[LeetCode]Counting Bits
Problem Statement
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array.
Example 1:
Input: 2
Output: [0,1,1]Example 2:
Input: 5
Output: [0,1,1,2,1,2]Follow up:
- It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
- Space complexity should be O(n).
- Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
Solution
This is one of the examples that involves a lot of analysis but little coding. Once you find out the rule/formula, the coding is straightforward and rudimentary.
How can we compute the number of 1s for a number j using the results for any number i where i<j?
- Base case: for number 0 (binary representation is 0) the number of 1s is zero; for number 1 (binary representation is 1) the number of 1s is one.
- For number 2 (binary representation is 10) and 3 (binary representation is 11), they have a leading 1 at the second position from right. We can generalize the formula as: the number of 1s for number 2 is…
