100DaysOfCode: Solving LeetCode 3100 Water Bottles II

Day 106 - LeetCode Journey Solved LeetCode 2073: Time Needed to Buy Tickets ✅ Looks like a queue simulation problem… but there’s a smarter math-based approach. Instead of simulating the whole process, directly calculate time using observations. Idea: For each person: • If before or at k → contribute min(tickets[i], tickets[k]) • If after k → contribute min(tickets[i], tickets[k] - 1) Key learnings: • Avoiding unnecessary simulation • Observational optimization 💡 • Turning O(n²) thinking into O(n) • Cleaner and faster approach ✅ All test cases passed ⚡ O(n) time | O(1) space Smart thinking > brute force 🚀 #LeetCode #DSA #Queue #MathLogic #Java #CodingJourney #ProblemSolving #InterviewPrep

🚀 Day 30 of My DSA Journey Solved LeetCode Problem #42 – Trapping Rain Water today! 🔍 This problem is a classic example of using the Two Pointer technique to efficiently calculate how much water can be trapped between bars. 💡 Key Learnings: Water trapped depends on the minimum of left and right boundaries Instead of using extra space, we can optimize using two pointers Understanding patterns helps reduce complexity from brute force to optimal 🧠 Approach: Use two pointers (left and right) Track leftMax and rightMax Move the pointer with smaller height: If left is smaller → calculate water using leftMax Else → use rightMax Accumulate total trapped water 💻 Time Complexity: O(n) 📦 Space Complexity: O(1) Consistency + pattern recognition = strong problem-solving 🚀 Day 30 completed — leveling up every day! #Day30 #LeetCode #DSA #Arrays #TwoPointers #Java #CodingJourney #ProblemSolving #TechGrowth

🚀 LeetCode Weekly Contest — Solved 3/4! Here's a quick breakdown of my approach for each problem: 🟢 Q1 — Traffic Signal Color (Easy) Straightforward conditional logic. Mapped timer values to signal colors using if-else checks — clean and concise. 🟡 Q2 — Count Digit Appearances (Medium) Wrote a function to extract digits one-by-one using modulo and division, then aggregated the count across the entire array. 🔴 Q3 — Minimum Operations to Transform Array into Alternating Prime (Medium) For each index, checked whether the element satisfies the prime/non-prime constraint. If not, calculated the cost to reach the next valid number (next prime or next composite) and added it to the total operation count. 🔁 Q4 — Will upsolve and share my learnings! #Consistency #LeetCode #CompetitiveProgramming #DSA #Java #ProblemSolving #CodingContest #SoftwareEngineering #TUF

🚀 Day 66 of #100DaysOfCode Solved 12. Integer to Roman on LeetCode 🔗 🧠 Key Insight: Roman numerals are built using fixed value-symbol pairs 👉 Including special cases like IV (4), IX (9), XL (40), CM (900) ⚙️ Approach (Greedy - Largest to Smallest): 1️⃣ Maintain two arrays: 🔹 Values → [1000, 900, 500, 400, ..., 1] 🔹 Symbols → ["M", "CM", "D", "CD", ..., "I"] 2️⃣ Traverse values from largest to smallest 3️⃣ For each value: 🔹 While num >= value[i]:   • Append corresponding symbol   • Subtract value from num 4️⃣ Stop when num == 0 ⏱️ Time Complexity: O(1) 📦 Space Complexity: O(1) 💡 Edge Cases: ✔️ 4 → "IV" ✔️ 9 → "IX" ✔️ 40 → "XL" ✔️ 900 → "CM" #100DaysOfCode #LeetCode #DSA #Greedy #Strings #Java #InterviewPrep #CodingJourney

47 of #100DaysOfCode 🚀 Solved the 20 Valid Parentheses problem using a Stack-based approach in C++. 🔍 Approach: Traverse the string character by character Push all opening brackets ( { [ onto the stack For every closing bracket: Check if the stack is empty → invalid case Match with the top element of the stack If valid → pop from stack Else → return false At the end, if the stack is empty → valid parentheses ✔️ 💡 Key Insight: Stack helps in maintaining the order (LIFO), making it perfect for matching nested structures. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) #DSA #Cpp #CodingJourney #LeetCode #Stack #ProblemSolving #TechGrowth

🚀 Day 64/100 – LeetCode Challenge. 🔍 Problem: Valid Parentheses. Today’s problem was all about checking whether a string of brackets is valid or not. Sounds simple… but the trick lies in handling it efficiently! 💡 Approach Used: Stack (LIFO) 👉 Idea: Push opening brackets → ( { [ On closing bracket → check top of stack If match → pop Else → invalid ❌ 🧠 Key Insight: Stack helps maintain the correct order of brackets. The last opened bracket must be the first one to close. ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(n) 📌 Takeaway: Whenever you see problems involving matching pairs / nested structures, think of using a stack! #100DaysOfCode #DSA #LeetCode #CodingJourney

🚀 LeetCode Challenge 15/50 💡 Approach: Dutch National Flag Algorithm The problem says no library sort — and even a two-pass counting sort feels like cheating when there's a legendary one-pass algorithm designed exactly for this! Edsger Dijkstra's Dutch National Flag algorithm sorts 3 distinct values in a single traversal! 🔍 Key Insight: → Maintain 3 pointers: low, mid, high → nums[mid] == 0 → swap with low, advance both low & mid → nums[mid] == 1 → already in place, just advance mid → nums[mid] == 2 → swap with high, shrink high only → Stop when mid crosses high 📈 Complexity: ❌ Two-pass Counting Sort → O(n) Time, O(1) Space (2 passes) ✅ Dutch National Flag → O(n) Time, O(1) Space (1 pass only!) Sometimes knowing the right algorithm is the solution itself. Dijkstra didn't just solve sorting — he solved elegance! 🇳🇱 #LeetCode #DSA #DutchNationalFlag #Java #ADA #PBL2 #LeetCodeChallenge #Day15of50 #CodingJourney #ComputerEngineering #AlgorithmDesign #SortColors

62 of #100DaysOfCode 🚀 Solved LeetCode 78 – Subsets using a pure recursive (include/exclude) approach in C++. 🔍 Approach: At each index, I make a decision: ✔️ Include the current element in the subset ❌ Exclude the current element This creates a recursion tree exploring all possibilities, and when we reach the end of the array, we store the current subset. 🧠 Core Idea: Pick → Recurse Backtrack → Skip → Recurse Store result when index reaches n #leetcode #cpp #recursion #backtracking #dsa #100DaysOfCode #codingjourney

🚀 LeetCode Challenge 31/50 💡 Approach: Backtracking Remember the old T9 keyboard on phones? This problem is exactly that — generate every possible letter combination from a sequence of digits using Backtracking! 🔍 Key Insight: → Map each digit (2-9) to its letters (just like a phone keypad) → Recursively build combinations digit by digit → At each step, try every letter mapped to the current digit → When index reaches end of digits → add combination to result → Backtrack by removing last character and try next letter 📈 Complexity: ✅ Time: O(4ⁿ) — at most 4 letters per digit (e.g. 7,9) ✅ Space: O(n) — recursion depth equals digits length Backtracking shines when you need ALL valid combinations — build, explore, undo, repeat! 📱 #LeetCode #DSA #Backtracking #Java #ADA #PBL2 #LeetCodeChallenge #Day31of50 #CodingJourney #ComputerEngineering #AlgorithmDesign #LetterCombinations