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1680. Concatenation of Consecutive Binary Numbers

Approach 1: Naive

  • Time: $O(n\log n)$
  • Space: $O(1)$
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class Solution {
 public:
  int concatenatedBinary(int n) {
    constexpr int kMod = 1'000'000'007;
    long ans = 0;

    for (int i = 1; i <= n; ++i)
      ans = ((ans << numberOfBits(i)) % kMod + i) % kMod;

    return ans;
  }

 private:
  int numberOfBits(int n) {
    return log2(n) + 1;
  }
};

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class Solution {
  public int concatenatedBinary(int n) {
    final int kMod = 1_000_000_007;
    long ans = 0;

    for (int i = 1; i <= n; ++i)
      ans = ((ans << numberOfBits(i)) % kMod + i) % kMod;

    return (int) ans;
  }

  private int numberOfBits(int n) {
    return (int) (Math.log(n) / Math.log(2)) + 1;
  }
}

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class Solution:
  def concatenatedBinary(self, n: int) -> int:
    kMod = 1_000_000_007
    ans = 0

    def numberOfBits(n: int) -> int:
      return int(math.log2(n)) + 1

    for i in range(1, n + 1):
      ans = ((ans << numberOfBits(i)) + i) % kMod

    return ans

Approach 2: Increase length gradually

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int concatenatedBinary(int n) {
    constexpr int kMod = 1'000'000'007;
    long ans = 0;
    int numberOfBits = 0;

    for (unsigned i = 1; i <= n; ++i) {
      if (popcount(i) == 1)
        ++numberOfBits;
      ans = ((ans << numberOfBits) % kMod + i) % kMod;
    }

    return ans;
  }
};

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class Solution {
  public int concatenatedBinary(int n) {
    final int kMod = 1_000_000_007;
    long ans = 0;
    int numberOfBits = 0;

    for (int i = 1; i <= n; ++i) {
      if (Integer.bitCount(i) == 1)
        ++numberOfBits;
      ans = ((ans << numberOfBits) % kMod + i) % kMod;
    }

    return (int) ans;
  }
}

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class Solution:
  def concatenatedBinary(self, n: int) -> int:
    kMod = 1_000_000_007
    ans = 0
    numberOfBits = 0

    for i in range(1, n + 1):
      if i.bit_count() == 1:
        numberOfBits += 1
      ans = ((ans << numberOfBits) + i) % kMod

    return ans