Counting Bits in Python

Suppose we have a non-negative integer number num. For each number i in the range 0 ≤ i ≤ num we have to calculate the number of 1's in their binary counterpart and return them as a list. So if the number is 5, then the numbers are [0, 1, 2, 3, 4, 5], and number of 1s in these numbers are [0, 1, 1, 2, 1, 2]

To solve this, we will follow these steps −

• res := an array which holds num + 1 number of 0s
• offset := 0
• for i in range 1 to num + 1
  • if i and i – 1 = 0, then res[i] := 1 and offset := 0
  • else increase offset by 1 and res[i] := 1 + res[offset]
• return res

Example(Python)

Let us see the following implementation to get a better understanding −

 Live Demo

class Solution:
   def countBits(self, num):
      result = [0] * (num+1)
      offset = 0
      for i in range(1,num+1):
         if i & i-1 == 0:
            result[i] = 1
            offset = 0
         else:
            offset+=1
            result[i] = 1 + result[offset]
      return result
ob1 = Solution()
print(ob1.countBits(6))

Input

6

Output

[0, 1, 1, 2, 1, 2, 2]

Arnab Chakraborty

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Updated on: 28-Apr-2020

1K+ Views

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