Solved LeetCode #389 in Java with ASCII-based solution

Day 92 of #100DaysOfLeetCode 💻✅ Solved #739. Daily Temperatures problem in Java. Approach: • Used Monotonic Stack to track indices • Compared current temperature with stack top • Updated result when a warmer day was found • Stored indices for future comparisons Performance: ✓ Runtime: 60 ms (Beats 79.01% submissions) 🚀 ✓ Memory: 107.89 MB (Beats 17.53% submissions) Key Learning: ✓ Learned Monotonic Stack technique ✓ Improved handling of next greater element problems ✓ Strengthened stack-based problem solving Learning one problem every single day 🚀 #Java #LeetCode #DSA #Stack #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Day 76 of #100DaysOfLeetCode 💻✅ Solved #33. Search in Rotated Sorted Array problem in Java. Approach: • Applied Binary Search on rotated array • Checked which half is sorted at each step • Reduced search space based on target position Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 43.80 MB (Beats 70.96% submissions) Key Learning: ✓ Mastered searching in rotated sorted arrays ✓ Improved Binary Search decision-making ✓ Learned efficient handling of edge cases Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

💻 Interesting Java problem: Find the closest target in a circular array Given a words[] array, a target, and a startIndex, the task is to find the shortest circular distance to the target. Leetcode Problem :https://lnkd.in/g3RbdChE Core idea: For every matching target: find direct distance → Math.abs(i - startIndex) find circular distance → n - dist take the minimum. This is a neat example of how circular arrays require us to think beyond normal linear traversal. Small twist, smart logic. 🚀 #Java #Algorithms #DSA #CodingInterview #ProblemSolving #LeetCode

Day 21 of #100DaysOfCode Today, I worked on the classic string problem: implementing the strStr() function in Java. The goal: Find the first occurrence of a substring (needle) inside another string (haystack). Approach I used: - Applied a sliding window technique - Compared substrings using "substring(i, j)" - Returned the starting index when a match is found Key learning: Understanding how index-based string operations work and how "substring()" helps in breaking down problems step-by-step. Also realized the importance of optimizing solutions to avoid unnecessary string creation. Consistency is slowly turning concepts into confidence! Looking forward to improving this further with more optimized approaches #Java #Coding #DSA #LeetCode #ProblemSolving #LearningJourney #Consistency

Day 77 of #100DaysOfLeetCode 💻✅ Solved #153. Find Minimum in Rotated Sorted Array problem in Java. Approach: • Used Binary Search to find minimum element • Compared mid with right to decide direction • Reduced search space until pointers converged Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 43.62 MB (Beats 78.18% submissions) Key Learning: ✓ Strengthened Binary Search on rotated arrays ✓ Learned how to identify unsorted portion efficiently ✓ Improved optimization over linear search Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

#CodeEveryday — My DSA Journey | Day 3 🧩 Problem Solved: Combination Sum III (LeetCode #216) 💭 What I Learned: Used backtracking to find all possible combinations of k numbers (1–9) that sum up to a target n. At each step: ✔️ Decided whether to include the current number ✔️ Reduced both the remaining sum (n) and count (k) ✔️ Ensured each number is used only once by moving to the previous index Applied constraints effectively to prune recursion when: Remaining sum becomes negative Required count becomes invalid This improved my understanding of handling multiple constraints (sum + size) simultaneously in recursion. ⏱ Time Complexity: O(C(9,k) 🧠 Space Complexity: O(k) (recursion depth) ⚡ Key Takeaways: ✔️ Backtracking with multiple constraints requires careful pruning ✔️ Fixed input size can significantly reduce complexity ✔️ Choosing + skipping pattern helps explore all valid combinations 💻 Language Used: Java ☕ 📘 Concepts: Backtracking · Recursion · Combinations · Constraint Handling #CodeEveryday #DSA #LeetCode #Java #Backtracking #ProblemSolving #Algorithms #CodingJourney #Consistency 🚀

Day 93 - LeetCode Journey Solved LeetCode 9: Palindrome Number in Java ✅ At first glance, it feels like a string problem… but the real challenge is solving it without converting to string. Instead of reversing the whole number, I reversed only half of it and compared both parts. This avoids overflow and keeps it efficient. Smart approach > brute force 💡 Key takeaways: • Handling edge cases (negative numbers, trailing zeroes) • Reversing only half of the number • Avoiding extra space (no string conversion) • Writing optimized mathematical logic ✅ All test cases passed ⚡ O(log n) time and O(1) space Sometimes the best solutions are the simplest ones, just a different way of thinking 🔥 #LeetCode #DSA #Java #Math #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode

🤯 Most developers get this wrong! The misconception: You might think the output is 2147483648. After all, Math.abs() should return a positive value… right? The reality: The output is actually -2147483648. Yes — still negative. Why does this happen? 🧠 • Integer limits: In Java, int is a 32-bit signed integer. Integer.MIN_VALUE = -2147483648 • Maximum value: Integer.MAX_VALUE = 2147483647 • The catch (overflow): The positive counterpart of Integer.MIN_VALUE doesn’t fit in an int. Because Java uses two’s complement, it overflows… and wraps right back to the same negative value. How to handle it? 🛠️ Use Math.absExact() if you want to catch this edge case. It throws an ArithmeticException instead of silently overflowing. Drop a 💻 if you knew this, or a 🤯 if this surprised you! #Java #Programming #Coding #DeveloperLife #TechTrivia #ComputerScience #SoftwareEngineering #LearnToCode #CodingMisconceptions

Day 74 of #100DaysOfLeetCode 💻✅ Solved #162. Find Peak Element problem in Java. Approach: • Used Binary Search technique to efficiently find the peak element • Set two pointers, left at start and right at end of the array • Calculated mid index using safe mid formula • Compared nums[mid] with nums[mid + 1] to determine direction • If mid element is smaller, moved search space to right half • Otherwise, moved search space to left half including mid • Continued until left and right pointers converged • Final position (left == right) represents the peak index Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 44.32 MB (Beats 25.49% submissions) Key Learning: ✓ Strengthened understanding of Binary Search on unsorted arrays ✓ Learned how to apply divide-and-conquer beyond traditional searching ✓ Improved intuition for peak finding using neighbor comparison ✓ Practiced optimizing search space instead of linear scanning Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Day 68 of #100DaysOfLeetCode 💻✅ Solved #977. Squares of a Sorted Array problem in Java. Approach: • Traversed the array and squared each element • Used Arrays.sort() to sort the squared values • Returned the sorted array as the result Performance: ✓ Runtime: 10 ms (Beats 37.38% submissions) ✓ Memory: 47.98 MB (Beats 18.78% submissions) Key Learning: ✓ Practiced array transformation and sorting ✓ Learned how squaring affects order in sorted arrays ✓ Understood the importance of optimizing from O(n log n) to O(n) using two pointers Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #Sorting #ProblemSolving #CodingJourney #100DaysOfCode