Mathworks

2020-05-28

## 338. Counting Bits

### Question:

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array.

#### Example 1:

``````Input: 2
Output: [0,1,1]
``````

#### Example 2:

``````Input: 5
Output: [0,1,1,2,1,2]
``````

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

### Solution:

Using DP to find the previous number and add back to the current one. There are two cases: if the number is odd, then its number of 1 equals to `[i/2] + 1`. If i is even, the number of 1 equals to `[i/2]`.

Using bit manipulation, since we know the result for previous number, the current one is remove the last 1 digit in binary and add it back. Similar idea at LeetCode Question 191.

More reference at CSDN.

Using DP:

``````class Solution {
public int[] countBits(int num) {
int[] dp = new int[num + 1];
for (int i = 1; i <= num; i++) {
dp[i] = i % 2 == 0 ? dp[i/2] : dp[i/2] + 1;
}
return dp;
}
}
``````
``````class Solution {
public int[] countBits(int num) {
int[] dp = new int[num + 1];
for (int i = 1; i <= num; i++) {
dp[i] = dp[i >> 1] + (i & 1);
}
return dp;
}
}
``````

Using Bit Manipulation:

``````class Solution {
public int[] countBits(int num) {
int[] dp = new int[num + 1];
for (int i = 1; i <= num; i++) {
// Remove the last bits that is 1, get the number of one and add back
dp[i] = dp[i & (i-1)] + 1;
}
return dp;
}
}
``````