🔗 Complete Guide to Hashing

📋 Table of Contents

1. What is Hashing?
2. How Hashing Works
3. Types of Hashing Techniques
4. Collision Handling Strategies
5. Hashing in Python
6. Time & Space Complexity
7. Common Interview Problems
8. Interview Tips and Tricks

1. What is Hashing? 🤔

            Definition: Hashing is a technique that uses a hash function to map keys to specific locations in a hash table, enabling fast data retrieval with average O(1) time complexity.
        

Key Components:

Hash Function: A mathematical function that converts keys into array indices
Hash Table: An array-based data structure that stores key-value pairs
Buckets: Individual slots in the hash table where data is stored

Real-World Examples:

📖 Dictionary/Phone Book:

Instead of searching through every name sequentially, you jump to a specific section (A-D, E-H, etc.) based on the first letter. The first letter acts as a "hash function" that maps names to sections.

🔐 Password Storage:

Websites don't store your actual password. They store a hashed version. When you login, your entered password is hashed and compared with the stored hash.

# Password: "mypassword123"
# Stored Hash: "5e884898da28047151d0e56f8dc6292773603d0d6aabbdd62a11ef721d1542d8"
            

💾 Caching Systems:

Web browsers use hashing to quickly check if a webpage is already cached. The URL is hashed to create a unique identifier for quick lookup.

🗄️ Database Indexing:

Databases use hash indices to quickly locate records based on primary keys without scanning the entire table.

2. How Hashing Works ⚙️

Key: "apple"

→

Hash Function

→

Index: 3

→

Table[3] = "apple"

Step-by-Step Process:

Input Key: Provide a key (string, number, object)
Hash Function: Apply mathematical transformation
Generate Index: Get array index within table bounds
Store/Retrieve: Access data at calculated index

Hash Function Properties:

Deterministic: Same input always produces same output
Uniform Distribution: Keys should be evenly distributed
Fast Computation: Should execute quickly
Minimize Collisions: Different keys should produce different indices

# Simple Hash Function Example
def simple_hash(key, table_size):
    """Convert string key to table index"""
    hash_value = 0
    for char in str(key):
        hash_value += ord(char)  # Sum ASCII values
    return hash_value % table_size  # Ensure within bounds

# Example
key = "hello"
table_size = 10
index = simple_hash(key, table_size)
print(f"Key '{key}' maps to index {index}")  # Output: Key 'hello' maps to index 2
        

3. Types of Hashing Techniques 🎯

3.1 Direct Addressing

            Concept: Use the key itself as the array index. Works when keys are small integers within a known range.
        

# Direct Addressing Example
class DirectAddressTable:
    def __init__(self, max_size):
        self.table = [None] * max_size
        self.max_size = max_size
    
    def insert(self, key, value):
        if 0 <= key < self.max_size:
            self.table[key] = value
    
    def search(self, key):
        if 0 <= key < self.max_size:
            return self.table[key]
        return None
    
    def delete(self, key):
        if 0 <= key < self.max_size:
            self.table[key] = None

# Usage
table = DirectAddressTable(1000)
table.insert(42, "Answer to everything")
table.insert(100, "Century")
print(table.search(42))  # Output: "Answer to everything"
        

Direct addressing is perfect when you have a small, known range of integer keys (like ages 0-120, or student IDs 1000-9999).

3.2 Division Method (Mod Operator)

            Formula: h(k) = k mod m

            Where: k = key, m = table size

def division_hash(key, table_size):
    """Division method hashing"""
    return hash(key) % table_size

# Example with different keys
table_size = 7  # Prime number (recommended)
keys = ["apple", "banana", "cherry", "date", 123, 456]

print("Division Method Results:")
for key in keys:
    index = division_hash(key, table_size)
    print(f"Key: {key} → Index: {index}")

# Output:
# Key: apple → Index: 2
# Key: banana → Index: 4
# Key: cherry → Index: 1
# Key: date → Index: 6
# Key: 123 → Index: 4  # Collision with banana!
# Key: 456 → Index: 1  # Collision with cherry!
        

Choose table size as a prime number to reduce clustering and improve distribution. Avoid powers of 2.

3.3 Multiplication Method

            Formula: h(k) = floor(m * (k * A mod 1))

            Where: A = constant (0 < A < 1), typically A = (√5 - 1)/2 ≈ 0.618

import math

def multiplication_hash(key, table_size):
    """Multiplication method hashing"""
    A = (math.sqrt(5) - 1) / 2  # Golden ratio - 1
    key_hash = hash(key) if isinstance(key, str) else key
    return int(table_size * ((key_hash * A) % 1))

# Example
table_size = 10
keys = [10, 22, 31, 4, 15, 28, 17, 88, 59]

print("Multiplication Method Results:")
for key in keys:
    index = multiplication_hash(key, table_size)
    print(f"Key: {key} → Index: {index}")
        

3.4 Universal Hashing

            Concept: Randomly select hash function from a family of functions to minimize worst-case collisions.
        

import random

class UniversalHashing:
    def __init__(self, table_size):
        self.table_size = table_size
        self.p = 1000000007  # Large prime
        self.a = random.randint(1, self.p - 1)
        self.b = random.randint(0, self.p - 1)
    
    def hash_function(self, key):
        """Universal hash function: ((a*k + b) mod p) mod m"""
        key_val = hash(key) if isinstance(key, str) else key
        return ((self.a * key_val + self.b) % self.p) % self.table_size

# Example
universal_hash = UniversalHashing(10)
keys = ["hello", "world", "python", "hashing"]

print("Universal Hashing Results:")
for key in keys:
    index = universal_hash.hash_function(key)
    print(f"Key: {key} → Index: {index}")
        

4. Collision Handling Strategies 🚧

Collision: When two different keys produce the same hash value/index.

4.1 Chaining (Separate Chaining)

Store multiple elements at the same index using linked lists or dynamic arrays.

class HashTableChaining:
    def __init__(self, size=10):
        self.size = size
        self.table = [[] for _ in range(size)]  # List of lists
    
    def _hash(self, key):
        return hash(key) % self.size
    
    def insert(self, key, value):
        index = self._hash(key)
        bucket = self.table[index]
        
        # Check if key already exists
        for i, (k, v) in enumerate(bucket):
            if k == key:
                bucket[i] = (key, value)  # Update existing
                return
        
        # Add new key-value pair
        bucket.append((key, value))
    
    def search(self, key):
        index = self._hash(key)
        bucket = self.table[index]
        
        for k, v in bucket:
            if k == key:
                return v
        return None
    
    def delete(self, key):
        index = self._hash(key)
        bucket = self.table[index]
        
        for i, (k, v) in enumerate(bucket):
            if k == key:
                del bucket[i]
                return True
        return False
    
    def display(self):
        for i, bucket in enumerate(self.table):
            print(f"Index {i}: {bucket}")

# Example usage
ht = HashTableChaining(7)
ht.insert("apple", 100)
ht.insert("banana", 200)
ht.insert("orange", 300)
ht.insert("grape", 400)  # May cause collision

ht.display()
print(f"Search 'apple': {ht.search('apple')}")  # Output: 100
        

Chaining Visualization:

Index 0: []
Index 1: [('apple', 100)]
Index 2: [('banana', 200), ('grape', 400)]  ← Collision handled
Index 3: [('orange', 300)]
Index 4: []
Index 5: []
Index 6: []

4.2 Open Addressing

Find alternative locations within the same table when collision occurs.

4.2.1 Linear Probing

            Formula: h'(k, i) = (h(k) + i) mod m

            Where: i = 0, 1, 2, ... (probe sequence)

class HashTableLinearProbing:
    def __init__(self, size=10):
        self.size = size
        self.keys = [None] * size
        self.values = [None] * size
        self.deleted = [False] * size  # Track deleted slots
    
    def _hash(self, key):
        return hash(key) % self.size
    
    def _probe(self, key):
        index = self._hash(key)
        
        while (self.keys[index] is not None and 
               self.keys[index] != key and 
               not self.deleted[index]):
            index = (index + 1) % self.size  # Linear probing
        
        return index
    
    def insert(self, key, value):
        index = self._probe(key)
        self.keys[index] = key
        self.values[index] = value
        self.deleted[index] = False
    
    def search(self, key):
        index = self._hash(key)
        
        while self.keys[index] is not None:
            if self.keys[index] == key and not self.deleted[index]:
                return self.values[index]
            index = (index + 1) % self.size
        
        return None
    
    def delete(self, key):
        index = self._hash(key)
        
        while self.keys[index] is not None:
            if self.keys[index] == key and not self.deleted[index]:
                self.deleted[index] = True
                return True
            index = (index + 1) % self.size
        
        return False
    
    def display(self):
        for i in range(self.size):
            status = " (DEL)" if self.deleted[i] else ""
            print(f"Index {i}: {self.keys[i]} -> {self.values[i]}{status}")
        

4.2.2 Quadratic Probing

            Formula: h'(k, i) = (h(k) + c₁i + c₂i²) mod m

            Where: c₁, c₂ are constants (commonly c₁=c₂=1)

class HashTableQuadraticProbing:
    def __init__(self, size=10):
        self.size = size
        self.keys = [None] * size
        self.values = [None] * size
    
    def _hash(self, key):
        return hash(key) % self.size
    
    def _quadratic_probe(self, key):
        index = self._hash(key)
        i = 0
        
        while (self.keys[index] is not None and 
               self.keys[index] != key):
            i += 1
            index = (self._hash(key) + i * i) % self.size
        
        return index
    
    def insert(self, key, value):
        index = self._quadratic_probe(key)
        self.keys[index] = key
        self.values[index] = value
    
    def search(self, key):
        index = self._hash(key)
        i = 0
        
        while self.keys[index] is not None:
            if self.keys[index] == key:
                return self.values[index]
            i += 1
            index = (self._hash(key) + i * i) % self.size
        
        return None
        

4.2.3 Double Hashing

            Formula: h'(k, i) = (h₁(k) + i * h₂(k)) mod m

            Where: h₁, h₂ are different hash functions

class HashTableDoubleHashing:
    def __init__(self, size=10):
        self.size = size
        self.keys = [None] * size
        self.values = [None] * size
    
    def _hash1(self, key):
        return hash(key) % self.size
    
    def _hash2(self, key):
        # Second hash function (should be coprime to table size)
        return 7 - (hash(key) % 7)
    
    def _double_hash_probe(self, key):
        index = self._hash1(key)
        step_size = self._hash2(key)
        i = 0
        
        while (self.keys[index] is not None and 
               self.keys[index] != key):
            i += 1
            index = (self._hash1(key) + i * step_size) % self.size
        
        return index
    
    def insert(self, key, value):
        index = self._double_hash_probe(key)
        self.keys[index] = key
        self.values[index] = value
        

Collision Resolution Comparison:
• Chaining: Simple, handles high load factors well
• Linear Probing: Good cache performance, but clustering issues
• Quadratic Probing: Reduces clustering, moderate complexity
• Double Hashing: Best distribution, more complex computation

5. Hashing in Python 🐍

5.1 Built-in `dict` Data Structure

# Dictionary (Hash Table) Operations
student_grades = {}

# Insertion - O(1) average
student_grades["Alice"] = 95
student_grades["Bob"] = 87
student_grades["Charlie"] = 92

# Access - O(1) average  
print(student_grades["Alice"])  # Output: 95

# Update - O(1) average
student_grades["Alice"] = 98

# Deletion - O(1) average
del student_grades["Bob"]

# Check existence - O(1) average
if "Charlie" in student_grades:
    print(f"Charlie's grade: {student_grades['Charlie']}")

# Dictionary methods
print(student_grades.keys())    # dict_keys(['Alice', 'Charlie'])
print(student_grades.values())  # dict_values([98, 92])
print(student_grades.items())   # dict_items([('Alice', 98), ('Charlie', 92)])

# Dictionary comprehension
squares = {x: x**2 for x in range(1, 6)}
print(squares)  # {1: 1, 2: 4, 3: 9, 4: 16, 5: 25}
        

5.2 Built-in `set` Data Structure

# Set (Hash Set) Operations
fruits = set()

# Addition - O(1) average
fruits.add("apple")
fruits.add("banana") 
fruits.add("cherry")
fruits.add("apple")  # Duplicate ignored

print(fruits)  # {'cherry', 'banana', 'apple'}

# Check membership - O(1) average
print("apple" in fruits)  # True

# Removal - O(1) average
fruits.remove("banana")
# fruits.discard("banana")  # No error if not found

# Set operations
set1 = {1, 2, 3, 4, 5}
set2 = {4, 5, 6, 7, 8}

print(set1 | set2)  # Union: {1, 2, 3, 4, 5, 6, 7, 8}
print(set1 & set2)  # Intersection: {4, 5}
print(set1 - set2)  # Difference: {1, 2, 3}
print(set1 ^ set2)  # Symmetric difference: {1, 2, 3, 6, 7, 8}

# Set comprehension
even_squares = {x**2 for x in range(10) if x % 2 == 0}
print(even_squares)  # {0, 4, 16, 36, 64}
        

5.3 Built-in `hash()` Function

# Using hash() function
print(hash("hello"))        # e.g., -5434435027328283325
print(hash(42))             # 42
print(hash(3.14))          # 322818021289917443
print(hash((1, 2, 3)))     # Tuples are hashable

# Hash of same object is consistent within session
text = "python"
print(hash(text) == hash(text))  # True

# Different objects may have same hash (collision)
print(hash("a") == hash("b"))  # Probably False, but not guaranteed

# Custom hashable class
class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
    
    def __hash__(self):
        return hash((self.x, self.y))
    
    def __eq__(self, other):
        return isinstance(other, Point) and self.x == other.x and self.y == other.y
    
    def __repr__(self):
        return f"Point({self.x}, {self.y})"

# Using custom hashable objects
points = {}
p1 = Point(1, 2)
p2 = Point(3, 4)

points[p1] = "Origin area"
points[p2] = "Upper right"

print(points[Point(1, 2)])  # "Origin area"
        

Remember: To use an object as a dictionary key, it must be hashable (implement __hash__ and __eq__ methods) and immutable.

6. Time & Space Complexity 📊

Operation	Average Case	Worst Case	Best Case	Notes
Search	O(1)	O(n)	O(1)	Worst case when all keys hash to same index
Insert	O(1)	O(n)	O(1)	May require table resizing
Delete	O(1)	O(n)	O(1)	Same as search complexity
Space	O(n)	O(n)	O(n)	n = number of key-value pairs

Load Factor Impact:

            Load Factor (α) = Number of elements / Table size

            • α < 0.7: Good performance

            • α > 0.8: Performance degrades

            • α = 1.0: Table full (open addressing impossible)

# Load factor calculation and resizing
class DynamicHashTable:
    def __init__(self, initial_size=8):
        self.size = initial_size
        self.count = 0
        self.keys = [None] * self.size
        self.values = [None] * self.size
        self.deleted = [False] * self.size
    
    def load_factor(self):
        return self.count / self.size
    
    def _needs_resize(self):
        return self.load_factor() > 0.7
    
    def _resize(self):
        # Save current data
        old_keys = self.keys[:]
        old_values = self.values[:]
        old_deleted = self.deleted[:]
        
        # Double the size
        self.size *= 2
        self.count = 0
        self.keys = [None] * self.size
        self.values = [None] * self.size
        self.deleted = [False] * self.size
        
        # Rehash all elements
        for i in range(len(old_keys)):
            if old_keys[i] is not None and not old_deleted[i]:
                self.insert(old_keys[i], old_values[i])
    
    def insert(self, key, value):
        if self._needs_resize():
            self._resize()
        
        index = self._probe(key)
        if self.keys[index] is None or self.deleted[index]:
            self.count += 1
        
        self.keys[index] = key
        self.values[index] = value
        self.deleted[index] = False
        

7. Common Interview Problems 💼

7.1 Two Sum Problem

Problem: Given an array of integers and a target sum, return indices of two numbers that add up to target.

def two_sum(nums, target):
    """
    Time: O(n), Space: O(n)
    """
    num_map = {}  # value -> index
    
    for i, num in enumerate(nums):
        complement = target - num
        if complement in num_map:
            return [num_map[complement], i]
        num_map[num] = i
    
    return []

# Example
nums = [2, 7, 11, 15]
target = 9
result = two_sum(nums, target)
print(f"Indices: {result}")  # [0, 1] because nums[0] + nums[1] = 2 + 7 = 9
        

7.2 Group Anagrams

Problem: Group strings that are anagrams of each other.

from collections import defaultdict

def group_anagrams(strs):
    """
    Time: O(n * k log k), Space: O(n * k)
    where n = number of strings, k = max length of string
    """
    anagram_map = defaultdict(list)
    
    for s in strs:
        # Sort characters to create key
        sorted_str = ''.join(sorted(s))
        anagram_map[sorted_str].append(s)
    
    return list(anagram_map.values())

# Example
strs = ["eat", "tea", "tan", "ate", "nat", "bat"]
result = group_anagrams(strs)
print(result)  # [['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]
        

7.3 First Non-Repeating Character

def first_non_repeating_char(s):
    """
    Time: O(n), Space: O(1) - at most 26 lowercase letters
    """
    char_count = {}
    
    # Count frequencies
    for char in s:
        char_count[char] = char_count.get(char, 0) + 1
    
    # Find first non-repeating
    for char in s:
        if char_count[char] == 1:
            return char
    
    return None

# Example
s = "leetcode"
result = first_non_repeating_char(s)
print(f"First non-repeating: '{result}'")  # 'l'
        

7.4 Subarray Sum Equals K

def subarray_sum(nums, k):
    """
    Count number of continuous subarrays whose sum equals k
    Time: O(n), Space: O(n)
    """
    count = 0
    prefix_sum = 0
    sum_map = {0: 1}  # prefix_sum -> frequency
    
    for num in nums:
        prefix_sum += num
        
        # Check if (prefix_sum - k) exists
        if (prefix_sum - k) in sum_map:
            count += sum_map[prefix_sum - k]
        
        # Update map
        sum_map[prefix_sum] = sum_map.get(prefix_sum, 0) + 1
    
    return count

# Example
nums = [1, 1, 1]
k = 2
result = subarray_sum(nums, k)
print(f"Number of subarrays: {result}")  # 2
        

7.5 Longest Consecutive Sequence

def longest_consecutive(nums):
    """
    Time: O(n), Space: O(n)
    """
    if not nums:
        return 0
    
    num_set = set(nums)
    longest = 0
    
    for num in num_set:
        # Check if this is start of sequence
        if num - 1 not in num_set:
            current_num = num
            current_length = 1
            
            # Count consecutive numbers
            while current_num + 1 in num_set:
                current_num += 1
                current_length += 1
            
            longest = max(longest, current_length)
    
    return longest

# Example
nums = [100, 4, 200, 1, 3, 2]
result = longest_consecutive(nums)
print(f"Longest consecutive length: {result}")  # 4 (sequence: 1,2,3,4)
        

7.6 Design HashMap

class MyHashMap:
    """
    Design a HashMap without using built-in hash table libraries
    """
    def __init__(self):
        self.size = 1000
        self.table = [[] for _ in range(self.size)]
    
    def _hash(self, key):
        return key % self.size
    
    def put(self, key, value):
        index = self._hash(key)
        bucket = self.table[index]
        
        for i, (k, v) in enumerate(bucket):
            if k == key:
                bucket[i] = (key, value)
                return
        
        bucket.append((key, value))
    
    def get(self, key):
        index = self._hash(key)
        bucket = self.table[index]
        
        for k, v in bucket:
            if k == key:
                return v
        return -1
    
    def remove(self, key):
        index = self._hash(key)
        bucket = self.table[index]
        
        for i, (k, v) in enumerate(bucket):
            if k == key:
                del bucket[i]
                return

# Usage
hashmap = MyHashMap()
hashmap.put(1, 1)
hashmap.put(2, 2)
print(hashmap.get(1))    # 1
print(hashmap.get(3))    # -1
hashmap.put(2, 1)        # Update existing key
print(hashmap.get(2))    # 1
hashmap.remove(2)
print(hashmap.get(2))    # -1
        

8. Interview Tips and Tricks 🎯

🔍 Pattern Recognition: Use hashing when you need to:
• Track what you've seen before
• Count frequencies
• Find complements or pairs
• Group related items
• Check membership quickly

8.1 Common Hash Table Applications

Frequency Counting: Count occurrences of elements
Caching/Memoization: Store computed results
Finding Duplicates: Track seen elements
Complement Search: Two Sum, Three Sum problems
Grouping: Group anagrams, similar items
Mapping: Character mappings, transformations

8.2 Key Interview Phrases

# When explaining hash tables in interviews, mention:

"I'll use a hash table to achieve O(1) average lookup time..."

"To handle collisions, I would recommend chaining with linked lists..."

"The space-time tradeoff here is O(n) extra space for O(1) operations..."

"We need to consider the load factor to maintain good performance..."

"Hash tables provide amortized O(1) operations..."

8.3 Common Pitfalls & How to Avoid Them

🚨 Collision Handling: Always mention how you'll handle collisions. Don't assume perfect hashing.

🚨 Key Mutability: Explain that keys must be immutable. You can't use lists as dictionary keys in Python.

🚨 Memory Usage: Hash tables trade space for time. Mention the O(n) space complexity.

🚨 Hash Function Quality: Poor hash functions can lead to clustering and worst-case O(n) performance.

8.4 Advanced Optimization Techniques

# 1. Counter for frequency problems
from collections import Counter

def find_anagrams(s, p):
    """Find all anagrams of p in s"""
    if len(p) > len(s):
        return []
    
    p_count = Counter(p)
    window_count = Counter()
    result = []
    
    for i in range(len(s)):
        # Add current character
        window_count[s[i]] += 1
        
        # Remove character outside window
        if i >= len(p):
            left_char = s[i - len(p)]
            window_count[left_char] -= 1
            if window_count[left_char] == 0:
                del window_count[left_char]
        
        # Check if current window is anagram
        if window_count == p_count:
            result.append(i - len(p) + 1)
    
    return result

# 2. defaultdict for cleaner code
from collections import defaultdict

def group_by_first_letter(words):
    """Group words by their first letter"""
    groups = defaultdict(list)
    for word in words:
        groups[word[0]].append(word)
    return dict(groups)

# 3. Using frozenset for hashable sets
def find_unique_combinations(lists):
    """Find unique combinations from list of lists"""
    unique_combos = set()
    for lst in lists:
        unique_combos.add(frozenset(lst))
    return [list(combo) for combo in unique_combos]
        

8.5 Step-by-Step Problem Solving Approach

            Identify the Pattern: Look for keywords like "count", "frequency", "seen before", "pairs", "duplicates"
Choose Data Structure: Dict for key-value mapping, Set for membership testing
Design Hash Strategy: What will be your key? What will be your value?
Handle Edge Cases: Empty input, single element, no solution
Optimize: Consider Counter, defaultdict, or other specialized tools

        

8.6 Practice Problems by Difficulty

📗 Easy Level:

Two Sum
Valid Anagram
First Unique Character
Contains Duplicate
Missing Number

📙 Medium Level:

Group Anagrams
Subarray Sum Equals K
Longest Consecutive Sequence
Design HashMap
Top K Frequent Elements

📕 Hard Level:

Substring with Concatenation of All Words
Design Twitter
Alien Dictionary
Insert Delete GetRandom O(1)

8.7 Code Templates for Common Patterns

# Template 1: Frequency Counter
def frequency_pattern(arr):
    freq = {}
    for item in arr:
        freq[item] = freq.get(item, 0) + 1
    return freq

# Template 2: Seen Before Pattern  
def seen_before_pattern(arr):
    seen = set()
    for item in arr:
        if item in seen:
            return True  # Found duplicate
        seen.add(item)
    return False

# Template 3: Complement Search Pattern
def complement_pattern(arr, target):
    complements = set()
    for item in arr:
        if target - item in complements:
            return True  # Found pair
        complements.add(item)
    return False

# Template 4: Grouping Pattern
def grouping_pattern(items, key_func):
    groups = {}
    for item in items:
        key = key_func(item)
        if key not in groups:
            groups[key] = []
        groups[key].append(item)
    return groups
        

🎤 What to Say in Interviews:
1. "Let me think about using a hash table for O(1) lookups..."
2. "I'll use a dictionary to map X to Y..."
3. "This looks like a frequency counting problem..."
4. "I can optimize this from O(n²) to O(n) using hashing..."
5. "The trade-off is O(n) extra space for faster time complexity..."

8.8 Final Interview Checklist

            ✅ Before submitting your solution:
            Explain time and space complexity
Handle edge cases (empty input, single element)
Consider collision scenarios
Mention load factor if relevant
Test with example inputs
Discuss alternative approaches

        

🌟 Sample Interview Response:

"For this problem, I notice we need to find pairs that sum to a target. This suggests using a hash table to store complements. As I iterate through the array, I'll check if the current element's complement exists in my hash table. If yes, I've found a pair. If no, I'll add the current element to the table. This gives us O(n) time complexity and O(n) space complexity, which is much better than the O(n²) brute force approach."

🎓 Master Hashing = Master Interviews! 🎓

Practice these concepts daily and you'll be ready for any hashing-related interview question!