🚀 Day 140 of #150DaysOfCode on LeetCode Solved: 1526. Minimum Number of Increments on Subarrays to Form a Target Array Language: Java The task was to determine the minimum number of subarray increment operations needed to build the target array from all zeros. 🔹 Intuition: Every time the current element is greater than the previous one, it means we need that many new operations to reach the next level. Decreases don’t add extra work since earlier increments already cover them. 🔹 Approach: Start with the first element’s value as the initial count. For each subsequent element, if it’s larger than the previous one, add the difference. The sum of all such differences gives the minimum number of operations required. 🔹 Complexity: Time: O(n) Space: O(1) #LeetCode #150DaysOfCode #Java #CodingChallenge #ProblemSolving #Greedy #Arrays #DSA #TechLearning #Programmer #WomenInTech #CodeJourney #DynamicProgramming
Solved LeetCode 1526: Minimum Increments to Target Array in Java
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✅Day 74 of #100DaysOfLeetCode 1.📌Problem: Reverse Vowels of a String 2.🟩 Difficulty: Easy 3.📍Topic: Two Pointers 4.🎯 Goal: Given a string, reverse only the vowels in the string and return the modified string. Vowels include 'a', 'e', 'i', 'o', 'u' in both lower and upper cases. 5.🧠key idea: Approach 1: Use two pointers, one starting from the beginning and one from the end. Move towards each other, swap the vowels found at each pointer, and continue until all vowels are reversed. #LeetCode #CodingChallenge #100DaysOfCode #Algorithms #TechCareers #CodeNewbie #WomenWhoCode #Java #Programming #DailyCoding #Developer #InterviewPrep #TwoPointers #DataStructures #LearnToCode #ProblemSolving
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Day 43 of #50DaysOfLeetCodeChallenge Problem: Valid Parentheses Approach: Used a stack to keep track of opening brackets. For each closing bracket, checked if it matches the top of the stack. Returned false if there was a mismatch or stack was empty. This ensures all brackets are properly nested and closed. Key Takeaways: Stacks are perfect for problems involving nested structures. Simple traversal with O(n) time complexity and O(n) space. Feeling more confident handling pointers and references in Java! Performance: Using a stack this way really helped me understand how to manage nested elements efficiently. #LeetCode #Java #DSA #CodingChallenge #ProblemSolving #Stack #Programming
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Day 47 of #100DaysOfLeetCode Today's problem: Reverse Words in a String Language used: Java What I did today: I implemented a solution to reverse the order of words in a given string while handling extra spaces efficiently. Used split("\\s+") to handle multiple spaces. Reconstructed the reversed string using StringBuilder. Applied trim() to clean up leading and trailing spaces. Key Takeaways: split("\\s+") is great for ignoring multiple spaces. Always use trim() to ensure a neat final output. StringBuilder makes string manipulation much faster than direct concatenation. #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving #Programming #DeveloperJourney #LearnToCode
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💥 Stop solving 500 random Leetcode questions! If you can master these 7 patterns, you can crack any DSA problem. Most developers don’t need more problems they need clarity on the patterns behind them. 🚀 Here’s what I wish I knew earlier: If you understand 1. Sliding Window 2. Two Pointers 3. Binary Search 4. Recursion 5. Dynamic Programming 6. Greedy 7. Backtracking ...then every new problem will start to look familiar. 💡 Save this post 🔖 and start with one pattern a week. Your consistency will matter more than your question count. 👇 Comment which pattern took you the longest to understand! #DSA #Java #BackendDevelopment #Leetcode #ProgrammersLife
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🚀 Day 129/160 – Total Decoding Messages | Dynamic Programming Today’s challenge was about decoding numeric messages into alphabets (like 1→A, 2→B...26→Z). Used a DP approach to count all valid decoding ways efficiently. 🧠 Key takeaway: Dynamic Programming shines when a problem’s state depends on previous results — here, each position depends on the previous one and the one before it. #100DaysOfCode #GeeksforGeeks #160DaysOfDSA #DynamicProgramming #ProblemSolving #Java #CodingJourney #LearnByDoing
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📌 Day 18/100 - Valid Palindrome (LeetCode 125) 🔹 Problem: Determine whether a string reads the same forward and backward, ignoring case and non-alphanumeric characters. 🔹 Approach: Implemented a two-pointer technique. Skipped all non-alphanumeric characters. Compared characters from both ends in lowercase. Returned true if all matched, otherwise false. 🔹 Key Learning: Two-pointer method keeps logic clean and efficient. Character handling is key when data isn’t uniform. Time complexity: O(n), Space complexity: O(1). Sometimes, solving elegantly is better than solving fast. ✨ #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingJourney
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🚀 Day-74 of #100DaysOfCodeChallenge 💡 LeetCode Problem: 3370. Smallest Number With All Set Bits (Easy) 🧠 Concepts Practiced: Bit Manipulation, Binary Representation, Logical Thinking Today’s challenge was short but conceptually sharp — finding the smallest number ≥ n whose binary representation consists only of 1’s (all set bits). This problem tested understanding of bit patterns and how to efficiently generate numbers with consecutive set bits using bitwise logic. 🔹 Key Idea: Keep generating numbers of the form (1 << k) - 1 (which gives 111...1 in binary) until we find one greater than or equal to n. It’s a neat example of how mathematical patterns meet binary logic in programming! ⚙️ ⚙️ Language: Java ⚡ Runtime: 0 ms (Beats 100.00%) 💾 Memory: 40.59 MB (Beats 87.17%) ✅ Result: 608 / 608 test cases passed — Accepted 🎯 Every problem, no matter how simple, reinforces that clarity in logic is the foundation of clean code 💪 #100DaysOfCode #LeetCode #Java #CodingChallenge #BitManipulation #BinaryLogic #ProblemSolving #Programming #DeveloperJourney #LearningEveryday #TechMindset #CleanCode
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🧠 Day 13: Understanding Loops and Operators in Java Today, I took a deep dive into loops and operators — the real backbone of logical thinking in programming. When I first saw loops, they felt tricky. But once I started solving problems, I realized they’re just smart ways to repeat tasks — and operators help us make decisions inside them. 💻 Here’s what I practiced today: Sum or Product of numbers Terms of an Arithmetic Progression (AP) Reverse of a number Binary ↔ Decimal conversions Square root calculation Checking number sequence Pattern problems: ➤ Mirror number pattern ➤ Inverted number pattern ➤ Star pattern ➤ Triangle of numbers ➤ Diamond of stars ✨ Each problem taught me how to control logic flow, use nested loops effectively, and think step by step like a programmer. 🚀 Every loop I wrote made me more confident about how code really works behind the scenes. 💬 What was the first programming concept that made things finally “click” for you? #Java #Loops #Operators #CodingJourney #ProblemSolving #100DaysOfCode #LearningDSA #CodeNewbie #TechLearning #NeverStopLearning #WomenInTech #BuildInPublic #DeveloperJourney #CodingCommunity #LearnJava #ProgrammingBasics #KeepCoding #CodeNewbie #TechJourney
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NeetCode 21 | LeetCode #76 | Minimum Window Substring Used two pointers and HashMaps to maintain a dynamic window tracking character frequencies. Achieved an optimized O(n) solution with careful window expansion and contraction logic. #NeetCode #LeetCode #Algorithms #Java #DataStructures #Optimization #Coding
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💻 Day 69 of #LeetCode100DaysChallenge Solved LeetCode 264: Ugly Number II a dynamic programming problem that improves sequence generation and pointer-based optimization skills. 🧩 Problem: An ugly number is a positive integer whose prime factors are limited to 2, 3, and 5. Given an integer n, return the n-th ugly number. 💡 Approach — Dynamic Programming with Three Pointers: 1️⃣ Maintain an array dp where dp[i] stores the i-th ugly number. 2️⃣ Use three pointers p2, p3, and p5 to track multiples of 2, 3, and 5 respectively. 3️⃣ At each step, choose the smallest of dp[p2]*2, dp[p3]*3, and dp[p5]*5 as the next ugly number. 4️⃣ Increment the corresponding pointer(s) to avoid duplicates. 5️⃣ Continue until the nth ugly number is found. ⚙️ Complexity: Time: O(N) — one pass to generate all ugly numbers Space: O(N) — for storing the sequence ✨ Key Takeaways: ✅ Strengthened understanding of pointer-based DP techniques. ✅ Learned how to efficiently generate sorted multiplicative sequences. ✅ Practiced optimization over brute-force factor checking. #LeetCode #100DaysOfCode #Java #DynamicProgramming #TwoPointers #Math #DSA #ProblemSolving #CodingJourney #WomenInTech
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