Codility 1.1 Iterations BinaryGap

Problem

A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.

For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps.

Write a function:

int solution(int N);

that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn’t contain a binary gap.

For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5.

Assume that:

• N is an integer within the range [1..2,147,483,647].

Complexity:

• expected worst-case time complexity is O(log(N));
• expected worst-case space complexity is O(1).

Solving Process

Every time the digit is 1, bin_counter will be updated and if it is bigger than the previous biggest one, then update it to bin_max. At last, return the bin_max.

Imagine a child recording a jump of a frog(0) without a stop(1). This is as easy as that as long as we know how to process a number to binary number, which is str(bin(N))[2:].

Code Solution(mine)

def solution(N):     
     bin_rep = str(bin(N))[2:]
     bin_gap = False
     bin_max = 0
     bin_counter = 0
     for symbol in bin_rep:
         if symbol == '1':
             if bin_max < bin_counter:
                 bin_max = bin_counter
             bin_gap = True
             bin_counter = 0
         elif bin_gap:
             bin_counter += 1
     return bin_max

Big-O Calculation

def solution(N):     
     bin_rep = str(bin(N))[2:]   # Constant 1
     bin_gap = False             
     bin_max = 0                 
     bin_counter = 0             
     for symbol in bin_rep:      # Constant N
         if symbol == '1':
             if bin_max < bin_counter:
                 bin_max = bin_counter
             bin_gap = True
             bin_counter = 0
         elif bin_gap:
             bin_counter += 1
     return bin_max
## O(N)

Other People’s code solution

def solution(N):
  return len(max(bin(N)[2:].strip('0').strip('1').split('1')))
# O(N)

Learning Points

• strip(str) and split(str) functions can be used for the string manipulation.