Coding Interview

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Sure! Here's the text with the asterisks replaced by double asterisks: --- 🚀 Coding Interview Questions with Answers — Part 4 🗂️ Stacks, Queues & Heaps 🚀 31. How do you implement a stack with a max-stack (O(1) max query)? A Max Stack supports: • Push • Pop • Get Maximum Element in O(1) 🔹 Idea Maintain: 1. Main stack 2. Max stack Max stack stores current maximums. 🔹 Python Solution

    
        class MaxStack:
    def __init__(self):
        self.stack = []
        self.max_stack = []

    def push(self, value):
        self.stack.append(value)

        if not self.max_stack or value >= self.max_stack[-1]:
            self.max_stack.append(value)

    def pop(self):
        if self.stack[-1] == self.max_stack[-1]:
            self.max_stack.pop()

        return self.stack.pop()

    def get_max(self):
        return self.max_stack[-1]
{}
    

🔹 Complexity Operation - Complexity Push - O(1)  Pop - O(1)  Get Max - O(1)  🔹 Interview Tip Very common design-based stack question. 🚀 32. How do you implement a queue using two stacks? Queues are FIFO. Stacks are LIFO.  We can combine two stacks. 🔹 Idea Stack1 → enqueue  Stack2 → dequeue 🔹 Python Solution

    
        class Queue:
    def __init__(self):
        self.s1 = []
        self.s2 = []

    def enqueue(self, value):
        self.s1.append(value)

    def dequeue(self):
        if not self.s2:
            while self.s1:
                self.s2.append(self.s1.pop())

        return self.s2.pop()
{}
    

🔹 Complexity Operation - Complexity Enqueue - O(1)  Dequeue - Amortized O(1)  🔹 Interview Tip Interviewers love this because it tests understanding of stack behavior. 🚀 33. How do you design a stack that supports getMin() in O(1)? Very similar to Max Stack. 🔹 Idea Maintain: • Main stack • Min stack 🔹 Python Solution

    
        class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []

    def push(self, value):
        self.stack.append(value)

        if not self.min_stack or value <= self.min_stack[-1]:
            self.min_stack.append(value)

    def pop(self):
        if self.stack[-1] == self.min_stack[-1]:
            self.min_stack.pop()

        return self.stack.pop()

    def get_min(self):
        return self.min_stack[-1]
{}
    

🔹 Complexity Operation - Complexity Push - O(1)  Pop - O(1)  Get Min - O(1)  🔹 Interview Tip This is one of the highest-frequency interview problems. 🚀 34. What is a monotonic stack and when is it useful? A monotonic stack maintains elements in: • Increasing order OR • Decreasing order 🔹 Uses ✅ Next Greater Element  ✅ Largest Rectangle in Histogram  ✅ Stock Span Problem  ✅ Daily Temperatures  🔹 Example

    
        arr = [2, 1, 3]

stack = []

for num in arr:
    while stack and stack[-1] > num:
        stack.pop()

    stack.append(num)
{}
    

🔹 Complexity Most monotonic stack problems:  O(n)  because every element is pushed and popped once. 🔹 Interview Tip Extremely important pattern for medium/hard problems. 🚀 35. How do you implement a priority queue / heap? A heap is a complete binary tree. Types: • Min Heap • Max Heap 🔹 Python Min Heap

    
        import heapq

heap = []

heapq.heappush(heap, 10)
heapq.heappush(heap, 5)
heapq.heappush(heap, 20)

print(heapq.heappop(heap))
{}
    

🔹 Output 5 🔹 Complexity Operation - Complexity Insert - O(log n)  Delete - O(log n)  Peek - O(1)  🔹 Uses ✅ Task scheduling  ✅ Dijkstra’s algorithm  ✅ Top K problems  ✅ Priority processing  🚀 36. How do you find the top K frequent elements? 🔹 Approach 1. Count frequency using hashmap 2. Use heap 🔹 Python Solution

    
        from collections import Counter
import heapq

def top_k(nums, k):
    freq = Counter(nums)

    return heapq.nlargest(k, freq.keys(), key=freq.get)

print(top_k([1, 1, 1, 2, 2, 3], 2))
{}
    

🔹 Output [1, 2] 🔹 Complexity Complexity - Value Time - O(n log k)  Space - O(n)  🔹 Interview Tip Heap + hashmap combination is frequently tested. 🚀 37. How do you merge K sorted lists? 🔹 Efficient Approach Use a Min Heap. Heap stores:  smallest current node 🔹 Python Idea

    
        import heapq{}
    

    
        heapq.heappush(heap, (node.val, node))
{}
    

Repeatedly: • Pop smallest node • Add next node from same list 🔹 Complexity  Complexity - Value  Time - O(n log k)  Space - O(k)  Where:  n = total nodes  k = number of lists  🔹 Interview Tip  Very common hard interview problem. 🚀 38. How do you implement LRU / LFU cache? 🔹 LRU Cache  LRU: Least Recently Used  Remove least recently accessed item. 🔹 Efficient Design  Use:  1. HashMap 2. Doubly Linked List 🔹 Python LRU Example

    
        from collections import OrderedDict

class LRUCache:
    def __init__(self, capacity):
        self.cache = OrderedDict()
        self.capacity = capacity

    def get(self, key):
        if key not in self.cache:
            return -1

        self.cache.move_to_end(key)

        return self.cache[key]

    def put(self, key, value):
        if key in self.cache:
            self.cache.move_to_end(key)

        self.cache[key] = value

        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)
{}
    

🔹 Complexity  Operation - Complexity  Get - O(1)  Put - O(1)  🔹 Interview Tip  LRU cache is a FAANG-favorite system design question. 🚀 39. How do you check for balanced parentheses? Use a stack. 🔹 Idea  • Push opening brackets. • When closing bracket appears: Check top of stack 🔹 Python Solution

    
        def is_valid(s):
    stack = []

    mapping = {
        ')': '(',
        '}': '{',
        ']': '['
    }

    for char in s:
        if char in mapping.values():
            stack.append(char)

        elif char in mapping:
            if not stack or stack.pop() != mapping[char]:
                return False

    return not stack

print(is_valid("({[]})"))
{}
    

🔹 Output  True  🔹 Complexity  Complexity - Value  Time - O(n)  Space - O(n)  🔹 Uses  ✅ Compilers  ✅ Expression parsing  ✅ Syntax validation  🚀 40. How do you implement a circular queue? Circular queue reuses empty spaces efficiently. 🔹 Visualization  Front → [1,2,3,_,_]  After dequeue + enqueue:  [,2,3,4,]  🔹 Python Implementation

    
        class CircularQueue:
    def __init__(self, size):
        self.queue = [None] * size
        self.front = 0
        self.rear = 0
        self.size = size
        self.count = 0

    def enqueue(self, value):
        if self.count == self.size:
            return "Full"

        self.queue[self.rear] = value
        self.rear = (self.rear + 1) % self.size
        self.count += 1

    def dequeue(self):
        if self.count == 0:
            return "Empty"

        value = self.queue[self.front]
        self.front = (self.front + 1) % self.size
        self.count -= 1

        return value
{}
    

🔹 Complexity  Operation - Complexity  Enqueue - O(1)  Dequeue - O(1)  🔹 Real-World Uses  ✅ CPU scheduling  ✅ Streaming systems  ✅ Buffers  ✅ Embedded systems  🔥 Double Tap ❤️ For Part-5

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🚀 Coding Interview Questions with Answers — Part 3  🔗 Linked Lists 🚀 21. How do you reverse a singly linked list?  A singly linked list can be reversed by changing the direction of pointers. 🔹 Example  Before:  1 → 2 → 3 → NULL  After:  3 → 2 → 1 → NULL  🔹 Iterative Solution

    
        class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def reverse(head):
    prev = None
    current = head

    while current:
        next_node = current.next
        current.next = prev

        prev = current
        current = next_node

    return prev
{}
    

🔹 Complexity  Time → O(n)  Space → O(1)  🔹 Interview Tip  This is one of the most important linked-list questions. 🚀 22. How do you detect a cycle in a linked list?  Use Floyd’s Cycle Detection Algorithm.  Also called: Tortoise and Hare Algorithm  🔹 Idea  • Slow pointer moves 1 step • Fast pointer moves 2 steps • If they meet → cycle exists 🔹 Python Solution

    
        def has_cycle(head):
    slow = fast = head

    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next

        if slow == fast:
            return True

    return False
{}
    

🔹 Complexity  Time → O(n)  Space → O(1)  🔹 Interview Tip  Very common interview question. 🚀 23. How do you find the middle node of a linked list?  Use two pointers. 🔹 Approach  • Slow pointer → moves 1 step • Fast pointer → moves 2 steps When fast reaches end:  slow = middle  🔹 Python Solution

    
        def middle_node(head):
    slow = fast = head

    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next

    return slow
{}
    

🔹 Complexity  Time → O(n)  Space → O(1)  🔹 Interview Tip  Two-pointer technique is heavily used in linked lists. 🚀 24. How do you merge two sorted linked lists?  🔹 Example  1 → 3 → 5  2 → 4 → 6  Merged:  1 → 2 → 3 → 4 → 5 → 6  🔹 Python Solution

    
        def merge_lists(l1, l2):
    dummy = Node(0)
    current = dummy

    while l1 and l2:
        if l1.data < l2.data:
            current.next = l1
            l1 = l1.next
        else:
            current.next = l2
            l2 = l2.next

        current = current.next

    current.next = l1 or l2

    return dummy.next
{}
    

🔹 Complexity  Time → O(n + m)  Space → O(1)  🔹 Interview Tip  This problem is the base concept behind merge sort on linked lists. 🚀 25. How do you find and remove a duplicate in a list?  🔹 Using HashSet

    
        def remove_duplicates(head):
    seen = set()

    current = head
    prev = None

    while current:
        if current.data in seen:
            prev.next = current.next
        else:
            seen.add(current.data)
            prev = current

        current = current.next

    return head
{}
    

🔹 Complexity  Time → O(n)  Space → O(n)  🔹 Without Extra Space  Can also be solved using nested loops: O(n²)  🔹 Interview Tip  Interviewers may ask: Can you solve it without extra memory? 🚀 26. How do you implement a dummy head in linked-list problems?  A dummy node simplifies edge cases. 🔹 Why Useful?  Without dummy node: Handling head insertion/deletion becomes complex  With dummy node: Logic becomes cleaner  🔹 Example

    
        dummy = Node(0)
dummy.next = head
{}
    

🔹 Use Cases  ✅ Remove nodes  ✅ Merge lists  ✅ Partition lists  ✅ Reverse sublists  🔹 Interview Tip  Using dummy nodes often makes solutions cleaner and bug-free. 🚀 27. How do you delete a node given only that node (no head)?  Important constraint: No access to head pointer  🔹 Trick  Copy next node value into current node. 🔹 Python Solution

    
        def delete_node(node):
    node.data = node.next.data
    node.next = node.next.next
{}
    

🔹 Limitation  Cannot delete last node because no next node exists. 🔹 Interview Tip  Classic interview trick question. 🚀 28. How do you implement a circular linked list? In a circular linked list: Last node → points to head instead of NULL. 🔹 Visualization  1 → 2 → 3  ↑       ↓  ← ← ← ← 🔹 Python Example class Node:   def init(self, data):     self.data = data     self.next = None

head = Node(1) second = Node(2) third = Node(3) head.next = second second.next = third third.next = head 🔹 Uses ✅ Round-robin scheduling ✅ Multiplayer games ✅ Music playlists ✅ CPU scheduling 🚀 29. How do you split a list into equal parts? 🔹 Approach 1. Count total nodes 2. Divide length 3. Break links carefully 🔹 Example 1 → 2 → 3 → 4 → 5 → 6 Split into 2 parts: 1 → 2 → 3 4 → 5 → 6 🔹 Python Idea length = count_nodes(head) part_size = length // k extra = length % k Distribute remaining nodes one by one. 🔹 Complexity Time → O(n) Space → O(1) 🔹 Interview Tip Frequently appears in partitioning problems. 🚀 30. How do you implement a doubly linked list? A doubly linked list stores: prev pointer + next pointer 🔹 Visualization NULL ← 1 ⇄ 2 ⇄ 3 → NULL 🔹 Python Implementation class Node:   def init(self, data):     self.data = data     self.prev = None     self.next = None 🔹 Advantages ✅ Bidirectional traversal ✅ Easier deletion ✅ Efficient backtracking 🔹 Disadvantages ❌ More memory ❌ Extra pointer management 🔹 Real-World Uses ✅ Browser history ✅ Undo/redo ✅ Navigation systems ✅ Music players 🔹 Complexity Insert/Delete → O(1) Search → O(n) 🔥 Double Tap ❤️ For Part-4

🚀 Coding Interview Questions with Answers — Part 2 🌱 Arrays, Strings & Two-Pointers 🚀 11. How do you remove duplicates from a sorted array? Since the array is already sorted, duplicates appear together. 🔹 Best Approach Use the Two-Pointer Technique. - One pointer tracks unique elements - Another scans the array 🔹 Python Solution def remove_duplicates(arr):     if not arr:         return 0     i = 0     for j in range(1, len(arr)):         if arr[j]!= arr[i]:             i += 1             arr[i] = arr[j]     return i + 1 arr = [1,1,2,2,3,4,4] length = remove_duplicates(arr) print(arr[:length]) 🔹 Output [1][2][3][4] 🔹 Complexity Time → O(n) Space → O(1) 🔹 Interview Tip This is one of the most common two-pointer interview problems. 🚀 12. How do you solve “Two Sum” efficiently? Problem: Find two numbers whose sum equals target. 🔹 Brute Force for i in range(len(arr)):     for j in range(i+1, len(arr)):         if arr[i] + arr[j] == target:             return [i, j] Complexity → O(n²) 🔹 Optimized HashMap Solution def two_sum(arr, target):     hashmap = {}     for i, num in enumerate(arr):         complement = target - num         if complement in hashmap:             return [hashmap[complement], i]         hashmap[num] = i print(two_sum([2,7,11,15], 9)) 🔹 Output [0][1] 🔹 Complexity Time → O(n) Space → O(n) 🔹 Interview Tip Hashing is the key optimization here. 🚀 13. How do you reverse a string or array? 🔹 Reverse String s = "hello" print(s[::-1]) Output → olleh 🔹 Two-Pointer Method def reverse_array(arr):     left = 0     right = len(arr) - 1     while left < right:         arr[left], arr[right] = arr[right], arr[left]         left += 1         right -= 1     return arr print(reverse_array([1,2,3,4])) 🔹 Complexity Time → O(n) Space → O(1) 🔹 Interview Tip Interviewers often prefer the two-pointer approach. 🚀 14. How do you find the maximum subarray sum (Kadane’s Algorithm)? Problem: Find contiguous subarray with maximum sum. 🔹 Kadane’s Algorithm def max_subarray(arr):     current_sum = arr[0]     max_sum = arr[0]     for num in arr[1:]:         current_sum = max(num, current_sum + num)         max_sum = max(max_sum, current_sum)     return max_sum print(max_subarray([-2,1,-3,4,-1,2,1,-5,4])) 🔹 Output 6 Subarray: [4, -1, 2, 1] 🔹 Complexity Time → O(n) Space → O(1) 🔹 Interview Tip Kadane’s Algorithm is a very high-frequency interview question. 🚀 15. How do you rotate an array? Rotate array by k positions. 🔹 Python Solution def rotate(arr, k):     k = k % len(arr)     return arr[-k:] + arr[:-k] print(rotate([1,2,3,4,5], 2)) 🔹 Output [4][5][1][2][3] 🔹 Complexity Time → O(n) Space → O(n) 🔹 In-Place Optimization Can be solved in O(1) extra space using reversal algorithm. 🚀 16. How do you find the first missing positive number? Problem: Find smallest missing positive integer. Example: [3,4,-1,1] Output: 2 🔹 Optimized Solution Idea Place each number at its correct index. 1 → index 0 2 → index 1 🔹 Python Solution def first_missing_positive(nums):     n = len(nums)     for i in range(n):         while 1 <= nums[i] <= n and nums[nums[i]-1]!= nums[i]:             nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]     for i in range(n):         if nums[i]!= i + 1:             return i + 1     return n + 1 print(first_missing_positive([3,4,-1,1])) 🔹 Complexity Time → O(n) Space → O(1) 🔹 Interview Tip This is considered a hard interview problem. 🚀 17. How do you implement sliding-window problems? Sliding window helps optimize subarray/substring problems. 🔹 Example Problem Maximum sum of subarray of size k. def max_sum(arr, k):     window_sum = sum(arr[:k])     max_sum = window_sum     for i in range(k, len(arr)):         window_sum += arr[i] - arr[i-k]         max_sum = max(max_sum, window_sum)     return max_sum print(max_sum([1,2,3,4,5], 3)) 🔹 Output 12