🚀 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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