Skip to content

2450. Number of Distinct Binary Strings After Applying Operations 👍

  • Time: $O(\log n)$
  • Space: $O(\log n)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
class Solution {
 public:
  int countDistinctStrings(string s, int k) {
    // Since content of `s` doesn't matter, for each i in [0, n - k], we can
    // flip s[i..i + k] or not flip it. Therefore, there's 2^(n - k + 1) ways.
    return myPow(2, s.length() - k + 1);
  }

 private:
  constexpr static int kMod = 1'000'000'007;

  int myPow(long x, int n) {
    if (n == 0)
      return 1;
    if (n & 1)
      return x * myPow(x, n - 1) % kMod;
    return myPow(x * x % kMod, n / 2);
  }
};
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution {
  public int countDistinctStrings(String s, int k) {
    // Since content of `s` doesn't matter, for each i in [0, n - k], we can
    // flip s[i..i + k] or not flip it. Therefore, there's 2^(n - k + 1) ways.
    return (int) myPow(2, s.length() - k + 1);
  }

  private final static int kMod = 1_000_000_007;

  private long myPow(long x, int n) {
    if (n == 0)
      return 1;
    if (n % 2 == 1)
      return x * myPow(x, n - 1) % kMod;
    return myPow(x * x % kMod, n / 2);
  }
}
1
2
3
4
5
class Solution:
  def countDistinctStrings(self, s: str, k: int) -> int:
    # Since content of `s` doesn't matter, for each i in [0, n - k], we can
    # flip s[i..i + k] or not flip it. Therefore, there's 2^(n - k + 1) ways.
    return pow(2, len(s) - k + 1, 1_000_000_007)