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Before we can compare data structures meaningfully, we need a precise vocabulary for what we do with data. This isn't just terminology—it's a framework for analysis.
Every data structure exists to support a set of operations. Understanding these operations in depth—what they mean, how they're measured, and how they interact—is the foundation of intelligent data structure selection.
In this page, we'll dissect each fundamental operation with surgical precision. By the end, you'll be able to analyze any problem in terms of its operation requirements and immediately understand which data structures are candidates and which are disqualified.
You will master the five fundamental operations (Access, Search, Insert, Delete, Update), understand variants within each category, and learn to analyze real problems by decomposing them into these operations. This vocabulary becomes the basis for all future data structure discussions.
Access refers to retrieving an element when you know its location (index or position), not its value. This is fundamentally different from searching.
The Key Question Access Answers:
"Give me the element at position X."
Examples:
array[42] — Give me the element at index 42list.get(100) — Give me the 100th elementbuffer.charAt(5) — Give me the 5th characterAccess does NOT answer:
Why Access Speed Varies:
The speed of access depends on how data is physically organized in memory:
Contiguous Memory (Arrays):
Memory: [elem₀][elem₁][elem₂][elem₃][elem₄]...
Address: 1000 1004 1008 1012 1016
To access element at index i:
address = base_address + (i × element_size)
Time: O(1) — Direct calculation, no traversal
Linked Structure (Linked Lists):
Memory: [elem₀|→]...[elem₁|→]...[elem₂|→]
↓ ↓ ↓
scattered across memory
To access element at index i:
Start at head, follow i pointers
Time: O(n) — Must traverse i nodes
| Data Structure | Access Time | Explanation |
|---|---|---|
| Static Array | O(1) | Direct address calculation |
| Dynamic Array (ArrayList, Vector) | O(1) | Same as static array (contiguous memory) |
| Singly Linked List | O(n) | Must traverse from head |
| Doubly Linked List | O(n) | Can traverse from either end, still O(n/2) = O(n) |
| Hash Table | N/A | No concept of 'index position' — access by key instead |
| Binary Search Tree | N/A | No concept of position — access by key |
| Stack | O(1) for top only | Only top element is accessible by design |
| Queue | O(1) for front only | Only front element is accessible by design |
Data structures that support O(1) access to any position are called 'random access' structures. This property is crucial for algorithms like binary search, which require jumping to arbitrary positions. If you need random access, linked lists are immediately disqualified.
Variants of Access:
Each variant has different characteristics and not all structures support all variants efficiently.
Search refers to finding an element based on its value or properties, not its position. This is one of the most common and performance-critical operations.
The Key Questions Search Answers:
"Is this value in the collection?" (Membership test) "Where is this value in the collection?" (Location finding) "What value(s) match these criteria?" (Filtered retrieval)
Examples:
if 42 in my_list — Membership testarray.indexOf("hello") — Find position of valueusers.filter(u => u.age > 30) — Find all matching elementsWhy Search Speed Varies Dramatically:
Search performance depends entirely on how data is organized and whether that organization can be exploited:
Unsorted Array — Linear Search:
[7, 2, 9, 4, 1, 8, 3, 6, 5]
Searching for 8:
Check 7? No. Check 2? No. Check 9? No. Check 4? No.
Check 1? No. Check 8? Yes! Found at index 5.
Worst case: Check all n elements → O(n)
Average case: Check n/2 elements → O(n)
Sorted Array — Binary Search:
[1, 2, 3, 4, 5, 6, 7, 8, 9]
Searching for 8:
Middle is 5. 8 > 5, search right half.
Middle is 7. 8 > 7, search right half.
Middle is 8. Found!
Each step eliminates half → O(log n)
Hash Table — Hash Lookup:
hash(8) = 3 (example)
Go directly to bucket 3, check if 8 is there.
Direct computation → O(1) average case
| Data Structure | Search Time | Condition |
|---|---|---|
| Unsorted Array | O(n) | Must check every element |
| Sorted Array | O(log n) | Binary search possible |
| Linked List (Unsorted) | O(n) | Must traverse sequentially |
| Linked List (Sorted) | O(n) | Still must traverse (no random access for binary search) |
| Hash Table / Hash Set | O(1) average, O(n) worst | Direct hash computation; worst case with collisions |
| Binary Search Tree (Balanced) | O(log n) | Divide-and-conquer through tree |
| Binary Search Tree (Unbalanced) | O(n) worst | Degenerates to linked list |
| Trie | O(k) where k = key length | Character-by-character traversal |
A common misconception: 'If I sort my linked list, I can use binary search.' This is FALSE. Binary search requires O(1) random access to find the middle element. Linked lists only support O(n) access. A sorted linked list still requires O(n) search time.
Variants of Search:
Critical Insight: The choice of data structure often depends on which type of search you need. A hash table gives O(1) exact match but cannot do range search. A BST gives O(log n) exact match AND efficient range search.
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# Demonstrating search performance differencesimport timeimport random def linear_search(arr, target): """O(n) search in unsorted array""" for i, val in enumerate(arr): if val == target: return i return -1 def binary_search(arr, target): """O(log n) search in sorted array""" left, right = 0, len(arr) - 1 while left <= right: mid = (left + right) // 2 if arr[mid] == target: return mid elif arr[mid] < target: left = mid + 1 else: right = mid - 1 return -1 def hash_search(hash_set, target): """O(1) average search using hash set""" return target in hash_set # Performance comparisonn = 1_000_000data = list(range(n))random.shuffle(data)sorted_data = sorted(data)hash_data = set(data)target = n - 1 # Worst case for linear search # Results (approximate):# Linear search: ~50,000 microseconds (50ms)# Binary search: ~20 microseconds (0.02ms) # Hash search: ~1 microsecond (0.001ms)# # Ratio: Linear is ~50,000x slower than hash lookup!Insert refers to adding a new element to a collection. The complexity of insertion varies dramatically based on where the element is inserted and how the structure maintains its invariants.
The Key Question Insert Answers:
"Add this element to the collection."
Critical Variants:
Dynamic Array (ArrayList, List, Vector) Insertion:
Insert at End (Append):
Before: [1, 2, 3, 4, 5, _, _, _] (size=5, capacity=8)
After: [1, 2, 3, 4, 5, 6, _, _] (size=6, capacity=8)
Time: O(1) — Just place at index 'size' and increment
Insert at Beginning:
Before: [1, 2, 3, 4, 5]
Shift: [_, 1, 2, 3, 4, 5] — Move all elements right
After: [0, 1, 2, 3, 4, 5] — Place new element
Time: O(n) — Must shift all n elements
Insert at Position k:
Before: [1, 2, 3, 4, 5] Insert 9 at position 2
Shift: [1, 2, _, 3, 4, 5] — Shift elements from k onward
After: [1, 2, 9, 3, 4, 5] — Place new element
Time: O(n-k) — Shift (n-k) elements
When a dynamic array is full, it must allocate a new larger array (typically 2x size) and copy all elements. This takes O(n) time. However, this happens infrequently enough that the AMORTIZED cost of append remains O(1).
| Data Structure | Insert at Front | Insert at End | Insert at Position k |
|---|---|---|---|
| Dynamic Array | O(n) | O(1) amortized | O(n-k) |
| Singly Linked List (with tail) | O(1) | O(1) | O(k) + O(1) |
| Singly Linked List (no tail) | O(1) | O(n) | O(k) + O(1) |
| Doubly Linked List | O(1) | O(1) | O(min(k, n-k)) + O(1) |
| Hash Table/Set | N/A | N/A | O(1) average |
| Balanced BST | N/A | N/A | O(log n) |
| Heap (Binary) | N/A | O(log n) | N/A (insert always at end, then heapify) |
Delete refers to removing an element from a collection. Like insertion, deletion complexity depends on the element's position and the structure's invariants.
The Key Questions Delete Addresses:
"Remove the element at position X." (Delete by position) "Remove element with value Y." (Delete by value — requires search first) "Remove and return the first/last element." (Pop operations)
Critical Consideration: Delete by value = Search + Delete by position. The total cost includes both.
Array Deletion — The Shifting Problem:
Delete element at index 2 from [A, B, C, D, E]:
Before: [A, B, C, D, E]
Remove: [A, B, _, D, E] — C is removed
Shift: [A, B, D, E, _] — D, E shift left to fill gap
After: [A, B, D, E] — Size decreases by 1
Time: O(n-k) — Must shift all elements after position k
Delete at end: O(1) — No shifting needed Delete at front: O(n) — Shift all n-1 elements Delete at middle: O(n/2) = O(n) — Shift half the elements on average
Linked List Deletion — The Pointer Surgery:
Delete node containing 'C' from A → B → C → D → E:
Before: [A|→] → [B|→] → [C|→] → [D|→] → [E|null]
↑ ↑
prev target
Operation: prev.next = target.next
After: [A|→] → [B|→] ------→ [D|→] → [E|null]
(C is now unreachable, garbage collected)
Time: O(1) IF you already have pointer to prev node
O(n) if you need to search for the element first
When someone says 'deletion in a linked list is O(1),' they mean the pointer manipulation is O(1). But if you only have the value (not a pointer to the node), you must first search — which is O(n). Total: O(n) + O(1) = O(n). Always clarify: delete by position or delete by value?
| Data Structure | Delete at Front | Delete at End | Delete by Value (includes search) |
|---|---|---|---|
| Dynamic Array | O(n) | O(1) | O(n) search + O(n) shift = O(n) |
| Singly Linked List | O(1) | O(n) | O(n) search + O(1) delete = O(n) |
| Doubly Linked List | O(1) | O(1) | O(n) search + O(1) delete = O(n) |
| Hash Table/Set | N/A | N/A | O(1) average |
| Balanced BST | O(log n)* | O(log n)* | O(log n) |
| Heap (Binary) | O(log n) | Special** | O(n) search + O(log n) = O(n) |
*BST delete-min and delete-max are O(log n) in balanced trees. **Heap delete-max (max-heap) or delete-min (min-heap) is O(log n); deleting arbitrary elements is O(n) due to search.
Special Case — Lazy Deletion:
Some systems use 'lazy deletion' where elements are marked as deleted but not physically removed. This makes deletion O(1) but complicates search and wastes space. Eventually, a cleanup pass removes all marked elements.
Update refers to changing the value of an existing element. This seems simple, but has subtle complexities depending on the data structure.
The Key Question Update Addresses:
"Change element X to value Y."
Update is fundamentally: Find element + Modify element
For most structures, update time = search time + O(1) modification.
The Update Trap in Ordered Structures:
In structures that maintain ordering (sorted arrays, BSTs, heaps), you cannot simply change a value in place. The modification might violate the ordering invariant.
Example — BST Update Gone Wrong:
10 10
/ \ / \
5 15 Update 5 25 15 ← INVALID BST!
/ \ to 25 / \
3 8 3 8
Correct BST Update:
Heap Update:
| Data Structure | Update by Index | Update by Key/Value | Notes |
|---|---|---|---|
| Array | O(1) | O(n) search + O(1) | Direct index update is instant |
| Linked List | O(n) traverse + O(1) | O(n) search + O(1) | Must find position first |
| Hash Table | N/A | O(1) | Key lookup + value modification |
| Sorted Array | O(1) if order preserved | O(n) | If order changes, need re-sort or delete+insert |
| BST | N/A | O(log n) × 2 | Delete + insert to maintain order |
| Heap | O(n) find + O(log n) | O(n) find + O(log n) | Need to find then restore heap |
In a hash table, you can update the VALUE associated with a key in O(1). But you CANNOT update the KEY itself. Changing a key changes its hash, meaning it would be stored in a different bucket. To 'update' a key, you must delete the old entry and insert a new one: O(1) + O(1) = O(1), but it's conceptually different from in-place update.
Beyond the core CRUD operations (Create/Insert, Read/Access, Update, Delete), several secondary operations are critical for data structure selection.
Traversal — Visiting All Elements:
Almost every structure supports O(n) traversal, but the order of traversal differs:
Important: If you need elements in sorted order, a hash table requires O(n log n) to sort, while a BST provides this for free (O(n) in-order traversal).
| Data Structure | Find Minimum | Find Maximum | Notes |
|---|---|---|---|
| Unsorted Array | O(n) | O(n) | Must check all elements |
| Sorted Array | O(1) | O(1) | First element / last element |
| Linked List | O(n) | O(n) | Must traverse all nodes |
| Hash Table | O(n) | O(n) | No ordering — must check all |
| BST | O(log n)* | O(log n)* | Go left/right to extreme |
| Min-Heap | O(1) | O(n) | Root is minimum; max could be anywhere in leaves |
| Max-Heap | O(n) | O(1) | Root is maximum; min could be anywhere in leaves |
*In a balanced BST, finding min/max is O(log n). In an unbalanced BST, it could be O(n).
Range Queries — Elements Between X and Y:
This operation highlights a critical difference between structures:
If your application needs range queries, hash tables are immediately disqualified despite their O(1) point queries.
When analyzing a problem, don't just consider insert/search/delete. Ask: Do I need min/max quickly? Do I need sorted traversal? Do I need range queries? These secondary operations often eliminate otherwise-attractive candidates.
Understanding individual operations isn't enough. You must analyze how frequently each operation occurs in your specific use case.
The Fundamental Principle:
Optimize for the common case, tolerate the rare case.
A data structure with O(n) insert but O(1) search is excellent if you insert once and search millions of times. It's terrible if you insert millions of times and search once.
Case Study: Spell Checker
Consider building a spell checker:
Operations:
Analysis:
Search dominates by ~1000x. Even if insert is O(n), total insert time = 100,000 × O(n) = O(n²). But with 6,000 searches per minute at O(n) each, that's O(6000n) per minute, ongoing.
Optimal Choice: Hash Set
Suboptimal Choice: Sorted Array with Binary Search
Terrible Choice: Unsorted Array
Often, 80% of operations are of one or two types. Optimize relentlessly for those, and accept reasonable (not catastrophic) performance for the remaining 20%. Trying to optimize everything equally usually means optimizing nothing well.
You now have the complete vocabulary for data structure analysis. Let's consolidate:
| Operation | What It Does | Key Complexity Factors |
|---|---|---|
| Access | Get element by position | Contiguous memory = O(1); Linked = O(n) |
| Search | Find element by value | Unsorted = O(n); Sorted = O(log n); Hash = O(1) |
| Insert | Add new element | Position matters; May require shifts or rebalancing |
| Delete | Remove element | Position matters; May require shifts; Includes search cost |
| Update | Modify existing element | = Search + Modify; May require restructuring if ordered |
| Traverse | Visit all elements | Usually O(n); Order depends on structure |
| Min/Max | Find extreme values | Unsorted = O(n); Heap = O(1) for one extreme |
| Range Query | Find elements in range | Hash = O(n); Sorted/Tree = O(log n + k) |
You now speak the language of data structure operations fluently. Next, we'll see how these operation costs combine across different structures — creating a comprehensive comparison that enables intelligent selection.