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Hash tables are extraordinary. O(1) average-time insert, lookup, and delete. Constant-time membership testing. Blazing fast key-value storage. With performance like this, why would you ever use anything else?
The answer reveals a fundamental limitation that no amount of clever hashing can overcome: hash tables destroy ordering information.
The very mechanism that makes hash tables fast—transforming keys into pseudo-random bucket indices—makes it impossible to efficiently answer questions like:
These operations require ordering, and hash tables fundamentally cannot provide it. For such requirements, balanced binary search trees (BSTs) aren't just an alternative—they're the correct choice.
By the end of this page, you will understand exactly which operations hash tables cannot efficiently support, why BSTs excel at ordered operations, how to recognize when your problem requires ordering, and how to make the right data structure choice for your specific requirements.
To understand why hash tables cannot support ordered operations, we need to revisit how they work at a fundamental level.
Hash Tables Destroy Order by Design:
A hash function is designed to spread keys uniformly across buckets, regardless of their natural ordering. Consider keys 1, 2, 3, 4, 5. In a sorted array, they'd be stored in positions 0, 1, 2, 3, 4. In a hash table with 10 buckets:
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NATURAL ORDER: 1 < 2 < 3 < 4 < 5 AFTER HASHING (example hash function):═══════════════════════════════════════════════════════════════════ Key │ hash(key) │ Bucket Index (hash % 10)─────┼────────────┼─────────────────────────── 1 │ 7839462 │ 2 2 │ 3284751 │ 1 3 │ 9571638 │ 8 4 │ 1426395 │ 5 5 │ 6847293 │ 3 Hash Table Layout:┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐│ 0 │ 1 │ 2 │ 3 │ 4 │ 5 │ 6 │ 7 │ 8 │ 9 │├────┼────┼────┼────┼────┼────┼────┼────┼────┼────┤│ │ 2 │ 1 │ 5 │ │ 4 │ │ │ 3 │ │└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘ PHYSICAL ORDER IN TABLE: 2, 1, 5, 4, 3 There is NO relationship between bucket position and key order!To find keys in sorted order, you must:1. Collect all keys: O(n)2. Sort them: O(n log n)3. Total: O(n log n) — much worse than BST's O(n) in-order traversalThe Core Issue: No Spatial Locality
In a hash table, keys that are "close" in value (like 100 and 101) can end up in completely different buckets. There's no way to find "nearby" keys without examining all keys.
Contrast this with a BST:
Operations That Require Order:
| Operation | Hash Table | Balanced BST |
|---|---|---|
| Find minimum | O(n) — scan all elements | O(log n) — leftmost node |
| Find maximum | O(n) — scan all elements | O(log n) — rightmost node |
| Find predecessor (next smaller) | O(n) — scan all elements | O(log n) — tree traversal |
| Find successor (next larger) | O(n) — scan all elements | O(log n) — tree traversal |
| Range query [a, b] | O(n) — scan all elements | O(log n + k) — k = result size |
| Sorted iteration | O(n log n) — collect + sort | O(n) — in-order traversal |
| Rank query (k-th element) | O(n log n) — sort first | O(log n) — with augmentation |
When you need ordered operations on hash table data, you're forced to scan all elements—O(n). This completely negates the O(1) advantage hash tables provide for point lookups. If your workload is dominated by ordered operations, a hash table is actively harmful to performance.
Let's examine specific use cases where balanced BSTs are the clearly superior choice, even though their O(log n) operations are theoretically slower than hash tables' O(1).
Use Case 1: Price Books in Trading Systems
Stock exchanges maintain "order books" that track buy and sell orders at various prices. Critical operations include:
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TRADING ORDER BOOK STRUCTURE:═══════════════════════════════════════════════════════════════════ BUY ORDERS (bids): SELL ORDERS (asks):├── $150.25: 1000 shares ├── $150.50: 500 shares ← Best ask (min)├── $150.20: 2500 shares ├── $150.55: 750 shares├── $150.15: 800 shares ├── $150.60: 1200 shares├── $150.10: 3000 shares └── $150.75: 2000 shares└── $150.00: 5000 shares ↑ Best bid (max) TYPICAL OPERATIONS:───────────────────────────────────────────────────────────────────── 1. GET BEST BID (highest buy price): Hash Table: Scan all bids → O(n) BST: Find rightmost node → O(log n) 2. GET BEST ASK (lowest sell price): Hash Table: Scan all asks → O(n) BST: Find leftmost node → O(log n) 3. MATCH ORDERS (when bid ≥ ask): Need to frequently access min/max → BST wins decisively 4. PRICE IMPROVEMENT SEARCH (find orders within $0.05 of target): Hash Table: Scan all, filter by price → O(n) BST: Range query [target-0.05, target+0.05] → O(log n + k) REAL-WORLD IMPACT:─────────────────────────────────────────────────────────────────────Exchanges process millions of orders per second.BST operations: millions × O(log n) = millions × ~20 comparisonsHash operations: millions × O(n) = millions × thousands of comparisons BST is 100-1000x faster for this workload!Use Case 2: Leaderboards and Rankings
Gaming leaderboards, performance dashboards, and ranked lists require:
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# HASH TABLE APPROACH (poor for leaderboards)class HashLeaderboard: def __init__(self): self.scores = {} # player_id -> score def update_score(self, player_id, score): self.scores[player_id] = score # O(1) ✓ def get_score(self, player_id): return self.scores.get(player_id) # O(1) ✓ def get_top_k(self, k): # Must sort ALL scores to find top k! sorted_scores = sorted(self.scores.items(), key=lambda x: x[1], reverse=True) return sorted_scores[:k] # O(n log n) ✗ def get_rank(self, player_id): player_score = self.scores.get(player_id) if player_score is None: return None # Must count ALL players with higher score! rank = sum(1 for s in self.scores.values() if s > player_score) + 1 return rank # O(n) ✗ # BST APPROACH (designed for leaderboards)# Using a balanced BST with augmented nodesclass BSTLeaderboard: def __init__(self): self.tree = BalancedBST() # e.g., Red-Black Tree self.player_nodes = {} # player_id -> tree node def update_score(self, player_id, new_score): if player_id in self.player_nodes: self.tree.delete(self.player_nodes[player_id]) node = self.tree.insert(new_score, player_id) self.player_nodes[player_id] = node # O(log n) def get_top_k(self, k): # In-order traversal (reverse) yields sorted order return self.tree.get_largest_k(k) # O(log n + k) ✓ def get_rank(self, player_id): node = self.player_nodes.get(player_id) if node is None: return None # With size augmentation, rank is computable via tree structure return self.tree.get_rank(node) # O(log n) ✓ # PERFORMANCE COMPARISON (1 million players, k=10):# ─────────────────────────────────────────────────────────────────# get_top_k(10):# Hash: Sort 1M elements → ~13 million comparisons# BST: Navigate to rightmost → ~20 comparisons + 10 traversal steps# # get_rank(player_id):# Hash: Count larger scores → ~1 million comparisons# BST: Tree traversal → ~20 comparisonsUse Case 3: Time-Series Data and Event Logs
Systems that store timestamped events often need:
Databases use B-trees (a generalized BST) for indexed columns precisely because queries need ordering: WHERE date BETWEEN x AND y, ORDER BY timestamp, MIN/MAX aggregations. Hash indexes exist but are rarely the default because so many queries require order.
Range queries are perhaps the clearest case where BSTs dominate. The operation "find all elements between A and B" appears constantly in real applications.
The Range Query Problem:
Given a data structure with n elements, find all elements x where low ≤ x ≤ high, where k is the number of results.
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RANGE QUERY: Find all x where 25 ≤ x ≤ 75═══════════════════════════════════════════════════════════════════ DATA: [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]EXPECTED RESULT: [30, 40, 50, 60, 70] (k=5 elements) HASH TABLE APPROACH:───────────────────────────────────────────────────────────────────── Hash Table (no ordering):┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐│ 60 │ │ 100│ 30 │ │ 80 │ 20 │ │ 40 │ 70 │└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘ ↑ ↑ ↑ ↑ ↑ | | | | | check check check check check 25≤60≤75?✓ 25≤30≤75?✓ 25≤20≤75?✗ .. 25≤70≤75?✓ Algorithm: for each element e in table: if low ≤ e ≤ high: add to result Time: O(n) — must check EVERY elementFor n=1,000,000, k=100: must examine 1,000,000 to find 100 BST APPROACH:───────────────────────────────────────────────────────────────────── Balanced BST: [50] / \ [30] [70] / \ / \ [20] [40] [60] [80] / \ [10] [90] \ [100] Algorithm: 1. Find first element ≥ 25: Navigate: 50→30→20→(20<25)→30 ✓ Comparisons: O(log n) = 4 2. In-order traverse until > 75: 30→40→50→60→70→(80>75)→stop Comparisons: O(k) = 5 Time: O(log n + k)For n=1,000,000, k=100: examine ~20 + 100 = 120 elements SPEEDUP: 1,000,000 / 120 ≈ 8,000×The O(log n + k) Guarantee:
The BST range query complexity deserves scrutiny:
This is optimal—you can't do better than O(k) because you must at least output k elements. The O(log n) overhead for finding where to start is negligible for most workloads.
Contrast with Hash Table:
A hash table must check every element because qualifying elements could be in any bucket. There's no structure to exploit—the hash function destroyed all ordering information.
| Result Size (k) | Hash Table (O(n)) | BST (O(log n + k)) | Speedup |
|---|---|---|---|
| k = 10 | 1,000,000 ops | 30 ops | 33,000× |
| k = 100 | 1,000,000 ops | 120 ops | 8,300× |
| k = 1,000 | 1,000,000 ops | 1,020 ops | 980× |
| k = 10,000 | 1,000,000 ops | 10,020 ops | 100× |
| k = 100,000 | 1,000,000 ops | 100,020 ops | 10× |
| k = 500,000 | 1,000,000 ops | 500,020 ops | 2× |
BST advantage is greatest when k << n (result set is small relative to total data). This is exactly the common case! Most range queries return a tiny fraction of total data. Finding this week's orders in a database of years of orders? k is tiny compared to n.
Another operation hash tables fundamentally cannot support efficiently is finding the predecessor (largest element smaller than X) or successor (smallest element larger than X).
Real-World Applications:
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# HASH TABLE: Predecessor requires scanning ALL elementsdef hash_predecessor(hash_table, target): """ Find largest element smaller than target. With a hash table, must check every element. """ predecessor = None for key in hash_table: if key < target: if predecessor is None or key > predecessor: predecessor = key return predecessor # O(n) — must scan everything # Usage: Find highest price < $100# prices = {99.99, 149.99, 79.99, 199.99, 49.99, 129.99}# hash_predecessor(prices, 100) # → Checks all 6 prices, returns 99.99 # BST: Predecessor is O(log n) via tree navigationclass BSTNode: def __init__(self, key, left=None, right=None, parent=None): self.key = key self.left = left self.right = right self.parent = parent def bst_predecessor(root, target): """ Find largest element smaller than target in O(log n). """ predecessor = None node = root while node is not None: if node.key < target: # This node qualifies; look for larger in right subtree predecessor = node.key node = node.right else: # Node too large; look in left subtree node = node.left return predecessor # O(log n) — follows single path # VISUALIZATION OF BST PREDECESSOR SEARCH:# ═══════════════════════════════════════════════════════════════════# # Find predecessor of 75 in BST:## [50] ← 50 < 75, save as candidate, go right# / \# [30] [70] ← 70 < 75, save as candidate, go right# / \ / \# [20] [40] [60] [80] ← 80 ≥ 75, go left# ↓# [null] ← done!## Predecessor = 70 (last saved candidate)# Comparisons: 3 (not 6)The Floor and Ceiling Operations:
Related to predecessor/successor are floor (largest element ≤ X) and ceiling (smallest element ≥ X):
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DATA: [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] FLOOR(45): Largest element ≤ 45 → 40 ✓CEILING(45): Smallest element ≥ 45 → 50 ✓ FLOOR(50): Largest element ≤ 50 → 50 ✓ (exact match)CEILING(50): Smallest element ≥ 50 → 50 ✓ (exact match) FLOOR(5): Largest element ≤ 5 → None (no element ≤ 5)CEILING(105): Smallest element ≥ 105 → None (no element ≥ 105) USE CASE: Price matching with constraints───────────────────────────────────────────────────────────────────User budget: $75Available prices: [29.99, 49.99, 69.99, 89.99, 109.99] FLOOR(75) = 69.99 → "Best option within your budget"CEILING(75) = 89.99 → "Cheapest option above your budget" Hash table: O(n) to find eachBST: O(log n) for each — crucial when prices change frequentlyBST operations like predecessor, successor, floor, and ceiling all work by navigating the tree structure. Each comparison eliminates half the remaining candidates. This is impossible with hash tables because the structure provides no navigational information—elements are scattered into buckets with no relationship to their values.
A common requirement is iterating through elements in sorted order. This arises in:
The Iteration Complexity Difference:
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REQUIREMENT: Iterate through all elements in sorted order HASH TABLE APPROACH:═══════════════════════════════════════════════════════════════════ Step 1: Collect all elements into a list Time: O(n) Step 2: Sort the list Time: O(n log n) Step 3: Iterate through sorted list Time: O(n) Total: O(n log n) — dominated by sorting step BST APPROACH (In-Order Traversal):═══════════════════════════════════════════════════════════════════ [50] / \ [30] [70] / \ / \ [20] [40] [60] [80] In-order traversal: Left → Root → Right Visit order: 20 → 30 → 40 → 50 → 60 → 70 → 80 def inorder(node): if node is None: return inorder(node.left) # Visit left subtree yield node.key # Process current node inorder(node.right) # Visit right subtree Each node is visited exactly once.Total: O(n) SPACE COMPARISON:═══════════════════════════════════════════════════════════════════ Hash Table sorted iteration:- Must allocate O(n) additional space for the sorted copy- Cannot iterate "in place" BST in-order traversal:- O(h) stack space for recursion (h = tree height)- For balanced tree: O(log n) space- Can be made O(1) with Morris traversal (modifies tree temporarily)Incremental Sorted Access:
Beyond one-time iteration, consider scenarios where you need sorted access incrementally:
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# HASH TABLE: No good option for incremental sorted accessclass HashTableSortedIterator: def __init__(self, hash_table): # Must sort EVERYTHING upfront even if we only need a few elements self.sorted_keys = sorted(hash_table.keys()) # O(n log n) self.index = 0 def has_next(self): return self.index < len(self.sorted_keys) def next(self): key = self.sorted_keys[self.index] self.index += 1 return key # O(1) per call, but O(n log n) setup # BST: True incremental sorted accessclass BSTSortedIterator: def __init__(self, root): # Only O(log n) setup: go to minimum self.stack = [] self._push_left(root) # O(log n) def _push_left(self, node): while node: self.stack.append(node) node = node.left def has_next(self): return len(self.stack) > 0 def next(self): node = self.stack.pop() self._push_left(node.right) return node.key # Amortized O(1) per call, O(log n) setup # USAGE PATTERN: "Get first 10 sorted elements"# ─────────────────────────────────────────────────────────────────── # Hash table: Pay O(n log n) upfront, then O(1) per element# Total for 10 elements: O(n log n) + O(10) = O(n log n) # BST: Pay O(log n) setup, then amortized O(1) per element# Total for 10 elements: O(log n) + O(10) = O(log n) # For n = 1,000,000 and k = 10:# Hash: ~20 million operations# BST: ~30 operations# # The BST is almost a MILLION times faster!A common mistake is using hash tables for data that needs pagination ('show next 20 items'). Each page request sorts all data to find the right 20 items: O(n log n) per page! With a BST, you navigate directly to the starting point and collect 20 items: O(log n + 20) per page.
With a clear understanding of when each data structure excels, let's formalize the decision process.
The Operation Mix Question:
The right choice depends on your operation mix—which operations dominate your workload?
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START: What operations does your application need?═══════════════════════════════════════════════════════════════════ ┌─────────────────────────────────────────────────────────┐ │ PRIMARY OPERATION ANALYSIS │ └────────────────────────────┬────────────────────────────┘ │ ┌────────────────────┴────────────────────┐ ▼ ▼ ┌───────────────────┐ ┌───────────────────┐ │ Need ordering? │ │ Only need exact │ │ - Range queries │ │ match lookup? │ │ - Min/Max │ │ - Contains key? │ │ - Sorted access │ │ - Get by key │ │ - Predecessor/ │ │ - Insert/delete │ │ Successor │ │ by key │ │ - k-th element │ │ - Count distinct │ └────────┬──────────┘ └────────┬──────────┘ │ │ ▼ ▼ ┌────────────────────┐ ┌────────────────────────┐ │ Use BST │ │ Use Hash Table │ │ (TreeMap/TreeSet)│ │ (HashMap/HashSet) │ │ │ │ │ │ O(log n) all ops │ │ O(1) avg all ops │ │ O(log n + k) │ │ O(n) ordered ops │ │ range queries │ │ │ │ O(n) sorted │ │ O(n log n) sorted │ │ iteration │ │ iteration │ └────────────────────┘ └────────────────────────┘ │ │ │ ┌────────────────────────┐ │ └─────►│ MIXED WORKLOAD? │◄──────┘ │ Both ordered AND │ │ point operations │ └───────────┬────────────┘ │ ▼ ┌────────────────────────┐ │ CONSIDER: │ │ │ │ • Use both structures │ │ • BST if ordering wins │ │ • SkipList as middle │ │ ground │ │ • Database indexes │ │ (they face this too) │ └────────────────────────┘Quick Reference Table:
| Use Case | Best Choice | Reason |
|---|---|---|
| User session lookup | Hash Table | Point lookup only, O(1) wins |
| Cache implementation | Hash Table | Fast lookup by key |
| Configuration storage | Hash Table | Key-value access pattern |
| Leaderboard/ranking | BST | Needs sorted access, rank queries |
| Time-series queries | BST | Range queries by timestamp |
| Order book (trading) | BST | Min/max, range queries |
| Database index | B-tree (BST family) | Range scans, sorted access |
| Spell checker | Either (depends) | Hash for existence, BST for suggestions |
| Priority queue | Heap or BST | Need min/max extraction |
| Unique ID set | Hash Table | Only existence checks |
If you're unsure, prototype both and measure with realistic data. The 'slower' O(log n) BST operations are still fast in absolute terms—often microseconds. For many applications, the difference between O(1) and O(log n) is imperceptible, but the difference between O(log n) and O(n) for ordered operations is dramatic.
Modern languages provide ordered containers alongside hash-based ones. Knowing which to use is essential.
Java:
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import java.util.*; // HASH-BASED: O(1) operations, no orderingHashMap<String, Integer> hashMap = new HashMap<>();HashSet<String> hashSet = new HashSet<>(); // TREE-BASED: O(log n) operations, full orderingTreeMap<String, Integer> treeMap = new TreeMap<>();TreeSet<String> treeSet = new TreeSet<>(); // TreeMap-specific ordered operations:treeMap.put("apple", 1);treeMap.put("banana", 2);treeMap.put("cherry", 3);treeMap.put("date", 4); treeMap.firstKey(); // → "apple" (minimum)treeMap.lastKey(); // → "date" (maximum)treeMap.lowerKey("cherry"); // → "banana" (predecessor)treeMap.higherKey("banana"); // → "cherry" (successor)treeMap.floorKey("cat"); // → "cherry" (floor)treeMap.ceilingKey("cat"); // → "cherry" (ceiling) // Range view (subMap):SortedMap<String, Integer> range = treeMap.subMap("banana", "date"); // ["banana", "cherry"]// Returns a VIEW, not a copy — O(log n) to create! // NavigableMap extends SortedMap with more methods:NavigableMap<String, Integer> nav = treeMap;nav.descendingMap(); // Reverse-order viewnav.headMap("cherry", true); // All keys ≤ "cherry"nav.tailMap("banana", true); // All keys ≥ "banana"Python:
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# HASH-BASED (built-in): O(1) operationshash_dict = {}hash_set = set() # Python has no built-in sorted container!# Use the 'sortedcontainers' package (pip install sortedcontainers) from sortedcontainers import SortedDict, SortedSet, SortedList # SORTED CONTAINERS: O(log n) operations with orderingsorted_dict = SortedDict()sorted_set = SortedSet()sorted_list = SortedList() # SortedDict operations:sorted_dict['apple'] = 1sorted_dict['banana'] = 2sorted_dict['cherry'] = 3sorted_dict['date'] = 4 sorted_dict.peekitem(0) # → ('apple', 1) - minimumsorted_dict.peekitem(-1) # → ('date', 4) - maximum # Index-based access (O(log n)):sorted_dict.peekitem(2) # → ('cherry', 3) - 3rd element # Range operations via slicing:list(sorted_dict.irange('banana', 'date')) # → ['banana', 'cherry', 'date'] # SortedList specific operations:sl = SortedList([30, 10, 50, 20, 40])sl.bisect_left(25) # → 2 (index where 25 would go)sl.bisect_right(30) # → 3 (index after existing 30s)sl[2] # → 30 (3rd smallest) # IMPORTANT: sortedcontainers uses different implementation# than traditional BST, but provides similar O(log n) guaranteesC++:
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#include <unordered_map>#include <unordered_set>#include <map>#include <set> // HASH-BASED: O(1) average operationsstd::unordered_map<std::string, int> hash_map;std::unordered_set<std::string> hash_set; // TREE-BASED (Red-Black Tree): O(log n) operations with orderingstd::map<std::string, int> tree_map;std::set<std::string> tree_set; // std::map ordered operations:tree_map["apple"] = 1;tree_map["banana"] = 2;tree_map["cherry"] = 3;tree_map["date"] = 4; tree_map.begin()->first; // "apple" (minimum)tree_map.rbegin()->first; // "date" (maximum) // lower_bound: iterator to first element ≥ keyauto it = tree_map.lower_bound("cat"); // points to "cherry" // upper_bound: iterator to first element > keyauto it2 = tree_map.upper_bound("banana"); // points to "cherry" // Range iteration:for (auto it = tree_map.lower_bound("banana"); it != tree_map.upper_bound("cherry"); ++it) { // Iterates: banana, cherry} // equal_range: pair of (lower_bound, upper_bound)auto [lo, hi] = tree_map.equal_range("banana");| Language | Hash-Based | Ordered (BST) |
|---|---|---|
| Java | HashMap, HashSet | TreeMap, TreeSet |
| Python | dict, set | sortedcontainers.SortedDict/Set |
| C++ | unordered_map, unordered_set | map, set |
| JavaScript | Map, Set, Object | No built-in (use libraries) |
| Go | map | No built-in (use btree package) |
| Rust | HashMap, HashSet | BTreeMap, BTreeSet |
While we use 'BST' as shorthand, actual implementations vary. Java uses Red-Black trees, C++ typically uses Red-Black trees, and some libraries use B-trees or skip lists. The interface and complexity guarantees are similar, but cache performance and constant factors differ.
Hash tables are powerful, but they cannot do everything. When your problem involves ordering, they are fundamentally the wrong tool.
What's Next:
We've covered security vulnerabilities (collision attacks), quality issues (poor hash functions), and capability limitations (ordered operations). The final consideration is practical: space overhead. Hash tables use more memory than their contents strictly require, and in some contexts, this overhead is unacceptable.
You now understand when hash tables are the wrong choice due to ordering requirements. This knowledge prevents the common mistake of forcing hash tables where BSTs belong, resulting in O(n) operations that should be O(log n). Next, we'll examine the space overhead of hash tables and when it becomes problematic.