🚀 Coding Interview Questions with Answers (Part 6)

2️⃣2️⃣4️⃣ Find the Height of a Binary Tree
👉 Recursively find the max depth of left and right subtrees.

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

def tree_height(root):
    if not root:
        return 0
    return 1 + max(tree_height(root.left), tree_height(root.right))

⏱ O(n)

2️⃣2️⃣5️⃣ Perform an Inorder Traversal of a Binary Tree
👉 Visit left subtree, then root, then right subtree.

def inorder(root, result=None):
    if result is None:
        result = []
    if root:
        inorder(root.left, result)
        result.append(root.data)
        inorder(root.right, result)
    return result

⏱ O(n)

2️⃣2️⃣6️⃣ Perform a Level Order Traversal (BFS) of a Binary Tree
👉 Use a queue to visit nodes level by level.

from collections import deque

def level_order(root):
    result = []
    queue = deque([root])
    while queue:
        node = queue.popleft()
        if node:
            result.append(node.data)
            queue.append(node.left)
            queue.append(node.right)
    return result

⏱ O(n)

2️⃣2️⃣7️⃣ Check if a Binary Tree is a Valid BST
👉 Recursively verify each node falls within a valid min/max range.

def is_valid_bst(root, low=float('-inf'), high=float('inf')):
    if not root:
        return True
    if not (low < root.data < high):
        return False
    return (is_valid_bst(root.left, low, root.data) and
            is_valid_bst(root.right, root.data, high))

⏱ O(n)

2️⃣2️⃣8️⃣ Find the Lowest Common Ancestor in a BST
👉 Traverse down; split point where paths diverge is the LCA.

def lowest_common_ancestor(root, p, q):
    while root:
        if p < root.data and q < root.data:
            root = root.left
        elif p > root.data and q > root.data:
            root = root.right
        else:
            return root.data

⏱ O(h)

2️⃣2️⃣9️⃣ Check if Two Binary Trees are Identical
👉 Compare values and recursively check both subtrees.

def is_identical(t1, t2):
    if not t1 and not t2:
        return True
    if not t1 or not t2:
        return False
    return (t1.data == t2.data and
            is_identical(t1.left, t2.left) and
            is_identical(t1.right, t2.right))

⏱ O(n)

2️⃣3️⃣0️⃣ Find the Diameter of a Binary Tree
👉 The longest path between any two nodes — may or may not pass through root.

def diameter(root):
    result = [0]
    def depth(node):
        if not node:
            return 0
        left = depth(node.left)
        right = depth(node.right)
        result[0] = max(result[0], left + right)
        return 1 + max(left, right)
    depth(root)
    return result[0]

⏱ O(n)

💬 Save this for your next interview prep! Should Part 7 cover Sorting Algorithms, Stacks & Queues, or Graphs? 👇

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🚀 Coding Interview Questions with Answers (Part 7)

2️⃣3️⃣1️⃣ Implement Bubble Sort
👉 Repeatedly compare adjacent elements and swap them if they are in the wrong order.

def bubble_sort(arr):
    n = len(arr)
    for i in range(n):
        for j in range(0, n - i - 1):
            if arr[j] > arr[j + 1]:
                arr[j], arr[j + 1] = arr[j + 1], arr[j]
    return arr

⏱ O(n²)

2️⃣3️⃣2️⃣ Implement Selection Sort
👉 Find the smallest element and place it at the correct position.

def selection_sort(arr):
    n = len(arr)
    for i in range(n):
        min_index = i
        for j in range(i + 1, n):
            if arr[j] < arr[min_index]:
                min_index = j
        arr[i], arr[min_index] = arr[min_index], arr[i]
    return arr

⏱ O(n²)

2️⃣3️⃣3️⃣ Implement Insertion Sort
👉 Build the sorted array one element at a time.

def insertion_sort(arr):
    for i in range(1, len(arr)):
        key = arr[i]
        j = i - 1

        while j >= 0 and arr[j] > key:
            arr[j + 1] = arr[j]
            j -= 1

        arr[j + 1] = key

    return arr

⏱ O(n²)

2️⃣3️⃣4️⃣ Implement Merge Sort
👉 Divide the array into smaller parts, sort them, and merge them.

def merge_sort(arr):
    if len(arr) <= 1:
        return arr

    mid = len(arr) // 2
    left = merge_sort(arr[:mid])
    right = merge_sort(arr[mid:])

    result = []
    i = j = 0

    while i < len(left) and j < len(right):
        if left[i] < right[j]:
            result.append(left[i])
            i += 1
        else:
            result.append(right[j])
            j += 1

    result.extend(left[i:])
    result.extend(right[j:])

    return result

⏱ O(n log n)

2️⃣3️⃣5️⃣ Implement Quick Sort
👉 Select a pivot and partition the array around it.

def quick_sort(arr):
    if len(arr) <= 1:
        return arr

    pivot = arr[-1]
    left = [x for x in arr[:-1] if x <= pivot]
    right = [x for x in arr[:-1] if x > pivot]

    return quick_sort(left) + [pivot] + quick_sort(right)

⏱ Average O(n log n) | Worst O(n²)

2️⃣3️⃣6️⃣ Implement a Stack Using a List
👉 Use the end of the list for efficient push and pop operations.

class Stack:
    def __init__(self):
        self.items = []

    def push(self, item):
        self.items.append(item)

    def pop(self):
        if self.items:
            return self.items.pop()
        return None

    def peek(self):
        return self.items[-1] if self.items else None

⏱ O(1) for push/pop

2️⃣3️⃣7️⃣ Implement a Queue Using deque
👉 Add elements from the rear and remove them from the front.

from collections import deque

class Queue:
    def __init__(self):
        self.items = deque()

    def enqueue(self, item):
        self.items.append(item)

    def dequeue(self):
        if self.items:
            return self.items.popleft()
        return None

⏱ O(1) for enqueue/dequeue

💬 Save this for your next interview prep!

🔥 Should Part 8 cover Graphs, Dynamic Programming, or Recursion & Backtracking? 👇

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🚀 Advanced Coding Interview Questions with Answers (Part 1)

1️⃣ Find the Longest Substring Without Repeating Characters

👉 Given a string, find the length of the longest substring containing no duplicate characters.

def longest_unique_substring(s):
    seen = set()
    left = 0
    max_length = 0

    for right in range(len(s)):
        while s[right] in seen:
            seen.remove(s[left])
            left += 1

        seen.add(s[right])
        max_length = max(max_length, right - left + 1)

    return max_length

print(longest_unique_substring("abcabcbb"))

📌 Output:

3

⏱ Time Complexity: O(n)
💾 Space Complexity: O(n)

---

2️⃣ Find the Kth Largest Element in an Array

👉 Find the Kth largest element without completely sorting the array.

import heapq

def kth_largest(nums, k):
    heap = nums[:k]
    heapq.heapify(heap)

    for num in nums[k:]:
        if num > heap[0]:
            heapq.heapreplace(heap, num)

    return heap[0]

print(kth_largest([3, 2, 1, 5, 6, 4], 2))

📌 Output:

5

⏱ Time Complexity: O(n log k)
💾 Space Complexity: O(k)

---

3️⃣ Detect a Cycle in a Linked List

👉 Determine whether a linked list contains a cycle using Floyd's Cycle Detection Algorithm.

def has_cycle(head):
    slow = head
    fast = head

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

        if slow == fast:
            return True

    return False

💡 The slow pointer moves one step while the fast pointer moves two steps.

⏱ Time Complexity: O(n)
💾 Space Complexity: O(1)

---

4️⃣ Find the Maximum Subarray Sum

👉 Find the contiguous subarray with the largest sum using Kadane's Algorithm.

def max_subarray_sum(nums):
    current = nums[0]
    maximum = nums[0]

    for num in nums[1:]:
        current = max(num, current + num)
        maximum = max(maximum, current)

    return maximum

print(max_subarray_sum([-2, 1, -3, 4, -1, 2, 1, -5, 4]))

📌 Output:

6

⏱ Time Complexity: O(n)
💾 Space Complexity: O(1)

---

5️⃣ Merge Overlapping Intervals

👉 Given a collection of intervals, merge all overlapping intervals.

def merge_intervals(intervals):
    intervals.sort(key=lambda x: x[0])
    merged = []

    for start, end in intervals:
        if not merged or start > merged[-1][1]:
            merged.append([start, end])
        else:
            merged[-1][1] = max(merged[-1][1], end)

    return merged

print(merge_intervals([[1, 3], [2, 6], [8, 10], [9, 12]]))

📌 Output:

[[1, 6], [8, 12]]

⏱ Time Complexity: O(n log n)
💾 Space Complexity: O(n)

---

💬 Save this for your advanced coding interview preparation!

🔥 Part 2 will cover 5 harder problems on Binary Search, Dynamic Programming, Graphs, Backtracking & Sliding Window.

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🚀 Advanced Coding Interview Questions with Answers (Part 3)
1️⃣1️⃣ Find the Top K Frequent Elements
👉 Given an array, return the k elements that appear most frequently.

from collections import Counter

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

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

📌 Output:

[1, 2]

⏱️ Time Complexity: O(n log n)
💾 Space Complexity: O(n)
1️⃣2️⃣ Generate All Permutations of a String
👉 Generate every possible arrangement of the characters in a string using Backtracking.

def permutations(s):
    result = []

    def backtrack(path, remaining):
        if not remaining:
            result.append("".join(path))
            return

        for i in range(len(remaining)):
            backtrack(
                path + [remaining[i]],
                remaining[:i] + remaining[i + 1:]
            )

    backtrack([], s)
    return result

print(permutations("ABC"))

📌 Output:

['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']

⏱️ Time Complexity: O(n × n!)
💾 Space Complexity: O(n × n!)
1️⃣3️⃣ Find the Minimum Coins for a Given Amount
👉 Given coin denominations, find the minimum number of coins required to make a target amount.

def min_coins(coins, amount):
    dp = [float("inf")] * (amount + 1)
    dp[0] = 0

    for current in range(1, amount + 1):
        for coin in coins:
            if coin <= current:
                dp[current] = min(
                    dp[current],
                    dp[current - coin] + 1
                )

    return dp[amount] if dp[amount] != float("inf") else -1

print(min_coins([1, 2, 5], 11))

📌 Output:

3

💡 5 + 5 + 1 = 11
⏱️ Time Complexity: O(amount × number of coins)
💾 Space Complexity: O(amount)
1️⃣4️⃣ Find the Maximum Product Subarray
👉 Find the contiguous subarray whose elements have the largest product.

def max_product_subarray(nums):
    current_max = nums[0]
    current_min = nums[0]
    result = nums[0]

    for num in nums[1:]:
        if num < 0:
            current_max, current_min = current_min, current_max

        current_max = max(num, current_max * num)
        current_min = min(num, current_min * num)

        result = max(result, current_max)

    return result

print(max_product_subarray([2, 3, -2, 4]))

📌 Output:

6

💡 The maximum product comes from [2, 3].
⏱️ Time Complexity: O(n)
💾 Space Complexity: O(1)
1️⃣5️⃣ Implement an LRU Cache
👉 An LRU (Least Recently Used) Cache removes the item that has not been accessed for the longest time when the cache reaches its capacity.

from collections import OrderedDict

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

    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)

📌 Example:

cache = LRUCache(2)

cache.put(1, "A")
cache.put(2, "B")

print(cache.get(1))

cache.put(3, "C")

print(cache.get(2))

📌 Output:

A
-1

⏱️ Average Time Complexity: O(1) for get() and put()
💾 Space Complexity: O(capacity)
💬 Save this for your advanced coding interview preparation!
🔥 Part 4 will cover 5 advanced problems on Dijkstra's Algorithm, Trie, Union-Find, Matrix & Dynamic Programming.
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🚀 Advanced Coding Interview Questions with Answers (Part 4)
1️⃣6️⃣ Find the Shortest Path Using Dijkstra's Algorithm
👉 Dijkstra's Algorithm finds the shortest path from a source node to other nodes in a graph with non-negative edge weights.

import heapq

def dijkstra(graph, start):
    distances = {node: float("inf") for node in graph}
    distances[start] = 0

    heap = [(0, start)]

    while heap:
        distance, node = heapq.heappop(heap)

        if distance > distances[node]:
            continue

        for neighbor, weight in graph[node]:
            new_distance = distance + weight

            if new_distance < distances[neighbor]:
                distances[neighbor] = new_distance
                heapq.heappush(heap, (new_distance, neighbor))

    return distances

⏱️ Time Complexity: O((V + E) log V)
1️⃣7️⃣ Implement a Trie
👉 A Trie is a tree-based data structure commonly used for prefix searching and autocomplete.

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end = False


class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        node = self.root

        for char in word:
            if char not in node.children:
                node.children[char] = TrieNode()

            node = node.children[char]

        node.is_end = True

    def search(self, word):
        node = self.root

        for char in word:
            if char not in node.children:
                return False
            node = node.children[char]

        return node.is_end

⏱️ Time Complexity: O(L) per operation
L = length of the word
1️⃣8️⃣ Find Connected Components Using Union-Find
👉 Union-Find, also called Disjoint Set Union (DSU), efficiently manages groups of connected elements.

class DSU:
    def __init__(self, n):
        self.parent = list(range(n))

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, a, b):
        root_a = self.find(a)
        root_b = self.find(b)

        if root_a != root_b:
            self.parent[root_b] = root_a

💡 It is commonly used in graph connectivity and Kruskal's algorithm.
⏱️ Amortized Time: Nearly O(1) per operation with path compression and union by rank/size.
1️⃣9️⃣ Rotate a Matrix 90 Degrees Clockwise
👉 Rotate an n × n matrix 90 degrees clockwise in place.

def rotate(matrix):
    n = len(matrix)

    for i in range(n):
        for j in range(i + 1, n):
            matrix[i][j], matrix[j][i] = (
                matrix[j][i],
                matrix[i][j]
            )

    for row in matrix:
        row.reverse()

    return matrix

matrix = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
]

print(rotate(matrix))

📌 Output:

[[7, 4, 1],
 [8, 5, 2],
 [9, 6, 3]]

⏱️ Time Complexity: O(n²)
💾 Space Complexity: O(1)
2️⃣0️⃣ Solve the 0/1 Knapsack Problem
👉 Given items with weights and values, find the maximum value that can be placed in a bag with limited capacity.

def knapsack(weights, values, capacity):
    dp = [0] * (capacity + 1)

    for i in range(len(weights)):
        for w in range(capacity, weights[i] - 1, -1):
            dp[w] = max(
                dp[w],
                dp[w - weights[i]] + values[i]
            )

    return dp[capacity]

print(knapsack([1, 3, 4], [15, 50, 60], 4))

📌 Output:

65

⏱️ Time Complexity: O(n × capacity)
💾 Space Complexity: O(capacity)
💬 Save this for your advanced coding interview preparation!
🔥 Next: Generative AI – Part 9
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