Counting Bits

Problem 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 1s in the binary representation of i.

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 <= 10^5

Follow-up

• Can you solve it in linear time O(n) and possibly in a single pass?
• Can you solve it without using any built-in function (i.e., like __builtin_popcount in C++)?

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Solution for Counting Bits Problem

Approach 1: Brute Force (Naive)

The brute force approach involves iterating through each number from 0 to n, converting each number to its binary form, and counting the number of 1s in that binary representation.

Code in Different Languages

C++

Written by @ImmidiSivani

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;
    }
};

Java

Written by @ImmidiSivani

class Solution {
    public int[] countBits(int n) {
        int[] ans = new int[n + 1];
        for (int i = 0; i <= n; ++i) {
            ans[i] = Integer.bitCount(i);
        }
        return ans;
    }
}

Python

Written by @ImmidiSivani

class Solution:
    def countBits(self, n: int) -> List[int]:
        return [bin(i).count('1') for i in range(n + 1)]

Complexity Analysis

• Time Complexity: $O(nlogn)$, as converting a number to binary and counting bits takes $O(\log n)$ time, and we do this for each number from 0 to n.
• Space Complexity: $O(n)$, for storing the result array.

Approach 2: Optimized Dynamic Programming

To achieve a linear time solution, we can use a dynamic programming approach. We use the property that the number of 1s in i can be derived from the number of 1s in i >> 1 (i.e., i divided by 2) plus 1 if the last bit of i is 1.

Code in Different Languages

C++

Written by @ImmidiSivani

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

Java

Written by @ImmidiSivani

class Solution {
    public int[] countBits(int n) {
        int[] ans = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            ans[i] = ans[i >> 1] + (i & 1);
        }
        return ans;
    }
}

Python

Written by @ImmidiSivani

class Solution:
    def countBits(self, n: int) -> List[int]:
        ans = [0] * (n + 1)
        for i in range(1, n + 1):
            ans[i] = ans[i >> 1] + (i & 1)
        return ans

Complexity Analysis

• Time Complexity: $O(n)$, as we calculate the number of bits for each number from 0 to n in constant time.
• Space Complexity: $O(n)$, for storing the result array.

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