Number of Distinct Binary Strings After Applying Operations Python/Java solution

The implementation is very short. Compute n = len(s), calculate exponent = n - k + 1, and return pow(2, exponent, 1_000_000_007) in Python or use fast modular exponentiation in Java. This directly gives the number of distinct reachable strings.

What is the best approach for Number of Distinct Binary Strings After Applying Operations?

The optimal approach uses a mathematical observation. There are exactly n - k + 1 possible positions where the k-length flip operation can start. Each position represents an independent binary decision (apply or not), so the number of reachable strings is 2^(n-k+1). The result is computed using modular exponentiation in O(log n) time and O(1) space.

Is Number of Distinct Binary Strings After Applying Operations asked at Google/Amazon/Meta?

Problems involving bit manipulation, combinatorics, and operation independence frequently appear in interviews at companies like Google, Amazon, and Meta. Variants that require recognizing XOR-style transformations or counting reachable states are common in mid-level interview rounds.

What data structure is used in Number of Distinct Binary Strings After Applying Operations?

The optimal solution does not require complex data structures. It relies on mathematical reasoning and modular exponentiation. Brute force approaches may use a hash set and queue for BFS to track visited strings.

What is the time complexity of Number of Distinct Binary Strings After Applying Operations?

The optimal solution runs in O(log n) time due to fast modular exponentiation used to compute 2^(n-k+1) mod 1e9+7. Space complexity is O(1) since only a few integer variables are required.

2450. Number of Distinct Binary Strings After Applying Operations - Practice Online

Q: Is Number of Distinct Binary Strings After Applying Operations easy or hard?

The problem is rated Medium because the key challenge is recognizing the mathematical pattern. Once you realize that each operation position contributes an independent binary choice, the implementation becomes trivial.

Q: How to solve Number of Distinct Binary Strings After Applying Operations in O(n)?

A full simulation would require exploring up to 2^n states, which is inefficient. Instead, observe that there are n-k+1 independent operations and each can be applied or skipped. This reduces the problem to computing 2^(n-k+1), which can be calculated using fast exponentiation in O(log n) time.

You are given a binary string s and a positive integer k.

You can apply the following operation on the string any number of times:

Choose any substring of size k from s and flip all its characters, that is, turn all 1's into 0's, and all 0's into 1's.

Return the number of distinct strings you can obtain. Since the answer may be too large, return it modulo 10⁹ + 7.

Note that:

A binary string is a string that consists only of the characters 0 and 1.
A substring is a contiguous part of a string.

Example 1:

Input: s = "1001", k = 3
Output: 4
Explanation: We can obtain the following strings:
- Applying no operation on the string gives s = "1001".
- Applying one operation on the substring starting at index 0 gives s = "0111".
- Applying one operation on the substring starting at index 1 gives s = "1110".
- Applying one operation on both the substrings starting at indices 0 and 1 gives s = "0000".
It can be shown that we cannot obtain any other string, so the answer is 4.

Example 2:

Input: s = "10110", k = 5
Output: 2
Explanation: We can obtain the following strings:
- Applying no operation on the string gives s = "10110".
- Applying one operation on the whole string gives s = "01001".
It can be shown that we cannot obtain any other string, so the answer is 2.

Constraints:

1 <= k <= s.length <= 10⁵
s[i] is either 0 or 1.