03 Counting Bits — Interview English Script (C++)
Source aligned with: docs/18_Bit_Manipulation/03_Counting_Bits.md
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1) 30-second problem restatement script
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me restate Counting Bits. | 我先重述 Counting Bits。 | Restatement |
| Given n, we return array ans of length n plus one. | 給定 n，要回傳長度 n+1 的陣列 ans。 | Restatement |
| ans[i] is number of one bits in i for all zero to n. | ans[i] 代表 i 的 1 位元數，i 從 0 到 n。 | Restatement |
| We should do better than calling popcount for each number. | 不能每個數都獨立 popcount 那麼慢。 | Restatement |
| I will use dynamic programming relation on bits. | 我會用位元關係做動態規劃。 | Restatement |
| Formula is dp[i] equals dp[i shifted right one] plus low bit. | 公式是 dp[i]=dp[i>>1]+(i&1)。 | Restatement |
2) Clarifying questions (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Is n non-negative and reasonably bounded? | n 是否為非負且在合理範圍？ | Clarify |
| Do you expect O(n) time overall? | 是否預期整體 O(n) 時間？ | Clarify |
| Is O(n) output array space acceptable? | O(n) 輸出陣列空間是否可接受？ | Clarify |
| Can I use either shift formula or offset formula? | 我可用 shift 公式或 offset 公式嗎？ | Clarify |
| Should ans[0] explicitly be zero? | ans[0] 是否要明確設為 0？ | Clarify |
3) Approach discussion
Brute force (3 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Brute force runs hammingWeight for each i from zero to n. | 暴力法是對 0..n 每個 i 都跑 hammingWeight。 | Approach |
| That leads to about O(n log n) bit work in practice. | 實務位元工作量約 O(n log n)。 | Approach |
| We can do linear DP by reusing previous answers. | 我們可重用前面結果做到線性 DP。 | Approach |
Optimized approach (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Create dp array size n plus one with dp[0] equals zero. | 建立 dp 長度 n+1，dp[0]=0。 | Approach |
| For i from one to n, compute dp[i] from smaller index. | i 從 1 到 n，由較小索引推得 dp[i]。 | Approach |
| Right shift removes least significant bit, so dp[i>>1] already known. | 右移會去掉最低位，因此 dp[i>>1] 已知。 | Approach |
| Add i and one to include removed low bit when it is one. | 再加上 i&1 補回被去掉的最低位。 | Approach |
| This gives O(n) time and O(n) space. | 這樣可達 O(n) 時間與 O(n) 空間。 | Approach |
4) Coding-and-speaking script (line-by-line, in coding order)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I initialize vector dp with n plus one zeros. | 我初始化長度 n+1 的 dp 全 0。 | Coding |
| dp[0] stays zero by definition. | dp[0] 依定義就是 0。 | Coding |
| I loop i from one to n. | 我把 i 從 1 迴圈到 n。 | Coding |
| Compute half as i shifted right one. | 先算 half=i>>1。 | Coding |
| Compute low as i and one. | 再算 low=i&1。 | Coding |
| Set dp[i] to dp[half] plus low. | 設 dp[i]=dp[half]+low。 | Coding |
| After loop, return dp. | 迴圈後回傳 dp。 | Coding |
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me dry-run n equals five. | 我用 n=5 手跑。 | Dry-run |
| dp[0] is zero. | dp[0]=0。 | Dry-run |
| i one gives dp[1]=dp[0]+1 equals one. | i=1，dp[1]=dp[0]+1=1。 | Dry-run |
| i two gives dp[2]=dp[1]+0 equals one. | i=2，dp[2]=dp[1]+0=1。 | Dry-run |
| i three gives dp[3]=dp[1]+1 equals two. | i=3，dp[3]=dp[1]+1=2。 | Dry-run |
| i four gives dp[4]=dp[2]+0 equals one. | i=4，dp[4]=dp[2]+0=1。 | Dry-run |
| i five gives dp[5]=dp[2]+1 equals two. | i=5，dp[5]=dp[2]+1=2。 | Dry-run |
| Final array is [0,1,1,2,1,2]. | 最終陣列是 [0,1,1,2,1,2]。 | Dry-run |
6) Edge/corner test script (at least 4 cases)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Case one: n equals zero. | 案例一：n=0。 | Edge test |
| Case two: n equals one. | 案例二：n=1。 | Edge test |
| Case three: powers of two boundaries like 7 to 8 transition. | 案例三：2 的冪邊界如 7 到 8 轉換。 | Edge test |
| Case four: larger n for performance sanity. | 案例四：較大 n 的效能檢查。 | Edge test |
| Case five: random spot checks against popcount. | 案例五：隨機點位與 popcount 對照。 | Edge test |
7) Complexity script
Short version (2 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Time complexity is O(n). | 時間複雜度是 O(n)。 | Complexity |
| Space complexity is O(n) for output array. | 空間複雜度是 O(n)（輸出陣列）。 | Complexity |
Full version (4 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| We compute each dp[i] once from O(1) operations. | 每個 dp[i] 只計算一次且是 O(1) 操作。 | Complexity |
| There are n values from one to n. | 需計算的值有 n 個（1 到 n）。 | Complexity |
| So runtime is linear O(n). | 因此時間是線性 O(n)。 | Complexity |
| DP array of size n plus one dominates memory usage. | 記憶體主要是長度 n+1 的 DP 陣列。 | Complexity |
8) If stuck rescue lines (10 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me derive relation from binary right shift. | 我先從二進位右移推導關係式。 | If stuck |
| i shifted right one removes the last bit. | i>>1 會移除最後一位。 | If stuck |
| The removed bit is exactly i and one. | 被移除那位正是 i&1。 | If stuck |
| So count(i) equals count(i>>1) plus i&1. | 所以 count(i)=count(i>>1)+(i&1)。 | If stuck |
| This means we can fill dp left to right. | 代表可由左到右填 dp。 | If stuck |
| Base case dp[0] is zero. | 基底 dp[0]=0。 | If stuck |
| Let me check i equals five quickly. | 我快速驗證 i=5。 | If stuck |
| five is one-zero-one so answer should be two. | 5 是 101，所以答案應是 2。 | If stuck |
| Formula gives dp[2]+1 which is two. | 公式給 dp[2]+1=2。 | If stuck |
| Great, recurrence is confirmed. | 很好，遞推式已確認。 | If stuck |
9) Final wrap-up lines (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I solved it with linear DP on bit relation. | 我用位元關係的線性 DP 解題。 | Wrap-up |
| Recurrence dp[i]=dp[i>>1]+(i&1) is the key. | 關鍵遞推是 dp[i]=dp[i>>1]+(i&1)。 | Wrap-up |
| It avoids repeated full popcount calls. | 這可避免重複完整 popcount。 | Wrap-up |
| Complexity is O(n) time and O(n) space. | 複雜度是 O(n) 時間、O(n) 空間。 | Wrap-up |
| This is the standard accepted interview approach. | 這是面試常見標準解法。 | Wrap-up |
10) Ultra-short cheat sheet (20 lines total)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Goal: output bit counts for all values zero..n. | 目標：輸出 0..n 每個數的 1 位元數。 | Cheat sheet |
| Use DP array size n+1. | 用長度 n+1 的 DP 陣列。 | Cheat sheet |
| Base dp[0]=0. | 基底 dp[0]=0。 | Cheat sheet |
| For i in 1..n. | i 從 1 到 n。 | Cheat sheet |
| dp[i]=dp[i>>1]+(i&1). | dp[i]=dp[i>>1]+(i&1)。 | Cheat sheet |
| i>>1 drops last bit. | i>>1 會去掉末位。 | Cheat sheet |
| i&1 tells if dropped bit is one. | i&1 告訴末位是否為 1。 | Cheat sheet |
| Return dp. | 回傳 dp。 | Cheat sheet |
| Example n=2 => [0,1,1]. | 範例 n=2 => [0,1,1]。 | Cheat sheet |
| Example n=5 => [0,1,1,2,1,2]. | 範例 n=5 => [0,1,1,2,1,2]。 | Cheat sheet |
| Time O(n). | 時間 O(n)。 | Cheat sheet |
| Space O(n). | 空間 O(n)。 | Cheat sheet |
| Common bug: wrong loop start at 0 causing self-reference confusion. | 常見錯誤：從 0 開始導致遞推混亂。 | Cheat sheet |
| Common bug: forgetting parentheses in bit ops. | 常見錯誤：位運算括號漏寫。 | Cheat sheet |
| Offset formula is equivalent alternative. | offset 公式是等價替代。 | Cheat sheet |
| Validate power-of-two boundaries. | 驗證 2 的冪邊界。 | Cheat sheet |
| Output itself costs O(n) memory. | 輸出本身就需 O(n) 記憶體。 | Cheat sheet |
| Keep explanation recurrence-first. | 解釋時先講遞推最清楚。 | Cheat sheet |
| No heavy bit tricks needed beyond shift and and. | 只需 shift 與 and，無需複雜技巧。 | Cheat sheet |
| Clean and interview-friendly solution. | 解法乾淨且面試友善。 | Cheat sheet |
Quality check
- Consistency with source solution: ✅ DP with shift/LSB relation.
- Constraint alignment: ✅ O(n) target achieved.
- Language simplicity: ✅ Structured spoken format.