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2450. Number of Distinct Binary Strings After Applying Operations 👍

  • Time: $O(\log n)$
  • Space: $O(\log n)$
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class Solution {
 public:
  int countDistinctStrings(string s, int k) {
    // Since the content of `s` doesn't matter, for each i in [0, n - k], we can
    // flip s[i..i + k] or don't flip it. Therefore, there's 2^(n - k + 1) ways.
    return modPow(2, s.length() - k + 1);
  }

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

  long modPow(long x, long n) {
    if (n == 0)
      return 1;
    if (n % 2 == 1)
      return x * modPow(x % kMod, (n - 1)) % kMod;
    return modPow(x * x % kMod, (n / 2)) % kMod;
  }
};

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class Solution {
  public int countDistinctStrings(String s, int k) {
    // Since the content of `s` doesn't matter, for each i in [0, n - k], we can
    // flip s[i..i + k] or don't flip it. Therefore, there's 2^(n - k + 1) ways.
    return (int) modPow(2, s.length() - k + 1);
  }

  private static final int kMod = 1_000_000_007;

  private long modPow(long x, long n) {
    if (n == 0)
      return 1;
    if (n % 2 == 1)
      return x * modPow(x, n - 1) % kMod;
    return modPow(x * x % kMod, n / 2);
  }
}

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class Solution:
  def countDistinctStrings(self, s: str, k: int) -> int:
    # Since the content of `s` doesn't matter, for each i in [0, n - k], we can
    # flip s[i..i + k] or don't flip it. Therefore, there's 2^(n - k + 1) ways.
    return pow(2, len(s) - k + 1, 1_000_000_007)