# 338. Counting Bits

### Problem:

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].

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?

### Solutions:

``````public class Solution {
public int[] countBits(int num) {
int[] result = new int[num + 1];
int[] base = new int[]{0, 1, 1, 2};
if (num <= 3) {
for (int i = 0; i <= num; i ++) {
result[i] = base[i];
}
return result;
}
int diff = 2;
int start = 4;
int j = start - diff;
for (int i = 0; i <= 3; i ++) {
result[i] = base[i];
}
for (int i = 4; i <= num; i ++) {
j ++;
if (j == start) {
j = start - diff;
}
else {
diff = diff << 1;
start = i + 1;
j = start - diff;
}
}
}
return result;
}
}
``````
``````public class Solution {
public int[] countBits(int num) {
int[] result = new int[num + 1];
int pow = 1;
int offset = 1;
for (int i = 1; i <= num; i ++) {
if (i == pow) {
result[i] = 1;
pow = pow << 1;
offset = 1;
}
else {
result[i] = result[offset] + 1;
offset ++;
}
}
return result;
}
}
``````