Top 10 Trickiest Programming Questions for Experienced Developers 🚀

 

Top 10 Trickiest Programming Questions for Experienced Developers 🚀

As you grow as a developer, you’re bound to face more complex coding challenges that test your problem-solving skills. These tricky programming questions go beyond the basics and push you to think algorithmically. Here’s a breakdown of the top 10 trickiest programming problems for experienced developers, along with example code snippets to help guide your solutions! 💻🔥

1. Find the Longest Palindromic Substring 🧩

Problem: Given a string, return the longest palindromic substring.

Approach: Use dynamic programming or the “expand around center” technique.

Example Code:

def longest_palindrome(s)
  return s if s.size < 2
  start, max_len = 0, 0
  (0...s.size).each do |i|
    len1 = expand_around_center(s, i, i)
    len2 = expand_around_center(s, i, i + 1)
    len = [len1, len2].max
    if len > max_len
      max_len = len
      start = i - (len - 1) / 2
    end
  end
  s[start, max_len]
end

def expand_around_center(s, left, right)
  while left >= 0 && right < s.size && s[left] == s[right]
    left -= 1
    right += 1
  end
  right - left - 1
end

📘 Explanation: The function expand_around_center checks all possible palindrome centers and returns the longest one.

2. Find Kth Largest Element in an Array 🔢

Problem: Given an unsorted array, find the k-th largest element.

Approach: Use the Quickselect algorithm or sort the array for a simple solution.

Example Code:

def find_kth_largest(nums, k)
  nums.sort[-k]
end

📘 Explanation: The array is sorted, and the k-th largest element is retrieved using the index -k.

3. Trapping Rain Water 🌧️

Problem: Calculate how much water can be trapped between bars given an array of heights.

Approach: Use two pointers to calculate the water trapped at each step.

Example Code:

def trap(height)
  left, right = 0, height.size - 1
  max_left, max_right, water = 0, 0, 0
  while left < right
    if height[left] < height[right]
      if height[left] >= max_left
        max_left = height[left]
      else
        water += max_left - height[left]
      end
      left += 1
    else
      if height[right] >= max_right
        max_right = height[right]
      else
        water += max_right - height[right]
      end
      right -= 1
    end
  end
  water
end

📘 Explanation: The two-pointer technique works by comparing the leftmost and rightmost bars and adjusting the boundaries while accumulating water.

4. Sudoku Solver 🧠

Problem: Solve a Sudoku puzzle by filling in the empty cells.

Approach: Use backtracking to recursively attempt placing numbers in valid positions.

Example Code:

def solve_sudoku(board)
  solve(board)
end

def solve(board)
  (0...9).each do |row|
    (0...9).each do |col|
      if board[row][col] == '.'
        ('1'..'9').each do |char|
          if valid_move?(board, row, col, char)
            board[row][col] = char
            return true if solve(board)
            board[row][col] = '.'
          end
        end
        return false
      end
    end
  end
  true
end

def valid_move?(board, row, col, char)
  (0...9).none? do |i|
    board[row][i] == char || board[i][col] == char || board[3*(row/3) + i/3][3*(col/3) + i%3] == char
  end
end

📘 Explanation: The function checks if a move is valid for each empty cell and backtracks if a solution can’t be found.

5. Word Break Problem 📝

Problem: Given a string and a dictionary of words, determine if the string can be segmented into valid words from the dictionary.

Approach: Use dynamic programming to check valid segmentation points.

Example Code:

def word_break(s, word_dict)
  dp = Array.new(s.size + 1, false)
  dp[0] = true
  (1..s.size).each do |i|
    (0...i).each do |j|
      if dp[j] && word_dict.include?(s[j...i])
        dp[i] = true
        break
      end
    end
  end
  dp[s.size]
end

📘 Explanation: The function creates a boolean dp array to track where valid word breaks occur.

6. Median of Two Sorted Arrays 📈

Problem: Find the median of two sorted arrays of different sizes.

Approach: Utilize binary search for an efficient solution.

Example Code:

def find_median_sorted_arrays(nums1, nums2)
  merged = (nums1 + nums2).sort
  len = merged.size
  len.even? ? (merged[len/2 - 1] + merged[len/2]) / 2.0 : merged[len/2]
end

📘 Explanation: The arrays are merged, sorted, and the median is calculated based on the length being even or odd.

7. Merge Intervals 📅

Problem: Merge overlapping intervals.

Approach: Sort the intervals by start time and merge them while iterating.

Example Code:

def merge(intervals)
  intervals.sort_by!(&:first)
  merged = []
  intervals.each do |interval|
    if merged.empty? || merged.last[1] < interval[0]
      merged << interval
    else
      merged.last[1] = [merged.last[1], interval[1]].max
    end
  end
  merged
end

📘 Explanation: After sorting, overlapping intervals are merged by checking if the current interval overlaps with the last merged one.

8. LRU Cache 📚

Problem: Design and implement a Least Recently Used (LRU) cache.

Approach: Use a hash map and a doubly linked list for O(1) operations.

Example Code:

class LRUCache
  def initialize(capacity)
    @cache = {}
    @capacity = capacity
    @order = []
  end
  
  def get(key)
    return -1 unless @cache.key?(key)
    @order.delete(key)
    @order.unshift(key)
    @cache[key]
  end

  def put(key, value)
    if @cache.key?(key)
      @order.delete(key)
    elsif @cache.size >= @capacity
      lru = @order.pop
      @cache.delete(lru)
    end
    @cache[key] = value
    @order.unshift(key)
  end
end

📘 Explanation: The LRUCache class tracks cache hits with an array for order, and manages a hash for quick access.

9. Reverse Nodes in k-Group 🔄

Problem: Reverse nodes in a linked list in groups of k.

Approach: Reverse k nodes at a time using recursion or iteration.

Example Code:

class ListNode
  attr_accessor :val, :next
  def initialize(val = 0, _next = nil)
    @val = val
    @next = _next
  end
end

def reverse_k_group(head, k)
  count = 0
  node = head
  while node && count < k
    node = node.next
    count += 1
  end
  if count == k
    reversed = reverse_list(head, k)
    head.next = reverse_k_group(node, k)
    return reversed
  end
  head
end

def reverse_list(head, k)
  prev = nil
  current = head
  while k > 0
    next_node = current.next
    current.next = prev
    prev = current
    current = next_node
    k -= 1
  end
  prev
end

📘 Explanation: The list is recursively reversed in groups of k, and the remaining part is handled accordingly.

10. Minimum Window Substring 🔍

Problem: Find the minimum window substring that contains all characters of a target string.

Approach: Use a sliding window technique with two pointers to adjust the window dynamically.

Example Code:

def min_window(s, t)
  return "" if t.empty? || s.empty?
  
  t_count = Hash.new(0)
  t.each_char { |c| t_count[c] += 1 }
  
  required = t_count.size
  l, r = 0, 0
  formed = 0
  window_counts = Hash.new(0)
  ans = [Float::INFINITY, nil, nil]
  
  while r < s.size
    c = s[r]
    window_counts[c] += 1
    
    formed += 1 if t_count[c] && window_counts[c] == t_count[c]
    
    while l <= r && formed == required
      c = s[l]
      
      if r - l + 1 < ans[0]
        ans = [r - l + 1, l, r]
      end
      
      window_counts[c] -= 1
      formed -= 1 if t_count[c] && window_counts[c] < t_count[c]
      l += 1
    end
    r += 1
  end
  
  ans[0] == Float::INFINITY ? "" : s[ans[1]..ans[2]]
end

📘 Explanation: The sliding window dynamically adjusts its size to find the minimum window containing all characters from the target string.

Final Thoughts 💡

Each of these programming problems tests a unique aspect of your coding expertise, from mastering algorithms to designing efficient solutions. Solving these problems isn’t just about finding the correct solution — it’s about thinking critically, optimizing for performance, and understanding the underlying principles.

Happy coding! 💻✨