05 Hand of Straights — Interview English Script (C++)

Source aligned with: docs/13_Greedy/05_Hand_of_Straights.md

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1) 30-second problem restatement script

English line	Traditional Chinese meaning (short)	Interview stage
Let me restate hand of straights.	我先重述 Hand of Straights。	Restatement
We have card values in array hand and an integer groupSize.	題目給手牌陣列 hand 與 groupSize。	Restatement
We must split all cards into groups of equal size.	要把所有牌分成固定大小群組。	Restatement
Each group must be consecutive values.	每組牌值必須連續。	Restatement
Return true if such partition is possible.	若可行回 true。	Restatement
I will use greedy with ordered frequency map.	我會用有序頻率表的貪心法。	Restatement

2) Clarifying questions (5 lines)

English line	Traditional Chinese meaning (short)	Interview stage
If hand length is not divisible by groupSize, can we return false immediately?	若手牌數不能整除 groupSize，能否直接 false？	Clarify
Are card values non-negative integers?	牌值是否為非負整數？	Clarify
Do we need exact partition of all cards with no leftovers?	是否必須完全分組且不可剩牌？	Clarify
Is sorted-map or min-heap approach both acceptable?	ordered map 或 min-heap 都可接受嗎？	Clarify
Is O(n log n) expected?	O(n log n) 是否預期？	Clarify

3) Approach discussion

Brute force (3 lines)

English line	Traditional Chinese meaning (short)	Interview stage
Brute force tries many grouping combinations.	暴力法會嘗試大量分組組合。	Approach
Search space explodes with repeated values.	重複值會讓搜尋空間爆炸。	Approach
We need deterministic greedy construction.	我們需要確定性的貪心構造。	Approach

Optimized approach (5 lines)

English line	Traditional Chinese meaning (short)	Interview stage
Count frequencies with ordered map keyed by card value.	用有序 map 依牌值統計頻率。	Approach
Always start from smallest card still remaining.	永遠從目前最小剩餘牌開始。	Approach
If smallest appears count times, we must open count groups from it.	若最小牌有 count 張，就必須開 count 組順子。	Approach
For every offset up to groupSize minus one, each required card must have at least count copies.	往後 groupSize-1 張每張都至少要有 count 張。	Approach
Deduct counts in batch; any shortage means false.	批次扣減頻率，任何不足即 false。	Approach

4) Coding-and-speaking script (line-by-line, in coding order)

English line	Traditional Chinese meaning (short)	Interview stage
First I check hand size divisible by groupSize.	先檢查手牌數可否整除 groupSize。	Coding
I build ordered map counts for each card value.	我建立每個牌值的有序頻率表。	Coding
I iterate map from smallest key upward.	我由最小 key 往上迭代 map。	Coding
Let startCard and count be current key and remaining copies.	令 startCard、count 為當前牌值與剩餘張數。	Coding
If count is zero, this key was already consumed.	若 count 為 0，表示已被前面組消耗。	Coding
Otherwise for i from zero to groupSize minus one, I check counts[startCard+i] >= count.	否則對每個偏移 i，檢查 counts[startCard+i] 是否至少 count。	Coding
I subtract count from each required consecutive card.	我把每張必需連續牌都扣掉 count。	Coding
If loop completes, return true.	若整輪完成，回傳 true。	Coding

5) Dry-run script using one sample input

English line	Traditional Chinese meaning (short)	Interview stage
Let me dry-run hand [1,2,3,6,2,3,4,7,8] and groupSize three.	我手跑 hand=[1,2,3,6,2,3,4,7,8]、groupSize=3。	Dry-run
Sorted counts start with one card one.	排序後頻率從牌 1 開始。	Dry-run
From startCard one and count one, consume one two three.	以 1 為起點 count=1，扣掉 1、2、3。	Dry-run
Next remaining smallest is two, consume two three four.	下一個最小剩餘是 2，再扣 2、3、4。	Dry-run
Next remaining smallest is six, consume six seven eight.	再來最小剩餘是 6，扣 6、7、8。	Dry-run
All counts become zero without shortage.	所有頻率都可扣到 0，無短缺。	Dry-run
Final answer is true.	最終答案為 true。	Dry-run

6) Edge/corner test script (at least 4 cases)

English line	Traditional Chinese meaning (short)	Interview stage
Case one: hand size not divisible by groupSize.	案例一：手牌數無法整除 groupSize。	Edge test
Case two: missing middle card in required sequence.	案例二：連續序列中間缺牌。	Edge test
Case three: groupSize equals one should always be true.	案例三：groupSize=1 應永遠 true。	Edge test
Case four: high duplicates requiring batch subtraction.	案例四：大量重複值需批次扣減。	Edge test
Case five: very large card values with sparse gaps.	案例五：牌值很大且分布稀疏。	Edge test

7) Complexity script

Short version (2 lines)

English line	Traditional Chinese meaning (short)	Interview stage
Time complexity is O(n log n).	時間複雜度是 O(n log n)。	Complexity
Space complexity is O(n).	空間複雜度是 O(n)。	Complexity

Full version (4 lines)

English line	Traditional Chinese meaning (short)	Interview stage
Building ordered frequency map costs O(n log n).	建立有序頻率表成本為 O(n log n)。	Complexity
Each card count is deducted a bounded number of times.	每張牌頻率只會被有限次扣減。	Complexity
Overall runtime remains O(n log n).	整體時間維持 O(n log n)。	Complexity
Frequency map storage is O(n) in worst case unique cards.	最壞唯一牌值數量下 map 空間為 O(n)。	Complexity

8) If stuck rescue lines (10 lines)

English line	Traditional Chinese meaning (short)	Interview stage
Let me use smallest-card-first greedy rule.	我先用最小牌優先的貪心規則。	If stuck
Smallest remaining card must be a group start.	最小剩餘牌一定是某組起點。	If stuck
I should process its full remaining count in batch.	我應一次處理它全部剩餘 count。	If stuck
For each offset, required card count must be enough.	每個偏移位置的牌數都必須足夠。	If stuck
If any required count is short, return false immediately.	任一需求短缺就立即 false。	If stuck
Otherwise subtract batch count from each consecutive value.	否則對連續值都扣掉 batch count。	If stuck
Let me test quickly with [1,2,3,4,5] and group four.	我快速測 [1,2,3,4,5]、group=4。	If stuck
Size check fails first, so false is correct.	先過不了整除檢查，false 正確。	If stuck
Batch deduction avoids repeated tiny operations.	批次扣減可避免重複小操作。	If stuck
Great, greedy proof and implementation align.	很好，貪心證明與實作一致。	If stuck

9) Final wrap-up lines (5 lines)

English line	Traditional Chinese meaning (short)	Interview stage
I solved hand of straights with ordered-frequency greedy.	我用有序頻率貪心解了 Hand of Straights。	Wrap-up
The smallest remaining card always starts new groups.	最小剩餘牌必定是新順子的起點。	Wrap-up
Batch subtraction enforces all required consecutive cards.	批次扣減可強制檢查所有必要連續牌。	Wrap-up
Complexity is O(n log n) time and O(n) space.	複雜度是 O(n log n) 時間、O(n) 空間。	Wrap-up
This is the standard interview-grade solution.	這是面試常用標準解法。	Wrap-up

10) Ultra-short cheat sheet (20 lines total)

English line	Traditional Chinese meaning (short)	Interview stage
Goal: split all cards into consecutive groups of groupSize.	目標：把所有牌分成長度 groupSize 的連續組。	Cheat sheet
If n % groupSize != 0 -> false.	若 n % groupSize != 0 -> false。	Cheat sheet
Build ordered frequency map.	建立有序頻率表。	Cheat sheet
Iterate keys from smallest upward.	由小到大遍歷 key。	Cheat sheet
Let count = remaining freq at startCard.	令 count 為 startCard 剩餘頻率。	Cheat sheet
If count==0, continue.	若 count==0 則略過。	Cheat sheet
Need cards startCard ... startCard+groupSize-1.	需要 startCard 到 startCard+groupSize-1。	Cheat sheet
Each required freq must be >= count.	每張需求牌頻率都要 >= count。	Cheat sheet
If any shortage -> false.	任一不足 -> false。	Cheat sheet
Deduct count from each required card.	對每張需求牌扣 count。	Cheat sheet
Finish all keys -> true.	全部完成 -> true。	Cheat sheet
Example [1,2,3,6,2,3,4,7,8],3 -> true.	範例 [1,2,3,6,2,3,4,7,8],3 -> true。	Cheat sheet
Example [1,2,3,4,5],4 -> false.	範例 [1,2,3,4,5],4 -> false。	Cheat sheet
Time O(n log n).	時間 O(n log n)。	Cheat sheet
Space O(n).	空間 O(n)。	Cheat sheet
Common bug: subtract one-by-one not by batch count.	常見錯誤：逐張扣而非批次扣 count。	Cheat sheet
Common bug: forgetting divisibility precheck.	常見錯誤：忘記整除前置檢查。	Cheat sheet
Keep smallest-first invariant explicit.	清楚說明最小牌優先不變量。	Cheat sheet
Ordered map simplifies correctness reasoning.	有序 map 可簡化正確性推理。	Cheat sheet
Validate with duplicated-card cases.	要用重複牌案例驗證。	Cheat sheet

Quality check

• Consistency with source solution: ✅ Ordered-map greedy with batch count deduction preserved.
• No hallucinated constraints: ✅ Exact partition and consecutive-group semantics maintained.
• Language simplicity: ✅ Clear interview explanation for invariant and failure condition.