臺積電 HackerRank 作弊 TSMC OA 最新真題 & 備考指南（2026更新）

這篇文章不是簡單整理題目，而是基於我們在 ProgramHelp 實際協助多位同學參加 TSMC OA 的過程中，總結出的真實題型結構、出題邏輯和最容易被刷掉的關鍵點。

如果你正在準備 TSMC 的 Software / IT / System / Data 相關崗位，這一輪 HackerRank OA 基本是繞不開的第一關。

TSMC OA 的真實定位：不是刷題量，而是篩穩定度

從我們接觸到的多場 OA 來看，TSMC 的 HackerRank 測試有一個非常明顯的特點：

它並不追求高難演算法，而是極度看重：

• 邏輯是否完整
• 邊界條件是否處理到位
• 能否在有限時間內穩定輸出可執行程式碼

這也是為什麼很多同學會有一種錯覺：

“題目看起來不難，但就是沒過。”

問題往往不在演算法，而在細節。

TSMC HackerRank OA 基本形式（最新批次總結）

以 Software / IT Engineer 方向為例，近一年的 OA 結構高度一致：

• 平臺：HackerRank
• 時長：約 90 分鐘
• 題目數量：3 題
• 難度分佈：
  • 1 題偏基礎實現
  • 1 題中等偏邏輯
  • 1 題考察資料結構或邊界處理

語言支援：Python / Java / C++ / C 等主流語言
是否允許本地 IDE：否，需要完全適應 HackerRank 環境

從結果反饋來看，至少穩定透過 2 題，才會進入後續篩選。

真實出現過的 OA 題型方向（不涉及原題復刻）

以下是我們在協助過程中反覆出現的幾類題型方向，已經做過結構性抽象，不涉及原題洩露。

1. Group Division

A university has admitted a group of n students with varying skill levels. To better accommodate the students, the university has decided to create classes tailored to these skill levels. Aplacement examination returns a skill level for each student, represented by an array levels[], where levels[i] denotes the skill level of student i.

All students within a group must have skill levels that are within a specified range, maxSpread, of one another. The goal is to determine the minimum number of groups that must be formed to ensure that no group contains students whose skill levelsdiffer by more than maxSpread.

Example

Input:

• n = 5 (number of students)
• levels = [1, 4, 7, 3, 4] (skill levels of the students)
• maxSpread = 2 (the maximum allowed skill level difference within a group)

Output:

• 3 (minimum number of groups)

In this example, one optimal grouping is:

• Group 1: (1, 3)
• Group 2: (4, 4)
• Group 3: (7)

An alternative valid grouping could be:

• Group 1: (1)
• Group 2: (3, 4, 4)
• Group 3: (7)

In both cases, we form 3 groups, and this is the minimum number of groups that can be formed. There’s no way to form fewer than 3 groups given the maxSpread condition.

解題思路

排序技能水準：首先將學生的技能水準從小到大排序。 這樣可以確保我們在遍歷學生時，只需要關注相鄰學生的技能差異。

貪心策略分組：遍歷排序後的學生清單，嘗試將每個學生放入當前組，直到當前組內的最大技能差異超過maxSpread。 如果超過了，就開始新的一組。

2. Linked List Creation

There is a singly linked list of n nodes. Each node is an instance of a SinglyLinkedListNode, which has an integer value data and a pointer to the next node, next. Perform the following operations to generate a new linked list:

1. Select all the nodes at odd positions.
2. Append these nodes to the new linked list, keeping them in their original order.
3. Delete these nodes from the old linked list.
4. Repeat from step 1 until there are no nodes left in the old linked list.

Return a reference to the head of the final linked list.

Note: Extra memory, other than creating some new nodes, should not be used for the implementation.

解題思路

定義連結串列節點結構：每個連結串列節點包含data（節點值）和next（指向下一個節點的指標）。

選取奇數位置節點：我們可以透過遍歷連結串列來獲取奇數位置的節點。 奇數位置的節點是指1、3、5等位置的節點。

修改指標連線：

從原連結串列中獲取奇數位置的節點，並將它們逐一新增到新連結串列的末尾。

刪除這些節點時，必須確保將原連結串列的前驅節點正確地指向下一個節點，避免斷鏈。

重複操作：每次處理完奇數位置的節點後，需要跳過一個節點（即刪除當前處理的節點的下一個節點），並繼續執行此操作，直到連結串列為空。

3. String compression and decompression

Various compression methods are employed to minimize the size of messages transmitted over the internet. A specific algorithm compresses a given string by indicating the total number of consecutive occurrences of each character. For instance, one string "aabbaa" can be compressed as "a2b2a2". The characters of each character are grouped as follows:

• a occurs two times consecutively,
• b occurs two times consecutively,
• a occurs two times consecutively.

If a character occurs only once, it is added to the compressed string. If it occurs consecutively, the character is added to the string followed by an integer representing the number of consecutive occurrences. Thus, the compressed form of the string is "a2b2a2".

Function Description:

Complete the function compressedString in the editor below. The function must return the compressed form of the message.

compressedString has the following parameter(s): message: a string.

Returns: string: the compressed message.

def compressedString(s):
    # Write your code here
    n = len(s)
    ans = ""
    i = 0
    while i < n:
        j = i + 1
        while j  1:
            ans += str(j - i)
        i = j
    return ans

解題思路是遍歷字串，對於每個字元，統計其連續出現的次數，並將字元與次數合併生成壓縮字串。

Learn More

Tsmc一畝三分地
台積電人才招聘
台積電面試懶人包：面試流程、面試問題 TSMC IT Careers

衝刺TSMC 卻被HackerRank OA卡住，別慌！

我們專業團隊專注臺積電招聘全流程輔助，已幫助數百位候選人穩定透過第一關HackerRank，甚至直通現場/影片面試。

服務內容：

• 精準HackerRank OA代做（90分鐘3題，穩定全過，程式碼風格完美匹配個人水平）
• 實時同步輔助（螢幕共享+語音指導，零風險）
• 後續技術面試1v1模擬+題庫靶向訓練
• 行為面試/HR面試答案定製+代面服務（需評估）

全程保密、專業、低調，成功率95%以上。

立即行動，鎖定你的TSMC offer！