LeetCode Solutions

868. Binary Gap

Time: $O(\log n)$

Space: $O(1)$

			

class Solution {
 public:
  int binaryGap(int n) {
    int ans = 0;

    // D := distance between any two 1's
    // Initialized to a reasonable small value
    for (int d = -32; n; n /= 2, ++d)
      if (n & 1) {
        ans = max(ans, d);
        d = 0;
      }

    return ans;
  }
};
			

class Solution {
  public int binaryGap(int n) {
    int ans = 0;

    // D := distance between any two 1's
    // Initialized to a reasonable small value
    for (int d = -32; n > 0; n /= 2, ++d)
      if ((n & 1) == 1) {
        ans = Math.max(ans, d);
        d = 0;
      }

    return ans;
  }
}
			

class Solution:
  def binaryGap(self, n: int) -> int:
    ans = 0
    d = -32  # Distance between any two 1's, initialized to a reasonable small value

    while n:
      if n & 1:
        ans = max(ans, d)
        d = 0
      n //= 2
      d += 1

    return ans