Counting Bits
Given an integer $n$, return an array `ans` of length $n+1$ such that for each $i$ ($0 \leq i \leq n$), `ans[i]` is the number of 1's in the binary representation of $i$. Solve in $O(n)$ time using DP/Bit Manipulation.
Why Interviewers Ask This
Amazon asks this question to evaluate a candidate's ability to optimize solutions beyond brute force, specifically testing their grasp of dynamic programming and bit manipulation patterns. They want to see if you can recognize that the number of set bits in an integer relates to smaller subproblems, demonstrating the 'Invent and Simplify' leadership principle by finding efficient algorithms.
How to Answer This Question
1. Clarify requirements: Confirm input constraints (n) and expected output format. 2. Discuss naive approaches first: Briefly mention iterating through each number and counting bits, noting the O(n log n) complexity. 3. Identify the pattern: Propose the key insight that i >> 1 is i shifted right, and i & 1 checks the last bit. 4. Define the recurrence: Explain that ans[i] = ans[i >> 1] + (i & 1). 5. Implement DP: Initialize array of size n+1, loop from 1 to n, apply formula. 6. Analyze: Confirm O(n) time and O(n) space. 7. Verify: Walk through a small example like n=5 to prove correctness.
Key Points to Cover
- Recognize the overlap between subproblems where ans[i] depends on ans[i >> 1]
- Explicitly state the recurrence relation: ans[i] = ans[i >> 1] + (i & 1)
- Demonstrate understanding of bitwise operators >> (shift) and & (AND)
- Prove the time complexity is strictly O(n) rather than O(n log n)
- Handle the base case ans[0] = 0 correctly before the loop
Sample Answer
To solve the Counting Bits problem efficiently for Amazon, I will use a dynamic programming approach leveraging bit manipulation. First, I acknowledge that a naive solution checking every bit for every number would be O(n log n), which isn't optimal. The goal is O(n).
The core insight relies on the relationship between a number and its half. If we look at any integer i, shifting it right by one bit (i >> 1) removes the least significant bit. The number of set bits in i is simply the count of set bits in (i >> 1) plus the value of that removed least significant bit. Mathematically, this is expressed as: ans[i] = ans[i >> 1] + (i & 1).
I initialize an array of size n+1 with zeros. Since the base case ans[0] is 0, I iterate from 1 to n. For each i, I calculate the result using the recurrence relation derived above. This ensures every calculation is O(1), leading to a total time complexity of O(n). Space complexity is also O(n) for the output array.
For example, if n=5: ans[1] uses ans[0]+1=1; ans[2] uses ans[1]+0=1; ans[3] uses ans[1]+1=2; ans[4] uses ans[2]+0=1; ans[5] uses ans[2]+1=2. This method is clean, efficient, and aligns well with Amazon's focus on scalable, optimized code.
Common Mistakes to Avoid
- Implementing a nested loop to count bits for each number individually, resulting in O(n log n) time
- Failing to initialize the output array with the correct size (n+1) instead of n
- Overlooking the edge case where n=0 or negative inputs are passed
- Not explaining the mathematical intuition behind why shifting works before coding
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