Counting Bits
Recognize the pattern
Brute force idea
If you approach Counting Bits in the most literal way possible, you get this: For each number, count its bits individually. Each number is processed independently, ignoring the relationship between consecutive numbers. It is a fair place to begin because it matches the surface of the question, yet it does not capture the deeper structure that makes the problem simpler.
Better approach
The real unlock in Counting Bits comes when you notice this: dp[i] = dp[i >> 1] + (i & 1). Each number's bit count equals its right-shifted version's count plus its last bit. This works because right-shifting removes the last bit, and we already know the answer for the smaller number. Instead of recomputing the world every time, you preserve just enough context to let the next decision become obvious.
Key invariant
The compass for Counting Bits is this: Every number's bit count can be derived from a smaller number's bit count plus 0 or 1. The recurrence dp[i] = dp[i/2] + (i % 2) captures this — we never need to count bits from scratch. As long as that statement keeps holding, you can trust the steps built on top of it.
Watch out for
The trap in Counting Bits usually looks like this: Not seeing the DP relationship — trying to count bits for each number independently. The key insight is that i >> 1 is always a number we've already computed. When the code becomes mechanical before the idea is clear, small edge cases start breaking the whole story.