LeetCode-in-All

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++)?

Solution

(define (near-bit n i)
    (if (and (<= i n) (< n (* i 2)))
        i
        (near-bit n (* i 2))))

(define/contract (count-bits n)
  (-> exact-integer? (listof exact-integer?))
  (match n
      [0 '(0)]
      [1 '(0 1)]
      [else 
        (let* ([b (near-bit n 1)]
               [prev (count-bits (- b 1))]
               [next (map (lambda (n) (+ n 1)) prev)]) 
               (take (append prev next) (+ n 1)))]
      ))

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