Welcome to Subscribe On Youtube

Question

Formatted question description: https://leetcode.ca/all/338.html

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++)?

Algorithm

DP: countAtNum[i] = countAtNum[i>>1] + (i&1);

Code

• Java
• C++
• Python
• Go

• 
  public class Counting_Bits {
      class Solution {
          public int[] countBits(int num) {
              int[] countAtNum = new int[num + 1];
  
              if (num <= 0) {
                  return countAtNum;
              }
  
              countAtNum[0] = 0; // to be explicit
              countAtNum[1] = 1;
  
              for (int i = 2; i <= num; i++) {
                  // @note @memorize : & operator lower than +, so should be in ()
                  countAtNum[i] = countAtNum[i>>1] + (i&1);
              }
  
              return countAtNum;
          }
      }
  }
  
  ############
  
  class Solution {
      public int[] countBits(int n) {
          int[] ans = new int[n + 1];
          for (int i = 1; i <= n; ++i) {
              ans[i] = ans[i & (i - 1)] + 1;
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/counting-bits
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      vector<int> countBits(int n) {
          vector<int> ans(n + 1);
          for (int i = 1; i <= n; i *= 2) {
              ans[i] = 1;
              for (int j = 1; j < i && i + j <= n; ++j) ans[i + j] = ans[i] + ans[j];
          }
          return ans;
      }
  };
  

• class Solution:
      def countBits(self, n: int) -> List[int]:
          ans = [0] * (n + 1)
          for i in range(1, n + 1):
              ans[i] = ans[i & (i - 1)] + 1
          return ans
  
  ############
  
  class Solution(object):
    def countBits(self, num):
      """
      :type num: int
      :rtype: List[int]
      """
      if num == 0:
        return [0]
      ans = [0, 1]
      j = 0
      for i in range(2, num + 1):
        ans.append(ans[i & (i - 1)] + 1)
      return ans
  
  

• func countBits(n int) []int {
  	ans := make([]int, n+1)
  	for i := 1; i <= n; i++ {
  		ans[i] = ans[i&(i-1)] + 1
  	}
  	return ans
  }
  

All Problems

All Solutions