1), where n is the length of s.
Space and Time Complexity
Time Complexity: O(log n) (using fast modular exponentiation) Space Complexity: O(1)
Solution
The problem can be analyzed in terms of linear algebra in GF(2). Label each valid flip decision by a binary variable f_i, where 0 ≤ i ≤ n - k. Flipping a substring starting at index i affects the bits from index i to i+k-1. The final bit at any position j is given by the original s[j] XOR the XOR-sum of all f_i such that the substring starting at i covers index j. This mapping from the flip decisions (vector of length n-k+1) to the final string is a linear function over GF(2). It can be shown that this transformation is injective (has full column rank), and hence every unique combination of flip decisions produces a distinct outcome.
There are exactly 2^(n-k+1) possible ways to choose the flip decisions. Since the original string s only acts as a fixed offset, the number of distinct outcomes is exactly 2^(n-k+1). To meet the modulus requirement, fast modular exponentiation is used for efficiency.