Counting Bits

Problem Statement

In this computational problem, you are tasked with generating a list called ans of size n + 1, where n is a non-negative integer provided as input. Each element ans[i] in this list should represent the count of '1's in the binary representation of the integer i. For example, for any index i, if you convert i to its binary form, the value at ans[i] should be the total number of '1' bits in that binary form.

Examples

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

Approach and Intuition

To tackle this problem, the key is understanding binary representation and bit manipulation. Here's a step-by-step breakdown of how one might approach this:

1. Initialize an Array: Start by creating an array ans of size n + 1 to hold the count of 1's for every number from 0 to n.

2. Iterate Through Numbers: Loop through all numbers from 0 to n. For each number:

   • Convert the number to its binary form or use bit manipulation to count the 1's directly.
   • Store the count of 1's in the respective index of the array ans.

3. Return the Array: Once all numbers are processed, return the ans array.

This method ensures that the problem constraints are adhered to, efficiently counting the number of 1's in the binary representation of each number up to n.

Analysis of Examples:

• In Example 1, with n = 2:

  • Binary of 0 is 0 —> 0 ones.
  • Binary of 1 is 1 —> 1 one.
  • Binary of 2 is 10 —> 1 one.
  • Thus, the output is [0, 1, 1].

• In Example 2, with n = 5:

  • We analyze each number from 0 to 5 and count the 1's as demonstrated for n=2.
  • Resulting output is [0, 1, 1, 2, 1, 2].

This strategy efficiently and correctly solves the problem using fundamental programming concepts and knowledge of binary systems.

Solutions

public class Solution {
    public int[] bitCounting(int maximum) {
        int[] result = new int[maximum + 1];
        for (int i = 1; i <= maximum; ++i) {
            result[i] = result[i & (i - 1)] + 1;
        }
        return result;
    }
}