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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0301_0400.s0338_counting_bits;

// #Easy #Top_100_Liked_Questions #Dynamic_Programming #Bit_Manipulation #Udemy_Bit_Manipulation
// #Big_O_Time_O(num)_Space_O(num) #2022_07_10_Time_2_ms_(86.73%)_Space_48.3_MB_(31.59%)

/**
 * 338 - Counting Bits\.
 *
 * Easy
 *
 * Given an integer `n`, return _an array_ `ans` _of length_ `n + 1` _such that for each_ `i` (`0 <= i <= n`)_,_ `ans[i]` _is the **number of**_ `1`_**'s** in the binary representation of_ `i`.
 *
 * **Example 1:**
 *
 * **Input:** n = 2
 *
 * **Output:** [0,1,1]
 *
 * **Explanation:**
 *
 *     0 --> 0
 *     1 --> 1
 *     2 --> 10 
 *
 * **Example 2:**
 *
 * **Input:** n = 5
 *
 * **Output:** [0,1,1,2,1,2]
 *
 * **Explanation:**
 *
 *     0 --> 0
 *     1 --> 1
 *     2 --> 10
 *     3 --> 11
 *     4 --> 100
 *     5 --> 101 
 *
 * **Constraints:**
 *
 * *   0 <= n <= 10⁵
 *
 * **Follow up:**
 *
 * *   It is very easy to come up with a solution with a runtime of `O(n log n)`. Can you do it in linear time `O(n)` and possibly in a single pass?
 * *   Can you do it without using any built-in function (i.e., like `__builtin_popcount` in C++)?
**/
public class Solution {
    public int[] countBits(int num) {
        int[] result = new int[num + 1];
        int borderPos = 1;
        int incrPos = 1;
        for (int i = 1; i < result.length; i++) {
            // when we reach pow of 2 ,  reset borderPos and incrPos
            if (incrPos == borderPos) {
                result[i] = 1;
                incrPos = 1;
                borderPos = i;
            } else {
                result[i] = 1 + result[incrPos++];
            }
        }
        return result;
    }
}