10 Redundant Connection — Interview English Script (C++)
Source aligned with: docs/15_Graphs/10_Redundant_Connection.md
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1) 30-second problem restatement script
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me restate redundant connection. | 我先重述 Redundant Connection。 | Restatement |
| We have an undirected graph that was a tree plus one extra edge. | 圖原本是樹，後來多加了一條邊。 | Restatement |
| Because of that extra edge, one cycle appears. | 因這條多餘邊而形成一個環。 | Restatement |
| We need to return the edge that can be removed to restore tree property. | 要回傳可移除並恢復樹性質的那條邊。 | Restatement |
| If multiple candidates exist, return the one appearing last in input. | 若有多候選，回輸入中最後出現者。 | Restatement |
| I will use Union Find to detect first cycle-forming edge in scan order. | 我會用 Union Find 掃描時抓成環那條邊。 | Restatement |
2) Clarifying questions (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Are node labels one-indexed from one to n? | 節點編號是否是 1 到 n？ | Clarify |
| Is graph undirected for every edge? | 每條邊是否都是無向邊？ | Clarify |
| Is there guaranteed at least one cycle due to extra edge? | 是否保證因額外邊至少有一個環？ | Clarify |
| Should we return edge values, not index position? | 是否要回傳邊的值，而非索引位置？ | Clarify |
| Is path compression enough without union-by-rank for constraints? | 在這個限制下只用路徑壓縮是否足夠？ | Clarify |
3) Approach discussion
Brute force (3 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Brute force can remove each edge and test if remaining graph is tree. | 暴力法可逐條刪邊再測剩餘圖是否為樹。 | Approach |
| That requires repeated graph traversals. | 這需要重複做圖遍歷。 | Approach |
| Complexity is much higher than necessary. | 複雜度明顯高於必要。 | Approach |
Optimized approach (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Use Union Find to maintain connected components dynamically. | 用 Union Find 動態維護連通分量。 | Approach |
| For each edge u v, check roots of u and v first. | 對每條邊 u,v 先查兩者根節點。 | Approach |
| If roots are already same, this edge closes a cycle and is redundant. | 若根已相同，該邊會閉合成環，即冗餘邊。 | Approach |
| Otherwise union the two sets. | 否則把兩集合合併。 | Approach |
| Scanning input order naturally satisfies return-last requirement under this setup. | 依輸入順序掃描可自然符合題目回傳要求。 | Approach |
4) Coding-and-speaking script (line-by-line, in coding order)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I set n as edges size and initialize parent array from zero to n. | 我令 n=edges.size() 並初始化 parent 0..n。 | Coding |
| I define find with path compression. | 我定義帶路徑壓縮的 find。 | Coding |
| I iterate edges in input order. | 我按輸入順序逐條掃描邊。 | Coding |
| For edge u v, compute rootU and rootV. | 對邊 u,v 計算 rootU 與 rootV。 | Coding |
| If rootU equals rootV, return this edge immediately. | 若 rootU==rootV，立即回傳該邊。 | Coding |
| Otherwise union by setting one root parent to the other. | 否則把一個根掛到另一個根完成 union。 | Coding |
| Continue scanning until cycle edge is found. | 持續掃描直到找到成環邊。 | Coding |
| Return empty only as fallback. | 空回傳僅作保底。 | Coding |
| Path compression keeps operations near constant. | 路徑壓縮讓操作接近常數。 | Coding |
| This avoids repeated DFS checks. | 這可避免反覆 DFS 驗證。 | Coding |
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me dry-run edges [[1,2],[1,3],[2,3]]. | 我手跑 edges=[[1,2],[1,3],[2,3]]。 | Dry-run |
| Edge one two connects two separate sets, so union. | 邊 1-2 連到不同集合，執行 union。 | Dry-run |
| Edge one three also connects separate sets, so union. | 邊 1-3 也連不同集合，執行 union。 | Dry-run |
| Now one and three and two are all connected. | 現在 1、2、3 已同屬一個連通集合。 | Dry-run |
| Edge two three has same root on both ends. | 邊 2-3 的兩端根節點相同。 | Dry-run |
| So this edge creates cycle and is returned. | 所以這條邊成環並被回傳。 | Dry-run |
| Final answer is [2,3]. | 最終答案是 [2,3]。 | Dry-run |
6) Edge/corner test script (at least 4 cases)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Case one: smallest cycle triangle graph. | 案例一：最小三角環圖。 | Edge test |
| Case two: cycle formed late after many tree-like edges. | 案例二：先是樹狀邊，最後才成環。 | Edge test |
| Case three: graph with two possible cycle-closing edges, verify last one by input order. | 案例三：有兩條候選成環邊，驗證輸入順序最後者。 | Edge test |
| Case four: long chain then one back edge to root. | 案例四：長鏈後回邊連回根。 | Edge test |
| Case five: ensure one-index parent sizing is correct. | 案例五：確認一索引 parent 大小正確。 | Edge test |
7) Complexity script
Short version (2 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Time complexity is O(n alpha n). | 時間複雜度是 O(n·α(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 |
| We process each of n edges once. | 我們對 n 條邊各處理一次。 | Complexity |
| Each find or union is almost constant with path compression. | 路徑壓縮下每次 find/union 近乎常數。 | Complexity |
| Formally this is O(n alpha n). | 形式上時間為 O(n·α(n))。 | Complexity |
| Parent array stores one entry per node, so O(n) memory. | parent 每節點一格，因此記憶體 O(n)。 | Complexity |
8) If stuck rescue lines (10 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Let me focus on cycle detection in undirected graph. | 我先聚焦無向圖環檢測。 | If stuck |
| Union Find is built exactly for dynamic connectivity. | Union Find 正是動態連通性的工具。 | If stuck |
| Same root before union means adding edge forms cycle. | union 前同根代表加邊會成環。 | If stuck |
| Different roots means safe to union. | 不同根表示可安全合併。 | If stuck |
| Path compression keeps find efficient. | 路徑壓縮讓 find 很高效。 | If stuck |
| Let me test quickly with triangle edges. | 我快速用三角邊測試。 | If stuck |
| Third edge should be returned as redundant. | 第三條邊應被回傳為冗餘。 | If stuck |
| This matches intuitive cycle closure. | 這與直覺閉環一致。 | If stuck |
| Parent array should be n plus one for one-index nodes. | 一索引節點時 parent 應開到 n+1。 | If stuck |
| Great, now implementation is straightforward. | 很好，現在實作很直觀。 | If stuck |
9) Final wrap-up lines (5 lines)
| English line | Traditional Chinese meaning (short) | Interview stage |
| I solved it with Union Find cycle detection. | 我用 Union Find 的環檢測解題。 | Wrap-up |
| While scanning edges, first same-root edge is redundant. | 掃描邊時，第一條同根邊即冗餘。 | Wrap-up |
| Path compression provides near-constant operations. | 路徑壓縮提供近常數操作效率。 | Wrap-up |
| Complexity is O(n alpha n) time and O(n) space. | 複雜度是 O(n·α(n)) 時間與 O(n) 空間。 | Wrap-up |
| This is the standard DSU pattern for extra-edge cycle problems. | 這是多餘邊成環題的標準 DSU 模式。 | Wrap-up |
10) Ultra-short cheat sheet (20 lines total)
| English line | Traditional Chinese meaning (short) | Interview stage |
| Goal: find redundant edge creating cycle. | 目標：找出造成環的冗餘邊。 | Cheat sheet |
| Graph is undirected. | 圖是無向圖。 | Cheat sheet |
| Use Union Find. | 使用 Union Find。 | Cheat sheet |
| parent size n plus one. | parent 大小開 n+1。 | Cheat sheet |
| Initialize parent[i]=i. | 初始化 parent[i]=i。 | Cheat sheet |
| find uses path compression. | find 使用路徑壓縮。 | Cheat sheet |
| For each edge u v in order. | 依序處理每條邊 u,v。 | Cheat sheet |
| ru=find(u), rv=find(v). | ru=find(u), rv=find(v)。 | Cheat sheet |
| If ru==rv return [u,v]. | 若 ru==rv 回 [u,v]。 | Cheat sheet |
| Else union roots. | 否則合併兩根。 | Cheat sheet |
| Continue until returned. | 持續直到回傳。 | Cheat sheet |
| Triangle test returns third edge. | 三角測試回第三邊。 | Cheat sheet |
| Time O(n alpha n). | 時間 O(n·α(n))。 | Cheat sheet |
| Space O(n). | 空間 O(n)。 | Cheat sheet |
| Common bug: forgetting one-index sizing. | 常見錯誤：忘記一索引開陣列。 | Cheat sheet |
| Common bug: no path compression. | 常見錯誤：沒做路徑壓縮。 | Cheat sheet |
| Input order matters for returned edge. | 輸入順序影響回傳邊。 | Cheat sheet |
| This is dynamic connectivity pattern. | 這是動態連通性模式。 | Cheat sheet |
| Alternative DFS is slower for repeated checks. | DFS 重複檢查通常較慢。 | Cheat sheet |
| Keep explanation focused on same-root logic. | 說明重點放在同根判斷。 | Cheat sheet |
Quality check
- Consistency with source solution: ✅ Union-Find with path compression preserved.
- No hallucinated constraints: ✅ Undirected cycle semantics and return behavior aligned.
- Language simplicity: ✅ Short clear interview lines.