LeetCode Problem 2450

Difficulty: 🟡 Medium
Topics: Math, String

Solution

LeetCode 2450 - Number of Distinct Binary Strings After Applying Operations

Problem Understanding

We are given a binary string s of length n and an integer k. An operation consists of selecting any contiguous substring of length k and flipping every bit in that substring. A 0 becomes 1, and a 1 becomes 0.

The operation may be applied any number of times, including zero times. Since we may repeatedly choose different substrings, many different binary strings can potentially be generated. The task is to determine how many distinct strings are reachable from the original string.

The answer can become very large, so it must be returned modulo 10^9 + 7.

The key observation is that the actual contents of s are less important than they initially appear. The problem is fundamentally asking how many different states can be reached through combinations of the allowed flip operations.

The constraints are large:

• 1 <= k <= n <= 100000

A state-space search over all possible strings is impossible because there are up to 2^100000 possible binary strings. Any solution that explicitly generates reachable strings will immediately become infeasible.

Some important edge cases include:

• k = 1, where every individual bit can be flipped independently.
• k = n, where the only possible operation flips the entire string.
• Very large strings near the maximum constraint.
• Strings consisting entirely of 0s or entirely of 1s.
• Cases where multiple sequences of operations produce the same final string.

Understanding the algebraic structure of the operations is the key to solving the problem efficiently.

Approaches

Brute Force

A natural first idea is to treat each binary string as a state in a graph.

From a given state, we can choose any substring of length k, flip it, and move to a new state. Starting from s, we could perform a BFS or DFS and count all reachable strings.

This approach is correct because it explicitly explores every reachable configuration.

However, it is completely impractical. A string of length n has up to 2^n possible states. Even for moderate values of n, the state space becomes astronomically large. Since n can reach 100000, exhaustive exploration is impossible.

Key Insight

Each operation is its own inverse.

If we represent the binary string as a vector over GF(2), flipping a length-k substring corresponds to XORing with a fixed vector:

• The chosen substring positions contain 1.
• All other positions contain 0.

Let there be m = n - k + 1 possible substrings of length k.

Each substring operation corresponds to one generator vector. Applying operations multiple times is equivalent to XORing together some subset of these generator vectors. Applying the same operation twice cancels out.

Therefore, every reachable string has the form:

original_string XOR linear_combination_of_generators

The number of distinct reachable strings equals the size of the vector space generated by these substring vectors.

If the rank of the generator vectors is r, then the number of distinct reachable strings is:

2^r

So the problem reduces to finding the rank of all length-k interval vectors.

A remarkable property is that these vectors always have rank:

n - k + 1

To see why, consider the vectors:

1110000...
0111000...
0011100...
...

Each vector begins at a different position. Looking from left to right, every new vector introduces a new leading position that cannot be produced by previous vectors. Therefore the vectors are linearly independent.

Since there are exactly n-k+1 such vectors, the rank is:

r = n - k + 1

Thus the answer is simply:

2^(n-k+1) mod (10^9+7)

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^n)	O(2^n)	Explores every reachable binary string
Optimal	O(log n)	O(1)	Uses the rank of the interval-flip vector space

Algorithm Walkthrough

1. Let n be the length of the string.
2. Compute the number of possible length-k substrings:

n - k + 1

Each such substring corresponds to one generator vector. 3. Observe that all generator vectors are linearly independent over GF(2). 4. Therefore the rank of the generated vector space is:

rank = n - k + 1

5. A vector space of rank r contains exactly:

2^r

distinct vectors. 6. Therefore the number of distinct reachable strings is:

2^(n-k+1)

7. Return this value modulo 10^9 + 7 using fast modular exponentiation.

Why it works

Every operation corresponds to XORing with a fixed length-k interval vector. Since XOR is addition in GF(2), the set of all reachable states forms an affine space generated by these interval vectors. The interval vectors are linearly independent because each one has a unique leftmost position containing a 1 that no later vector can produce. Consequently, the rank is exactly n-k+1. A vector space of rank r contains 2^r elements, giving the final answer 2^(n-k+1).

Python Solution

class Solution:
    def countDistinctStrings(self, s: str, k: int) -> int:
        MOD = 10**9 + 7
        return pow(2, len(s) - k + 1, MOD)

The implementation directly applies the mathematical result derived above.

First, we compute the rank of the interval-flip vector space, which is len(s) - k + 1. Since every independent generator doubles the number of reachable states, the answer is 2^(len(s)-k+1).

Python's built-in pow(base, exponent, mod) performs fast modular exponentiation in logarithmic time, making it ideal for handling very large exponents efficiently.

Go Solution

func countDistinctStrings(s string, k int) int {
	const mod int64 = 1000000007

	exp := len(s) - k + 1
	var result int64 = 1
	var base int64 = 2

	for exp > 0 {
		if exp&1 == 1 {
			result = (result * base) % mod
		}
		base = (base * base) % mod
		exp >>= 1
	}

	return int(result)
}

The Go version implements binary exponentiation manually because Go does not provide a built-in modular exponentiation function equivalent to Python's three-argument pow.

All arithmetic is performed using int64 to avoid overflow during multiplication before taking the modulus.

Worked Examples

Example 1

Input:

s = "1001"
k = 3

Compute:

Variable	Value
n	4
k	3
n-k+1	2

The rank is:

2

The number of reachable strings is:

2^2 = 4

Modulo 10^9+7:

4

Result:

4

This matches the strings listed in the problem statement.

Example 2

Input:

s = "10110"
k = 5

Compute:

Variable	Value
n	5
k	5
n-k+1	1

The rank is:

1

The number of reachable strings is:

2^1 = 2

Result:

2

This corresponds to either performing no operation or flipping the entire string once.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(log(n-k+1))	Fast modular exponentiation
Space	O(1)	Only a few variables are stored

The solution never examines the contents of the string because the answer depends only on its length and on k. The dominant cost is computing 2^(n-k+1) modulo 10^9+7, which requires logarithmic time through binary exponentiation.

Test Cases

sol = Solution()

assert sol.countDistinctStrings("1001", 3) == 4          # example 1
assert sol.countDistinctStrings("10110", 5) == 2         # example 2

assert sol.countDistinctStrings("0", 1) == 2            # minimum size
assert sol.countDistinctStrings("1", 1) == 2            # minimum size, different bit

assert sol.countDistinctStrings("0000", 1) == 16        # every bit independent
assert sol.countDistinctStrings("1111", 1) == 16        # all ones

assert sol.countDistinctStrings("00000", 5) == 2        # whole-string flip only
assert sol.countDistinctStrings("10101", 5) == 2        # whole-string flip only

assert sol.countDistinctStrings("00000", 4) == 4        # rank = 2
assert sol.countDistinctStrings("000000", 3) == 16      # rank = 4

MOD = 10**9 + 7
assert sol.countDistinctStrings("0" * 100000, 100000) == 2  # maximum k
assert sol.countDistinctStrings("0" * 100000, 1) == pow(2, 100000, MOD)  # maximum rank

Test Summary

Test	Why
"1001", 3	Official example 1
"10110", 5	Official example 2
"0", 1	Smallest valid input
"1", 1	Smallest input with different bit value
"0000", 1	Every bit independently flippable
"1111", 1	Same structure with all ones
"00000", 5	Only one possible operation
"10101", 5	Whole-string flip on mixed bits
"00000", 4	Small rank greater than one
"000000", 3	Multiple overlapping intervals
n=100000, k=100000	Maximum size, minimum rank
n=100000, k=1	Maximum size, maximum rank

Edge Cases

Edge Case 1: k = 1

When k = 1, every operation flips exactly one bit. Each position can therefore be toggled independently of all others. The rank becomes:

n - 1 + 1 = n

and the answer is 2^n. A common mistake is trying to model interactions between operations when none exist. The formula naturally handles this case.

Edge Case 2: k = n

When k equals the entire string length, there is only one possible operation, flipping every bit simultaneously. The rank is:

n - n + 1 = 1

so only two strings are reachable: the original string and its complement. The implementation correctly returns 2.

Edge Case 3: Maximum Constraint Values

For n = 100000, any simulation-based approach would be impossible. The solution avoids constructing states entirely and computes a single modular exponentiation. This keeps both memory usage and runtime extremely small even at the maximum input size.

Edge Case 4: String Contents Do Not Matter

It is easy to assume that the pattern of 0s and 1s influences the answer. In reality, all reachable states are obtained by XORing the original string with vectors from the generated subspace. The size of that subspace depends only on n and k, not on the initial bits. The implementation therefore never inspects individual characters of s.