# Find 1’s how many times repeated in bits of given number

Problem:
Given a non-negative integer number num. For every number 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].

1. Native Solution:
We can simply count the bits for each number like below,

``````package com.algorithm.bitcount;

public class NativeSolution {

public static void main(String[] args) {
for (int i : countBits(5)) {
System.out.println(i);
}
}

public static int[] countBits(int num) {
int[] result = new int[num + 1];
for (int i = 0; i <= num; i++) {
result[i] = countEach(i);
}
return result;
}

public static int countEach(int num) {
int result = 0;
while (num != 0) {
if (num % 2 == 1) {
result++;
}
num = num / 2;
}
return result;
}
}``````

2. Improved Solution:
For number 2(10), 4(100),6(1000),8(10000), etc the number of 1’s is 1. Any other can be converted to be 2m+x. For example, 9=8+1,10=8+2. The number of 1’s for any other number is 1+ # of 1’s in x.

``````package com.algorithm.bitcount;

public class ImprovedSolution {

public static void main(String[] args) {
for (int i : countBits(5)) {
System.out.println(i);
}
}

public static int[] countBits(int num) {
int[] result = new int[num + 1];
int p = 1; // p tracks the index for number x
int pow = 1;
for (int i = 1; i <= num; i++) {
if (i == pow) {
result[i] = 1;
pow <<= 1;
p = 1;
} else {
result[i] = result[p] + 1;
p++;
}
}
return result;
}
}``````