algorithms/cpp/countingBits/CountingBits.cpp

// Source : https://leetcode.com/problems/counting-bits/
// Author : Hao Chen
// Date   : 2016-05-30

/*************************************************************************************** 
 *
 * 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.
 * 
 *   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?
 * 
 * Credits:Special thanks to @ syedee  for adding this problem and creating all test 
 * cases.
 ***************************************************************************************/

class Solution {
public:
    /* 
     * Initialization
     * 
     *   bits_cnt[0] => 0000 => 0
     *   bits_cnt[1] => 0001 => 1
     * 
     * if the number has 2 bits (2, 3), then we can split the binary to two parts, 
     * 2 = 10 + 0  and  3= 10 + 1, then we can reuse the bits_cnt[0] and bits_cnt[1] 
     *
     *   bits_cnt[2] => 0010 => 0010 + 0 => 1 + bits_cnt[0];
     *   bits_cnt[3] => 0011 => 0010 + 1 => 1 + bits_cnt[1];
     *
     * if the number has 3 bits (4,5,6,7), then we can split the binary to two parts, 
     * 4 = 100 + 0,  5 = 100 + 01, 6= 100 + 10, 7 = 100 +11
     * then we can reuse the bits_cnt[0] and bits_cnt[1] 
     * 
     *   bits_cnt[4] => 0110 => 0100 + 00 => 1 + bits_cnt[0];
     *   bits_cnt[5] => 0101 => 0100 + 01 => 1 + bits_cnt[1];
     *   bits_cnt[6] => 0110 => 0100 + 10 => 1 + bits_cnt[2];
     *   bits_cnt[7] => 0111 => 0100 + 11 => 1 + bits_cnt[3];
     *
     * so, we can have the solution:
     *
     *   bits_cnt[x] = bits_cnt[x & (x-1) ] + 1;
     *
     */
    vector<int> countBits(int num) {
        vector<int> bits_cnt(num+1, 0); 
        
        for (int i=1; i<=num; i++) {
            bits_cnt[i] = bits_cnt[i & (i-1)] + 1;
        }
        return bits_cnt;
    }
};