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 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! 💻✨

Lakhveer Singh Rajput