Monotonic Stack Solution for Next Greater Element

🚀 50 Important Coding Questions – Question 37/50 🔹 Next Greater Element II | LeetCode An extension of the Next Greater Element problem, but with a circular array twist. 📌 Problem Statement Given a circular integer array, return the next greater number for every element. The next greater number of a number x is the first greater number to its right in the array. Since the array is circular, the search may continue from the beginning. If no greater element exists → return -1. Example: Input nums = [1,2,1] Output [2,-1,2] 💡 Approach We use a Monotonic Decreasing Stack. Key idea: 1️⃣ Traverse the array twice (2n) to simulate circular behavior 2️⃣ Use a stack to store indices 3️⃣ Remove elements from stack while they are ≤ current element 4️⃣ The stack top becomes the next greater element 5️⃣ Store results in the answer array This avoids checking the array repeatedly. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 📌 LeetCode Result: ✔ Accepted ⚡ Efficient stack-based implementation 🧠 Concepts Strengthened ✔ Monotonic Stack ✔ Circular array handling ✔ Stack optimization ✔ Efficient traversal techniques 📍 Question 37 of 50 in my “50 Important Coding Questions” series. Step by step building stronger DSA fundamentals 💯 👉 Question 38 coming next! #DSA #LeetCode #Stack #MonotonicStack #CodingInterview #ProblemSolving #CPlusPlus #TechJourney

🚀 50 Important Coding Questions – Question 46/50 🔹 Decode String | LeetCode A classic Stack + String manipulation problem involving nested encoding patterns. 📌 Problem Statement Given an encoded string s, decode it using the rule: k[encoded_string] → repeat encoded_string k times The encoding may be nested. Example: Input s = "3[a2[c]]" Output "accaccacc" Explanation: a2[c] -> acc 3[acc] -> accaccacc 💡 Approach We use a stack to handle nested patterns. Steps: 1️⃣ Traverse the string 2️⃣ Push characters until ] is found 3️⃣ When ] appears: • Pop characters until [ → get substring • Extract the number (k) • Repeat the substring k times • Push back into stack 4️⃣ Continue until string ends 5️⃣ Build final result from stack ⏱ Time Complexity: O(n * k) (due to repeated strings) 📦 Space Complexity: O(n) 📌 LeetCode Result ✔ Accepted ⚡ Efficient stack-based decoding 🧠 Concepts Strengthened ✔ Stack for nested structures ✔ String parsing ✔ Handling multi-digit numbers ✔ Recursion-like behavior using stack 📍 Question 46 of 50 in my “50 Important Coding Questions” series. Only 4 problems left — final stretch! 💯🔥 👉 Question 47 coming next! #DSA #LeetCode #Stack #Strings #CodingInterview #ProblemSolving #CPlusPlus #TechJourney

🚀 **268 / 268 Test Cases Passed — But the Real Win Was Understanding the Pattern** Today I solved **Minimum Window Substring** on LeetCode using the **Sliding Window technique**. ✅ 268/268 test cases passed ⚡ Runtime: 9 ms 💾 Memory: 11.52 MB But the most important thing I learned wasn’t just the solution. It was the **pattern behind the problem.** Many coding interview questions that look completely different actually rely on the **same underlying idea** — the **Sliding Window pattern**. Problems like: • Longest Substring Without Repeating Characters • Minimum Window Substring • Find All Anagrams in a String • Maximum Sum Subarray All follow the same strategy: 👉 **Expand the window** to include new elements 👉 **Track frequency / conditions** 👉 **Shrink the window** when constraints break Once you recognize this pattern, problems that initially look **O(n²)** often become **O(n)** solutions. That realization completely changed how I approach **DSA problems and coding interviews**. So I decided to write a **detailed guide explaining Sliding Window from scratch**, including: ✔ The intuition behind the technique ✔ Fixed vs Variable window patterns ✔ The universal sliding window template ✔ Common interview questions ✔ How to master it step-by-step If you’re preparing for **coding interviews or practicing LeetCode**, mastering this technique alone can help you solve **dozens of problems more efficiently.** 📖 I explained everything here: 👉 https://lnkd.in/gigV7Fg3 Would love to know: **What was the first problem that made Sliding Window “click” for you?** #LeetCode #DSA #Algorithms #CodingInterview #Programming #SoftwareEngineering #ProblemSolving #LearningInPublic

🚀 50 Important Coding Questions – Question 44/50 🔹 Evaluate Reverse Polish Notation | LeetCode A classic Stack problem used to evaluate expressions written in postfix notation. 📌 Problem Statement You are given an array of strings tokens representing an arithmetic expression in Reverse Polish Notation (RPN). Evaluate the expression and return the result. Allowed operators: ➕ Addition ➖ Subtraction ✖ Multiplication ➗ Division Example: Input tokens = ["2","1","+","3","*"] Output 9 Explanation: (2 + 1) * 3 = 9 💡 Approach We use a stack to evaluate the expression. Algorithm: 1️⃣ Traverse the tokens array 2️⃣ If the token is a number → push it onto the stack 3️⃣ If the token is an operator • Pop two numbers from the stack • Apply the operation • Push the result back into the stack 4️⃣ The final stack element is the answer ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 📌 LeetCode Result ✔ Accepted ⚡ Efficient stack-based solution 🧠 Concepts Strengthened ✔ Stack operations ✔ Expression evaluation ✔ Postfix (RPN) notation ✔ Arithmetic operation handling 📍 Question 44 of 50 in my “50 Important Coding Questions” series. Only 6 problems left to complete the challenge 💯 👉 Question 45 coming next! #DSA #LeetCode #Stack #Algorithms #CodingInterview #ProblemSolving #CPlusPlus #TechJourney

🚀 Recursion Basics – Quick Revision for Coding Interviews If you're starting DSA or preparing for placements, mastering recursion is a must. Here’s a crisp revision 👇 🧠 What is Recursion? Recursion is when a function calls itself to solve smaller instances of the same problem. ⚙️ Core Components 1️⃣ Base Case (Stopping Condition) Without this → infinite loop 🚫 2️⃣ Recursive Case (Break the Problem) Solve a smaller version of the same problem. 📌 Golden Pattern 👉 f(n) = f(n-1) 👉 f(n) = f(n-1) + f(n-2) 💡 Classic Examples ✔️ Factorial 👉 n! = n × (n-1)! ✔️ Fibonacci 👉 fib(n) = fib(n-1) + fib(n-2) ✔️ Reverse String 👉 reverse(s) = reverse(s[1:]) + s[0] 🔍 How to Think in Recursion ✅ Break problem into smaller parts ✅ Trust recursion (don’t overthink every call) ✅ Focus on base case + relation ⚡ Common Mistakes ❌ Forgetting base case ❌ Wrong recursive relation ❌ Stack overflow due to infinite calls ❌ Using recursion where iteration is better 📊 Complexity Insight ⏱ Time → depends on number of calls 🧠 Space → recursion stack (O(n)) 🔥 Pro Tip Recursion = foundation of: ✔️ Backtracking ✔️ Dynamic Programming ✔️ Tree/Graph problems 💪 Currently practicing: 📅 Day X of DSA Journey 📌 Focus: Recursion → Building strong fundamentals #DSA #Recursion #CodingInterview #LeetCode #Programming #PlacementPrep #100DaysOfCode

A Nested Loop Is Not Always O(n²)! (Don't make this mistake) When evaluating the time complexity, you might be tempted to say that a nested loop runs in quadratic time. This is not always the case, because for it to run in quadratic time, there are specific requirements, and actually, it might even have a different complexity! Let's take a look at some examples (top to bottom): 1) The inner loop runs n times on each iteration of the outer loop, which evaluates O(n²). This is the simplest and a common case you've probably encountered. 2) In the sliding window algorithm, the outer for loop runs 𝗲𝘅𝗮𝗰𝘁𝗹𝘆 n times, where n = arr.size(), and the while loop can run 𝗮𝘁 𝗺𝗼𝘀𝘁 n times in total, meaning if it executed n times for one iteration, it won't run at all for other iterations. - O(n). This is also known as amortized analysis. 3) The outer loop runs from i = 0 to i < n, while the inner loop runs from j = i to j < n, meaning for each i, it executes (n - i). If we expand this summation, the leading term becomes O(n²) . 4) The inner loop multiples j by 2 at each iteration, meaning it follows the sequence 1, 2, 4, 8, 16. The number of iterations is: j = 2^k ≤ n, which is k = log₂n <-- by the definition of the logarithm. This means that the inner loop runs log n times, which is why the nested loops run in O(n log n) time. --- Liked this post? Check out neetcode.io/li for a better way to prepare for coding interviews.

After repeatedly asking Claude Code to rewrite things in my coding style, I decided to write a skill to do it automatically. Articulating my coding style made me realize that beyond aesthetics, it actually serves an important function: it minimizes the distance between variable references, which means I read code much faster. This is especially useful in the new era of agentic engineering, where the bottleneck is no longer writing code but reading and reviewing it. The skill is called Tighten. Check it out and let me know what you think! Blog post: https://lnkd.in/eHzBHSAp Skill: https://lnkd.in/eg-_y3QV #claudecode #agenticengineering #codex #codingagents

🚀 50 Important Coding Questions – Question 42/50 🔹 Sliding Window Maximum | LeetCode A classic Sliding Window + Monotonic Deque problem widely asked in coding interviews. 📌 Problem Statement Given an array nums and a window size k, return the maximum element in every sliding window of size k. Example: Input nums = [1,3,-1,-3,5,3,6,7] k = 3 Output [3,3,5,5,6,7] Explanation: Each window of size 3 produces the maximum value: [1 3 -1] -> 3 [3 -1 -3] -> 3 [-1 -3 5] -> 5 [-3 5 3] -> 5 [5 3 6] -> 6 [3 6 7] -> 7 💡 Approach We use a Monotonic Deque (Double Ended Queue). Key idea: 1️⃣ Maintain indices in deque 2️⃣ Remove elements from back if they are smaller than current element 3️⃣ Remove elements from front if they are outside the window 4️⃣ The front of deque always holds the maximum element index This keeps the deque sorted in decreasing order. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(k) 📌 LeetCode Result ✔ Accepted ⚡ Efficient O(n) solution 🧠 Concepts Strengthened ✔ Sliding Window Technique ✔ Monotonic Deque ✔ Queue optimization ✔ Efficient window-based computation 📍 Question 42 of 50 in my “50 Important Coding Questions” series. Just 8 problems left to finish the challenge 💯 👉 Question 43 coming next! #DSA #LeetCode #SlidingWindow #Deque #CodingInterview #ProblemSolving #CPlusPlus #TechJourney

I’ve been getting back into solving algorithm problems recently, and I worked through a pretty challenging one: “Delete Columns to Make Sorted III.” At first glance, it looks like a simple string problem. But once you dig in, it becomes a lot more interesting — it’s really about finding a pattern across multiple sequences. 💡 Key idea I learned: Instead of thinking about deleting columns directly, you can reframe the problem as: ➡️ “What is the longest sequence of columns that keeps all strings in sorted order?” Once you see it that way, it turns into a dynamic programming problem — similar to finding a longest increasing subsequence, but across multiple strings. 🧠 What this reinforced for me: Many “hard” problems become easier when you reframe the question Patterns like DP and subsequences show up everywhere Explaining the solution clearly is just as important as solving it I wrote a full breakdown of my approach here: 👉 https://lnkd.in/giN69zAn I’m currently focused on sharpening my problem-solving and backend thinking again. If you’re working through similar problems or preparing for interviews, I’d be interested to hear how you approach these kinds of challenges. #Java #LeetCode #Algorithms #SoftwareEngineering #ProblemSolving

🚀 Day 17/50 – LeetCode Challenge 🧩 Problem: Longest Common Subsequence (LCS) Today I worked on the Longest Common Subsequence problem, which is a classic example of Dynamic Programming and appears frequently in coding interviews. 📌 Problem Summary: Given two strings, the goal is to find the length of the longest subsequence that appears in both strings in the same order (not necessarily contiguous). Example: String 1 → "abcde" String 2 → "ace" Output → 3, because "ace" is the longest common subsequence. 🔍 Approach Used ✔ Used Dynamic Programming (DP) ✔ Created a 2D table to store intermediate results ✔ Compared characters of both strings step by step ✔ If characters matched → increase the subsequence length ✔ Otherwise → take the maximum value from previous results ⏱ Time Complexity: O(m × n) 📦 Space Complexity: O(m × n) 💡 Key Learning ✔ Understanding dynamic programming patterns ✔ Solving problems with overlapping subproblems ✔ Improving logical thinking for sequence comparison problems Problems like LCS are important because they form the foundation for many advanced algorithms used in bioinformatics, text comparison, and version control systems. Consistency in problem solving is the key to improvement 🚀 #LeetCode #DSA #DynamicProgramming #ProblemSolving #CodingJourney #FutureAIEngineer #Consistency