HASHING AND HASH TABLES
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A hash table stores key-value pairs. Uses a hash function to map keys to indices.
Average time: O(1) for insert, delete, search. Worst case: O(n) with collisions.

Python's dict and set are hash tables.

--- Hash Table Basics ---
# Python dict = hash map
phone_book = {}
phone_book["Alice"] = "555-1234"
phone_book["Bob"]   = "555-5678"

print(phone_book["Alice"])          # "555-1234"
print("Alice" in phone_book)        # True
del phone_book["Bob"]
print(phone_book.get("Bob", "N/A")) # "N/A"

# Iterate
for name, number in phone_book.items():
    print(name, number)

--- Build a Hash Map from Scratch ---
class HashMap:
    def __init__(self, size=1000):
        self.size = size
        self.buckets = [[] for _ in range(size)]

    def _hash(self, key):
        return hash(key) % self.size

    def put(self, key, value):
        idx = self._hash(key)
        for i, (k, v) in enumerate(self.buckets[idx]):
            if k == key:
                self.buckets[idx][i] = (key, value)
                return
        self.buckets[idx].append((key, value))

    def get(self, key):
        idx = self._hash(key)
        for k, v in self.buckets[idx]:
            if k == key:
                return v
        return None

    def remove(self, key):
        idx = self._hash(key)
        self.buckets[idx] = [(k, v) for k, v in self.buckets[idx] if k != key]

hm = HashMap()
hm.put("name", "Alice")
hm.put("age", 25)
print(hm.get("name"))   # Alice
hm.remove("age")
print(hm.get("age"))    # None

--- Two Sum Problem ---
Find two numbers that add up to target. Classic hash map problem.
Time: O(n). Space: O(n).

def two_sum(nums, target):
    seen = {}  # value → index
    for i, num in enumerate(nums):
        complement = target - num
        if complement in seen:
            return [seen[complement], i]
        seen[num] = i
    return []

print(two_sum([2, 7, 11, 15], 9))   # [0, 1]
print(two_sum([3, 2, 4], 6))        # [1, 2]

--- Group Anagrams ---
Group words that are anagrams of each other.
Time: O(n * k log k) where k = max word length.

from collections import defaultdict

def group_anagrams(words):
    groups = defaultdict(list)
    for word in words:
        key = ''.join(sorted(word))  # sorted chars as key
        groups[key].append(word)
    return list(groups.values())

print(group_anagrams(["eat","tea","tan","ate","nat","bat"]))
# [["eat","tea","ate"], ["tan","nat"], ["bat"]]

--- Frequency Counter ---
Count frequency of each element.

from collections import Counter

def top_k_frequent(nums, k):
    count = Counter(nums)
    return [x for x, _ in count.most_common(k)]

print(top_k_frequent([1,1,1,2,2,3], 2))  # [1, 2]

--- Longest Consecutive Sequence ---
Find length of longest consecutive elements sequence.
Time: O(n). Space: O(n).

def longest_consecutive(nums):
    num_set = set(nums)
    best = 0
    for num in num_set:
        if num - 1 not in num_set:  # start of sequence
            length = 1
            while num + length in num_set:
                length += 1
            best = max(best, length)
    return best

print(longest_consecutive([100,4,200,1,3,2]))  # 4 (1,2,3,4)

--- Subarray Sum Equals K ---
Count subarrays with sum equal to k. Uses prefix sum + hash map.
Time: O(n). Space: O(n).

def subarray_sum(nums, k):
    count = 0
    prefix = 0
    freq = {0: 1}  # prefix_sum → frequency
    for num in nums:
        prefix += num
        if prefix - k in freq:
            count += freq[prefix - k]
        freq[prefix] = freq.get(prefix, 0) + 1
    return count

print(subarray_sum([1, 1, 1], 2))   # 2
print(subarray_sum([1, 2, 3], 3))   # 2

--- Check if Two Strings are Anagram ---
def is_anagram(s, t):
    return Counter(s) == Counter(t)

print(is_anagram("anagram", "nagaram"))  # True
print(is_anagram("rat", "car"))          # False

--- First Non-Repeating Character ---
def first_unique(s):
    count = Counter(s)
    for i, ch in enumerate(s):
        if count[ch] == 1:
            return i
    return -1

print(first_unique("leetcode"))   # 0 (l)
print(first_unique("aabb"))       # -1

--- Hash Set Uses ---
# O(1) membership test
visited = set()
visited.add(5)
print(5 in visited)   # True
print(10 in visited)  # False

# Remove duplicates
arr = [1, 2, 2, 3, 3, 4]
unique = list(set(arr))
print(unique)  # [1, 2, 3, 4]

# Intersection, Union
a = {1, 2, 3, 4}
b = {3, 4, 5, 6}
print(a & b)  # {3, 4}  intersection
print(a | b)  # {1,2,3,4,5,6} union
print(a - b)  # {1, 2}  difference