package com.fishercoder.solutions;

/**338. Counting Bits
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?

*
*
*/
public class _338 {
    public static class Solution1 {
        //use the most regular method to get it AC'ed first
        public int[] countBits(int num) {
            int[] ones = new int[num + 1];
            for (int i = 0; i <= num; i++) {
                ones[i] = countOnes(i);
            }
            return ones;
        }

        private int countOnes(int i) {
            int ones = 0;
            while (i != 0) {
                ones++;
                i &= (i - 1);
            }
            return ones;
        }
    }

    private class Solution2 {
        /**lixx2100's post is cool:https://discuss.leetcode.com/topic/40162/three-line-java-solution
        An easy recurrence for this problem is f[i] = f[i / 2] + i % 2
        and then we'll use bit manipulation to express the above recursion function
         right shift by 1 means to divide by 2
        AND with 1 means to modulo 2
        this is so cool!*/
        public int[] countBits(int num) {
            int[] ones = new int[num + 1];
            for (int i = 1; i <= num; i++) {
                ones[i] = ones[i >> 1] + (i & 1);
            }
            return ones;
        }
    }
}

@@ -33,841 +33,8 @@
 s;%0A%0A
-/**338. Counting Bits%0AGiven a non negative integer number num. For every numbers i in the range 0 %E2%89%A4 i %E2%89%A4 num calculate the number of 1's in their binary representation and return them as an array.%0A%0AExample:%0AFor num = 5 you should return %5B0,1,1,2,1,2%5D.%0A%0AFollow up:%0A%0AIt 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?%0ASpace complexity should be O(n).%0ACan you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.%0A%0AHint:%0A%0AYou should make use of what you have produced already.%0ADivide the numbers in ranges like %5B2-3%5D, %5B4-7%5D, %5B8-15%5D and so on. And try to generate new range from previous.%0AOr does the odd/even status of the number help you in calculating the number of 1s?%0A%0A*%0A*%0A*/%0A
 publ
@@ -610,16 +610,28 @@
      /**
+%0A         * 
 lixx2100
@@ -719,16 +719,19 @@
 %0A       
+  *
  An easy
@@ -793,16 +793,19 @@
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+  *
  and the
@@ -881,16 +881,18 @@
         
+ *
  right s
@@ -929,16 +929,19 @@
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+  *
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@@ -969,16 +969,19 @@
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+  *
  this is
@@ -989,16 +989,26 @@
 so cool!
+%0A         
 */%0A     

refactor 338