07 Palindrome Partitioning — Interview English Script (C++)
Source aligned with: docs/10_Backtracking/07_Palindrome_Partitioning.md
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1) 30-second problem restatement script
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me restate the palindrome partitioning problem. | 我先重述回文切割題。 | Restatement |
| We are given a string s. | 題目給一個字串 s。 | Restatement |
| We must split it into substrings. | 要把它切成多段子字串。 | Restatement |
| Every substring in a partition must be a palindrome. | 每一段都必須是回文。 | Restatement |
| We need to return all valid partition lists. | 要回傳所有合法切割方案。 | Restatement |
| I will use backtracking with palindrome checking on each cut. | 我會用回溯搭配每段回文檢查。 | Restatement |
2) Clarifying questions (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Is output order of partition lists irrelevant? | 切割結果順序是否不重要？ | Clarify |
| Should each single character be treated as palindrome by default? | 單字元是否視為回文？ | Clarify |
| Can I use two-pointer palindrome check each time? | 每次可用雙指標檢查回文嗎？ | Clarify |
| Is precomputed DP palindrome table optional discussion? | 可否把 DP 回文表當可選優化討論？ | Clarify |
| Are we returning all partitions, not just count? | 我們是回傳全部方案，不是只回數量嗎？ | Clarify |
3) Approach discussion
Brute force (3 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Brute force tries all cut positions of the string. | 暴力法嘗試所有切割位置組合。 | Approach |
| Then it filters partitions where every piece is palindrome. | 再篩選每段都回文的方案。 | Approach |
| Backtracking with early check is more practical. | 改用回溯加早檢查更實用。 | Approach |
Optimized approach (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| DFS state is start index of the next segment. | DFS 狀態是下一段的起始索引。 | Approach |
| For each end index i, test substring s[start..i] palindrome. | 對每個終點 i，檢查 s[start..i] 是否回文。 | Approach |
| If palindrome, push segment and recurse from i plus one. | 若為回文就 push 該段並遞迴到 i+1。 | Approach |
| When start reaches string length, record one partition. | 當 start 到字串尾端就記錄一組解。 | Approach |
| Backtrack by popping last segment after recursion. | 遞迴後 pop 最後一段完成回溯。 | Approach |
4) Coding-and-speaking script (line-by-line, in coding order)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I initialize result list and current partition path. | 我先初始化結果與當前切割路徑。 | Coding |
| I write dfs(start) recursion. | 我撰寫 dfs(start) 遞迴。 | Coding |
| Base case is start equals s length. | 基底條件是 start 等於字串長度。 | Coding |
| At base I append a copy of current partition path. | 基底時加入路徑拷貝。 | Coding |
| I loop end index from start to s length minus one. | end 從 start 迭代到字尾。 | Coding |
| I call isPalindrome on s from start to end. | 我呼叫 isPalindrome(start,end)。 | Coding |
| If true, I push substring, recurse, then pop. | 若為真就 push 子字串、遞迴、再 pop。 | Coding |
| This explores all valid palindrome partitions. | 這會遍歷所有合法回文切割。 | Coding |
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me dry-run s equals aab. | 我手跑 s="aab"。 | Dry-run |
| Start zero, choose a then recurse from one. | 從 0 開始，先選 a，再遞迴到 1。 | Dry-run |
| From one, choose a then b, giving partition [a,a,b]. | 在 1 再選 a，再選 b，得到 [a,a,b]。 | Dry-run |
| Backtrack to start zero, now try aa as first segment. | 回溯到 0，改試 aa 當第一段。 | Dry-run |
| Remaining b is palindrome, giving [aa,b]. | 剩餘 b 也是回文，得到 [aa,b]。 | Dry-run |
| Segment aab is not palindrome, so that branch stops. | aab 不是回文，該分支停止。 | Dry-run |
| Final outputs are [a,a,b] and [aa,b]. | 最終輸出是 [a,a,b] 與 [aa,b]。 | Dry-run |
6) Edge/corner test script (at least 4 cases)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Case one: single character string. | 案例一：單字元字串。 | Edge test |
| Case two: all same letters like aaaa with many partitions. | 案例二：全同字元如 aaaa，分割很多。 | Edge test |
| Case three: no multi-char palindrome like abc. | 案例三：無多字元回文如 abc。 | Edge test |
| Case four: whole string itself is palindrome. | 案例四：整串本身就是回文。 | Edge test |
| Case five: mixed pattern with overlapping palindrome choices. | 案例五：有重疊回文選擇的混合字串。 | Edge test |
7) Complexity script
Short version (2 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Worst-case time is O(n times two to the n). | 最壞時間是 O(n*2^n)。 | Complexity |
| Recursion stack space is O(n), excluding output. | 不含輸出時遞迴空間 O(n)。 | Complexity |
Full version (4 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Number of partition choices is exponential in string length. | 切割選擇數對字串長度呈指數。 | Complexity |
| For each candidate segment we run palindrome check. | 每個候選切段都要做回文檢查。 | Complexity |
| This leads to roughly O(n times two to the n) worst-case runtime. | 因此最壞約為 O(n*2^n)。 | Complexity |
| Stack depth is at most n, while output storage can dominate. | 堆疊深度最多 n，輸出儲存可能主導。 | Complexity |
8) If stuck rescue lines (10 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me focus on where to cut next, not the full global plan. | 我先專注下一刀切哪裡，不先想全局。 | If stuck |
| DFS state is just current start index. | DFS 狀態只需當前 start 索引。 | If stuck |
| I try every end index from start onward. | 我嘗試所有從 start 開始的 end。 | If stuck |
| Only palindrome segments are allowed to branch deeper. | 只有回文片段才可繼續分支。 | If stuck |
| So palindrome check is the key gatekeeper. | 因此回文檢查是關鍵閘門。 | If stuck |
| If start reaches n, one valid partition is complete. | 若 start 到 n，就完成一組合法切割。 | If stuck |
| Push before recurse and pop after recurse. | 遞迴前 push，遞迴後 pop。 | If stuck |
| Let me test quickly with aab. | 我快速用 aab 測試。 | If stuck |
| I obtain [a,a,b] and [aa,b], exactly expected. | 我得到 [a,a,b] 與 [aa,b]，完全符合。 | If stuck |
| Great, recursion flow and palindrome gate are correct. | 很好，遞迴流程與回文閘門正確。 | If stuck |
9) Final wrap-up lines (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I solved palindrome partitioning with DFS backtracking. | 我用 DFS 回溯解出回文切割。 | Wrap-up |
| Each step chooses a palindrome prefix segment. | 每一步選一段回文前綴。 | Wrap-up |
| Invalid non-palindrome segments are pruned immediately. | 非回文段會立即被剪掉。 | Wrap-up |
| We record a partition when start reaches string end. | 當 start 到字尾時記錄一組答案。 | Wrap-up |
| I can also discuss DP table acceleration if asked. | 若被問我也可補充 DP 回文表加速。 | Wrap-up |
10) Ultra-short cheat sheet (20 lines total)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Problem type: string partition enumeration. | 題型：字串切割列舉。 | Cheat sheet |
| Constraint: every piece must be palindrome. | 約束：每段都要回文。 | Cheat sheet |
| Return all valid partition lists. | 回傳所有合法切割。 | Cheat sheet |
| DFS state is start index. | DFS 狀態是 start 索引。 | Cheat sheet |
| Base when start == n. | start==n 為基底。 | Cheat sheet |
| Record current partition at base. | 基底時記錄當前切割。 | Cheat sheet |
| Loop end from start to n-1. | end 從 start 走到 n-1。 | Cheat sheet |
| Check palindrome(start,end). | 檢查 palindrome(start,end)。 | Cheat sheet |
| If false, skip branch. | 若否則跳過分支。 | Cheat sheet |
| If true, push substring. | 若為真則 push 子字串。 | Cheat sheet |
| Recurse with end+1. | 以 end+1 遞迴。 | Cheat sheet |
| Pop after recursion. | 遞迴後 pop。 | Cheat sheet |
| Example aab -> [a,a,b], [aa,b]. | 例 aab -> [a,a,b]、[aa,b]。 | Cheat sheet |
| Single char is palindrome. | 單字元必是回文。 | Cheat sheet |
| Worst time exponential. | 最壞時間指數級。 | Cheat sheet |
| Common bound O(n*2^n). | 常見上界 O(n*2^n)。 | Cheat sheet |
| Stack depth O(n). | 堆疊深度 O(n)。 | Cheat sheet |
| Output size may dominate. | 輸出大小可能主導。 | Cheat sheet |
| Optional optimization: DP palindrome table. | 可選優化：DP 回文表。 | Cheat sheet |
| Core idea: cut only on palindrome boundaries. | 核心：只在回文邊界切割。 | Cheat sheet |
Quality check
- Consistency with source solution: ✅ DFS partitioning with palindrome gate is preserved.
- No hallucinated constraints: ✅ Uses exact partition semantics from source.
- Language simplicity: ✅ Clear, spoken, and tightly tied to coding flow.