Write an efficient program to count number of 1s in binary representation of an integer.
Input : n = 6 Output : 2 Binary representation of 6 is 110 and has 2 set bits Input : n = 13 Output : 3 Binary representation of 11 is 1101 and has 3 set bits
In the previous post we had seen different method that solved this problem in O(log n) time. In this post we solve in O(1) using lookup table. Here we assume that the size of INT is 32-bits. It’s hard to count all 32 bits in one go using lookup table (” because it’s infeasible to create lookup table of size 232-1 “). So we break 32 bits into 8 bits of chunks( ow lookup table of size (28-1 ) index : 0-255 ).
In lookup tale, we store count of set_bit of every
number that are in a range (0-255)
LookupTable = 0 | binary 00000000 CountSetBits 0
LookupTable = 1 | binary 00000001 CountSetBits 1
LookupTable = 1 | binary 00000010 CountSetBits 1
LookupTanle = 2 | binary 00000011 CountSetBits 2
LookupTable = 1 | binary 00000100 CountSetBits 1
and so…on upto LookupTable.
Let’s take an Example How lookup table work.
Let's number be : 354 in Binary : 0000000000000000000000101100010 Split it into 8 bits chunks : In Binary : 00000000 | 00000000 | 00000001 | 01100010 In decimal : 0 0 1 98 Now Count Set_bits using LookupTable LookupTable = 0 LookupTable = 1 LookupTable = 3 so Total bits count : 4
Time Complexity : O(1)
- Count trailing zero bits using lookup table
- Reverse bits using lookup table in O(1) time
- Count set bits in an integer
- Check if bits of a number has count of consecutive set bits in increasing order
- How to swap two bits in a given integer?
- Next greater integer having one more number of set bits
- Previous smaller integer having one less number of set bits
- Count set bits in a range
- Range query for count of set bits
- Count total set bits in all numbers from 1 to n | Set 2
- Count total bits in a number
- Count unset bits in a range
- Count Set-bits of number using Recursion
- Count total set bits in all numbers from 1 to n
- Count unset bits of a number
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