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338. Counting Bits

Description

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++)?

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• class Solution {
  public:
      vector<int> countBits(int n) {
          vector<int> ans(n + 1);
          for (int i = 0; i <= n; ++i) {
              ans[i] = __builtin_popcount(i);
          }
          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
  
  

• 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
  }
  

• function countBits(n: number): number[] {
      const ans: number[] = Array(n + 1).fill(0);
      for (let i = 1; i <= n; ++i) {
          ans[i] = ans[i & (i - 1)] + 1;
      }
      return ans;
  }
  
  

All Problems

All Solutions