# 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]

Updated on: 28-Apr-2020

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