🚀 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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