LeetCode Solution: Find Unique Binary String in Java

Day 94 - LeetCode Journey Solved LeetCode 35: Search Insert Position in Java ✅ Classic Binary Search problem that tests your fundamentals. Instead of just finding the element, the twist is to return the correct insert position if it’s not present. The key idea is simple: keep narrowing the search space and finally return low, which represents the right position. Clean logic, high impact 💡 Key takeaways: • Strong grip on Binary Search fundamentals • Understanding search space boundaries • Returning correct insertion index • Writing efficient O(log n) solutions ✅ All test cases passed ⚡ O(log n) time and O(1) space Mastering basics like Binary Search is what builds real problem-solving strength 🔥 #LeetCode #DSA #Java #BinarySearch #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode

Day 94/100: #LeetCodeChallenge – Grid Partitioning in Java 🧩💻 Another day, another algorithmic deep dive! Today’s problem was about determining whether a grid can be partitioned in a specific way. While the problem may seem straightforward at first, the real challenge lies in handling edge cases, optimizing for efficiency, and ensuring clean, maintainable code. 🔍 Key takeaways from today’s solution: Understanding 2D array traversal and prefix sums Handling edge cases like 1x1 grids early Writing readable code that can scale with larger test cases Even though the sample output shows "You must run your code first," the process of thinking through the logic, testing edge cases, and refining the approach is where the real growth happens. Every problem adds another tool to the problem-solving toolkit. 🚀 Consistency > Intensity. Day 94 is in the books. On to the final sprint! #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving#GridPartitioning #DataStructuresAndAlgorithms #TechJourney#SoftwareEngineering #CodeNewbie #DeveloperLife #AlgorithmDesign#ConsistencyIsKey

🚀 Day 41/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 3427. Sum of Variable Length Subarrays Used a nested loop approach to calculate subarray sums for each index by dynamically determining the starting point using max(0, i - nums[i]). ⏱️ Time Complexity: O(n²) 📦 Space Complexity: O(1) Strengthening understanding of subarray problems and index-based range calculations. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 98 - #100DaysOfCode Today’s problem was all about string comparison and operations simulation in Java. 💡 Problem Insight: Given a list of operations like "++X", "X++", "--X", "X--", we need to compute the final value of X after performing all operations. ⚠️ One key learning today: In Java, always use .equals() for string comparison instead of ==. Using == compares references, not actual content — a very common mistake! 🧠 Approach: Initialize x = 0 Traverse through each operation Increment or decrement based on the operation string 📌 What I Improved Today: Better understanding of string handling in Java Avoiding common pitfalls in comparisons Writing cleaner conditional logic #Java #CodingJourney #LeetCode #100DaysOfCode #ProblemSolving #Consistency

Refactoring for Clarity: Array Manipulation in Java 👨💻 I’ve just finished a practical exercise on array manipulation. While the goal was simple (summing two arrays), I used it as an opportunity to apply improvements. - Method decomposition: Separated concerns into specialized methods (Read, Calculate, Display). - Input validation: Built a robust loop to handle invalid inputs and prevent crashes. - Data Formatting: Used printf to create a clear, readable results table. Small improvements in logic and organization make a huge difference in software quality. #Java #Algorithms #CleanCode #Backend #LearningToCode

Day 14/100 – LeetCode Challenge 🚀 Problem: #169 Majority Element   Difficulty: Easy   Language: Java   Approach: Sorting + Middle Element   Time Complexity: O(n log n)   Space Complexity: O(1) 🔍 Key Insight: The majority element appears **more than ⌊n / 2⌋ times** in the array. If we **sort the array**, the majority element must occupy the **middle position** because it appears more than half of the time. Therefore, the element at index **n/2** will always be the majority element. 🧠 Solution Brief: First sorted the array using `Arrays.sort()`. Since the majority element appears more than half of the array length, it will always be positioned at the middle index. Finally returned `nums[nums.length / 2]` as the majority element. 📌 What I Learned: Understanding problem constraints can simplify the solution significantly. Sometimes a simple observation (like majority occupying the middle after sorting) can avoid more complex implementations. #LeetCode #Day14 #100DaysOfCode #Java #DSA #Arrays #ProblemSolving #CodingJourney

Day 56 of #100DaysOfLeetCode 💻✅ Solved #205. Isomorphic Strings problem in Java. Approach: • Used two arrays to track character mappings for both strings • Traversed both strings simultaneously • Checked if the mapping values at current characters are equal • If not equal, returned false (mapping mismatch) • Updated both arrays with the current index + 1 • This ensures consistent one-to-one mapping between characters Performance: ✓ Runtime: 8 ms (Beats 72.19% submissions) ✓ Memory: 44.16 MB (Beats 22.15% submissions) Key Learning: ✓ Learned how to maintain mapping consistency between two strings ✓ Practiced using arrays for character indexing ✓ Strengthened understanding of string pattern matching problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #ProblemSolving #CodingJourney #100DaysOfCode

Today I solved “Fibonacci Number” on LeetCode using Java. 💡 Problem Summary The Fibonacci sequence is defined as: F(0) = 0 F(1) = 1 F(n) = F(n-1) + F(n-2) We need to compute F(n). 🧠 Approach I Used (Iterative Optimization) Instead of recursion (which is slow and redundant), I used an iterative approach: Maintain two variables → previous two values Keep updating them in a loop Build the answer step by step ⚙️ Why Not Recursion? Recursion leads to repeated calculations Time complexity becomes O(2ⁿ) ❌ Iterative solution reduces it to O(n) ✅ 📊 Complexity Analysis ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) ⚡ Result ✅ Accepted ⚡ Runtime: 0 ms (Beats 100%) 📚 Key Learning Always try to optimize recursion → iteration Use variables smartly to reduce space Even simple problems teach important optimization patterns Slowly building strong fundamentals, one problem at a time 💯 Day 15 done. Let’s keep the streak alive 🔥 #DSA #LeetCode #Java #100DaysOfCode #Algorithms #ProblemSolving #CodingJourney #Consistency #LearningInPublic

1980. LeetCode Daily Challenge Solved | Find Unique Binary String Solved today’s Find Unique Binary String problem using an elegant Diagonal Construction (Cantor’s Argument) approach. 🔹 Key Idea: Instead of generating all possible binary strings, we construct a new string by flipping the diagonal bits of the given strings. This guarantees the new string differs from every string in the array at least in one position. 🔹 Approach: Traverse each string i Check the i-th character of nums[i] Flip it (0 → 1, 1 → 0) Build a new binary string from these flipped bits This ensures the generated string cannot match any existing string in the input. 🔹 Time Complexity: O(n) 🔹 Space Complexity: O(n) 💻 Java Implementation: class Solution { public String findDifferentBinaryString(String[] nums) { char[] res = new char[nums.length]; for (int i = 0; i < nums.length; i++) { char c = nums[i].charAt(i); res[i] = (c == '0') ? '1' : '0'; } return new String(res); } } ⚡ Result: ✔ Runtime: 0 ms (Beats 100%) ✔ Memory: 42.86 MB Consistency with daily problems builds strong problem-solving intuition. On to the next challenge! #LeetCode #DSA #Java #CodingPractice #ProblemSolving #100DaysOfCode

100 Days of Coding Challenge – Day 24 📌 Problem: Two Sum II – Input Array Is Sorted 💻 Language: Java 🧠 Concept Used: Two Pointer Technique 🔍 Platform: LeetCode Today’s challenge was to find two numbers in a sorted array that add up to a given target and return their 1-indexed positions. The problem guarantees exactly one valid solution and requires constant extra space. Example: Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Approach: ✔ Use two pointers — one at the start (left) and one at the end (right) ✔ Calculate the sum of both elements ✔ If the sum equals the target → return their indices ✔ If the sum is smaller → move left forward ✔ If the sum is larger → move right backward Time Complexity: O(n) Space Complexity: O(1) 🔗 Problem Link: https://lnkd.in/gcM_dBA7 🔗 Code: https://lnkd.in/gaUrS-Ne #100DaysOfCode #Day24 #Java #DSA #LeetCode #TwoPointers #Arrays #ProblemSolving #CodingJourney