Leetcode 338. Counting Bits (Medium) (cpp)

Leetcode 338. Counting Bits (Medium) (cpp)

Tag: Dynamic Programming

Difficulty: Medium

/*338. Counting Bits (Medium)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:For num = 5 you should return [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.Hint:You should make use of what you have produced already.Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.Or does the odd/even status of the number help you in calculating the number of 1s?*/class Solution {public:vector<int> countBits(int num) {vector<int> table(num + 1);for (int i = 1; i <= num; i++)table[i] = 1 + table[i & (i - 1)];return table;}};