LeetCode 799: Champagne Tower Simulation in Java

🎯 Day 67 of #100DaysOfCode 📌 Problem: Combination Sum Today's challenge was all about finding all unique combinations of candidates where the chosen numbers sum to a given target. The same number can be used unlimited times, and combinations must be unique. 🧠 Approach: Used dynamic programming with a 3D list structure dp[t] stores all combinations that sum to target t Iterated through candidates and built combinations from smaller sums Extended each valid combination by adding current candidate 📊 Stats: ✅ 160/160 test cases passed ⚡ Runtime: 5 ms | Beats 10.10% 💾 Memory: 46.71 MB | Beats 5.88% 📝 Takeaway: This problem highlighted the trade-off between intuitive DP solutions and optimization. While the approach works, there's room for improvement in both runtime and memory efficiency. Backtracking might be a more elegant solution here! 🔗 Problem: Combination Sum 🏷️ #LeetCode #CodingChallenge #Java #DynamicProgramming #Backtracking #Algorithms #TechJourney #ProblemSolving

LeetCode Problem || Find All Possible Stable Binary Arrays II (3130)🚀 🔹 Problem Idea: We are given a number of 0s and 1s and a constraint called limit. The goal is to count the number of possible binary arrays such that no more than limit consecutive identical elements appear. 🔹 Approach: I used Dynamic Programming to track the number of valid arrays formed with: i zeros j ones and the last placed element (0 or 1). The DP state helps ensure that the consecutive limit condition is maintained while building the array. Always enjoying the process of learning and improving problem-solving skills! 💡 #LeetCode #DynamicProgramming #ProblemSolving #Java #CodingPractice #SoftwareEngineering

🚀 Day 522 of #750DaysOfCode 🚀 Today I solved the LeetCode problem “Minimum Changes To Make Alternating Binary String.” 🔹 Problem: Given a binary string consisting of 0s and 1s, we need to make the string alternating (no two adjacent characters are the same). In one operation, we can flip any character (0 → 1 or 1 → 0). The goal is to find the minimum number of operations required. 🔹 Key Idea: An alternating string can only follow two possible patterns: 1️⃣ 010101... 2️⃣ 101010... So we: Count mismatches assuming the string starts with '0'. The other pattern mismatches will be n - mismatch. The minimum of these two values gives the answer. 🔹 Time Complexity: O(n) – we traverse the string once. 📌 Example Input: "1111" Possible alternating strings → "0101" or "1010" Minimum operations → 2 Problems like this highlight how pattern observation can simplify the solution. #leetcode #programming #coding #java #softwaredevelopment #dsa #problemSolving #750DaysOfCode

Day 65: The "One-Liner" Win 🎯 Problem 1784: Check if Binary String Has at Most One Segment of Ones Today was a lesson in simplifying logic. The challenge: check if a binary string contains more than one contiguous segment of '1's, given that the string starts with '1'. The Strategy: • Observation: If there are multiple segments of '1's, they must be separated by at least one '0'. • The Pattern: In a string starting with '1', any "new" segment of ones would look like "01" somewhere in the string. • The Execution: A simple !s.contains("01") handles the entire check. Sometimes we hunt for complex algorithms when a single string method is the ultimate counter. Clean, readable, and passed all test cases. 🚀 #LeetCode #Java #StringManipulation #Coding #Efficiency #DailyCode

🚀 Solved LeetCode 643 – Maximum Average Subarray I Today I worked on strengthening my understanding of the Sliding Window pattern. At first, it looked like a simple subarray problem. But the real challenge was solving it efficiently. The key was: • Use a fixed-size sliding window • Avoid recalculating the full sum every time • Reuse the previous window’s computation • Track the maximum sum carefully This helped me understand: ✔ How fixed sliding window works internally ✔ Why initialization of the first window matters ✔ How optimization improves time complexity to O(n) ✔ Importance of recognizing common DSA patterns Small optimizations make a big difference in performance. Step by step improving problem-solving skills. 💻 #LeetCode #Java #ProblemSolving #DataStructures #CodingJourney

🚀 Day 55 Out of #365DaysOfCode -LeetCode Today I solved the classic Climbing Stairs problem on LeetCode using an optimized Dynamic Programming approach. 🔎 Problem Summary: Given n stairs, you can climb either 1 or 2 steps at a time. The task is to calculate how many distinct ways you can reach the top. 💡 Approach: Recognized the pattern as a Fibonacci sequence Used an iterative DP solution Optimized space complexity to O(1) by storing only the last two computed values Avoided recursion to prevent stack overflow and reduce overhead ⚡ Time Complexity: O(n) 📦 Space Complexity: O(1) Consistency in solving DSA problems is helping me improve logical thinking and problem-solving efficiency. #LeetCode #DataStructures #Algorithms #DynamicProgramming #Java #ProblemSolving #CodingJourney Github link: https://lnkd.in/gGUy_MKZ

Day 37 Implemented a Queue using two Stacks to maintain FIFO order by transferring elements when needed. This approach ensures amortized O(1) time complexity for push, pop, and peek operations. #LeetCode #Java #DataStructures #Stack #Queue #CodingPractice #ProblemSolving #Algorithms #DSA #Programming

#Day32 of #365DaysOfCode Today’s LeetCode Practice: 🔹 Container With Most Water (LeetCode 11) Solved a classic two-pointer optimization problem where the goal is to find two lines that together form a container holding the maximum water. 💡 Key Insight: Start with two pointers at both ends. Calculate area using: width × min(height[left], height[right]) Move the pointer with the smaller height inward, because the smaller height limits the water capacity. This guarantees exploring better possibilities efficiently. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) Consistency > Motivation. Day by day, improving problem-solving and logical thinking skills #LeetCode #ProblemSolving #Java #CodingJourney #FutureEngineer #Consistency #LearningEveryday

🚀 Day 119/500 – DSA LeetCode Challenge Today I solved three problems focused on string manipulation and greedy logic. ✅ Zigzag Conversion – Implemented row traversal using a cycle pattern. TC: O(n) | SC: O(n) ✅ Maximum Frequency Sum of Vowels & Consonants – Used HashMap to track character frequencies and find max vowel & consonant counts. TC: O(n) | SC: O(1) ✅ Integer to Roman – Applied Greedy approach using predefined Roman numeral values. TC: O(1) | SC: O(1) 💡 Key Learning: Many string problems become easier when we identify patterns like cycles, frequency counting, or greedy selection. 👉 Day 119/500 #DSA #Java #500DaysChallenge #ProblemSolving #CodingJourney

Day 15 of Programming Multiple Arrays Today I worked on Multiple Arrays and finding the Largest Repeating Element in a Sorted Array. 🔹 Multiple Arrays Learned how to handle and traverse more than one array in a program. This helps when comparing data, merging arrays, or solving problems involving multiple datasets. 🔹 Largest Repeating Element in a Sorted Array Since the array is sorted, repeating elements appear next to each other. By iterating through the array and counting occurrences, we can efficiently identify the largest element that appears more than once. 💡 What I Practiced Traversing multiple arrays Comparing elements across arrays Counting occurrences in sorted arrays Identifying repeating elements using loops and conditions 🧠 Example Problem Practiced Find the largest repeating element in a sorted array. Example: Array → 1 2 2 3 3 3 4 5 Output → 3 (largest element that repeats) #Day15 #ProgrammingJourney #Java #CodingPractice #ProblemSolving