CareerCup

I think the idea is using a method that has an O(1) in finding the number of ones ....
So may be we should have precomputed 256 values in of a 8bits byte and inserted inside a Hash or Lookup Table. then if the array has a size of N, we can do it in O(N).

int count=0
for(byte i : array){
for(int i=0;i<8;i++){
if (i&1<<i==1) count++;
}
}

public static int counteOne(int[] arr){
if(arr.length < 0){
return 0;
}
int counter = 0;
for(int i = 0; i < arr.length; i++){
if(arr[i] ==1){
counter ++;
}
}
return counter;
}

long count_bits(byte n) {
int c; // c accumulates the total bits set in n
for (c = 0; n > 0; c++)
n &= n - 1; // clear the least significant bit set
return c;
}

int count=0
for(byte i : array){
for(int j=0;j<8;++){
if (i>>>j==1) count++;
}
}

int getCountOfSetBits(byte byteArr[]) {
int count = 0;
for (byte b : byteArr) {
while (b != 0) {
b = (byte) (b & (b - 1));
++count;
}
}
return count;
}

byte[] array = {0,0,0,0,1,1,0,1,0,1,1,1,1,0,0,1,0,1,0};
int count = 0;
for(int i=0;i<array.length;i++) {
if(new Byte(array[i]).intValue() == 1) {
count++;
}
}

unsigned char table[256];
table[0] = 0;
for(int i = 0; i < 256; i++) {
table[i] = (i & 1) + table[i/2];
}

int countOne(byte[] arr)
{
int count=0;

for (int i = 0; i < arr.length; i++)
{
byte b = 1;
if(arr[i]==b)
{
count++;

}



}

return count;
}

public static void main(String[] args){
		int count = 0;
		byte[] arr = {3,4,1,5,3,9};
		for(byte single : arr){
			while(single!=0){
				if((single & 1 ) == 1)
					count++;
				single = (byte)((single & 0xff) >>> 1);
			}
		}
			
		System.out.println(count);

}

import java.util.ArrayList;
import java.util.List;

public class ArrayBytes {
public static void main(String[] args) {
List<String> arrayOfBytes = new ArrayList<String>(5);
// sample binary number starting from 10 to 15
// will be initially stored in array-
for (int j = 10; j < 15; j++) {
arrayOfBytes.add(Integer.toBinaryString(j));
}
long count = 0;
for (String string : arrayOfBytes) {
System.out.println(string);

char[] charArray = string.toCharArray();

for (char c : charArray) {
if (c == '1') {
count++;
}
}
}
// total count of 1 in all the elements of the array printed here
System.out.println(count);
}
}

In java i have:

public static int numOfOnes(byte[] byteArr){
	int count = 0;
	for(byte b: byteArr)
		count += countOnes(b);
	return count;
}

private static int countOnes(byte b){
	int count=0;
	for (int i=0; i<8; i++)
		count+= (b>>i)&1;
	return count;
}

This should take up O(1) space and O(n) time complexity.