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1922. Count Good Numbers

  • Time: $O(\log n)$
  • Space: $O(\log n)$
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class Solution {
 public:
  int countGoodNumbers(long long n) {
    return modPow(4 * 5, n / 2) * (n % 2 == 0 ? 1 : 5) % kMod;
  }

 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, n - 1) % kMod;
    return modPow(x * x % kMod, n / 2);
  }
};

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class Solution {
  public int countGoodNumbers(long n) {
    return (int) (modPow(4 * 5, n / 2) * (n % 2 == 0 ? 1 : 5) % kMod);
  }

  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 countGoodNumbers(self, n: int) -> int:
    kMod = 1_000_000_007

    def modPow(x: int, n: int) -> int:
      if n == 0:
        return 1
      if n % 2 == 1:
        return x * modPow(x, n - 1) % kMod
      return modPow(x * x % kMod, n // 2)

    return modPow(4 * 5, n // 2) * (1 if n % 2 == 0 else 5) % kMod