LeetCode Hard: Merging k Sorted Linked Lists

🚀 Day 63 of #LeetCode Practice Solved: Count Equal and Divisible Pairs in an Array (Easy) 🔹 Problem Summary: Given an array and an integer k, find the number of pairs (i, j) such that: • nums[i] == nums[j] • (i * j) is divisible by k 🔹 Approach: • Used two nested loops to check all pairs • Compared elements to find equal values • Checked if (i * j) % k == 0 • Counted valid pairs 🔹 Key Learning: Brute-force helps build a strong foundation and improves problem understanding before optimization. 💡 Consistency, patience, and small daily improvements lead to big results over time. #LeetCode #DSA #Java #CodingJourney #Consistency #KeepLearning

#Day72 Of Problem Solving Solved today’s LeetCode Daily Challenge ✅. Problem: Check if Binary String Has at Most One Segment of Ones Sometimes, the simplest logic gives the best results. Instead of overthinking, I went with a clean approach—just checking if "01" exists in the string. If it does, that means more than one segment of 1’s… and that’s it! ✔️ All test cases passed ⚡ Runtime: 0 ms 📊 Beat 100% in performance This problem was a good reminder that not every solution needs complexity—clarity matters more. Consistency > Complexity 🚀 On to the next challenge! #LeetCode #DailyChallenge #Java #ProblemSolving #CodingJourney #100DaysOfCode #LinkedIn

🚀 Day 11 of #100DaysOfCode Solved LeetCode Problem #27 – Remove Element today! 💻 🔹 Approach: Used the Two Pointer Technique One pointer (i) iterates through the array Another pointer (ptr) keeps track of valid elements If nums[i] != val, assign it to nums[ptr] and increment ptr 🔹 Time Complexity: O(n) 🔹 Space Complexity: O(1) 💡 Key Learning: In-place array modification using two pointers helps optimize space and improves efficiency. 📌 Consistency is the key — one step closer every day! #LeetCode #DSA #Java #CodingJourney #Day11 #100DaysOfCode

🚀 Day 50 – LeetCode Challenge 🔍 Problem: Guess Number Higher or Lower Today’s problem is a classic example of Binary Search. Instead of checking every number, we reduce the search space by half each time using hints like higher or lower. 💡 Key Idea: Start with range 1 to n Pick the middle number Use the guess API to decide direction Repeat until we find the correct number ⚡ Time Complexity: O(log n) ⚡ Space Complexity: O(1) 📌 What I learned: Binary Search is very powerful when the problem involves a sorted range or directional hints. Consistency is the key 🔁 Halfway through the journey 💪 #Day50 #LeetCode #BinarySearch #Java #CodingJourney

🚀 Day 94 -#100DaysOfCode Today I worked on a problem that looks simple at first but really tests your understanding of indexing and conditions. 💡 Problem Insight: Given an array, we need to calculate the sum of squares of elements whose indices (1-based) divide the length of the array. 🧠 Key Learning: Always be careful with 1-based vs 0-based indexing The condition n % i == 0 ensures we only pick valid positions Accessing elements correctly using nums[i-1] is crucial ⚡ What I Improved Today: Better understanding of index manipulation Writing cleaner loops with proper conditions Avoiding off-by-one errors #Day94 #CodingJourney #Java #DSA #LeetCode #Consistency #ProblemSolving

🚀 Day 8 of #100DaysOfCode Solved: Maximum Product of Two Elements in an Array 💻 Today’s problem was all about optimizing logic and thinking smart instead of brute force. Instead of checking every pair, I focused on finding the two largest elements efficiently and used them to compute the result in a single pass 🔥 ✅ Time Complexity: O(n) ✅ Space Complexity: O(1) Small problems like these really sharpen problem-solving skills and reinforce the importance of clean, efficient code. Consistency is key — showing up every day, learning something new, and getting 1% better 💯 #DSA #Java #CodingJourney #LeetCode #ProblemSolving #Consistency #Day8#DSAwithEdSlash

Day 98 - LeetCode Journey Solved LeetCode 20: Valid Parentheses in Java ✅ A classic stack problem that looks simple but tests your attention to detail. The idea is straightforward: Push opening brackets into stack and match them correctly when a closing bracket appears. If anything mismatches → invalid. Clean logic, zero confusion 💡 Key takeaways: • Stack fundamentals • Matching pairs efficiently • Handling edge cases (empty stack, wrong order) • Writing clean conditional logic ✅ All test cases passed ⚡ O(n) time and O(n) space Sometimes basics like this are the real building blocks of strong DSA 🔥 #LeetCode #DSA #Java #Stack #ProblemSolving #CodingJourney #InterviewPrep #Consistency #100DaysOfCode

Day 20/100 🚀 | LeetCode Grind Cracked “Guess Number Higher or Lower” using Binary Search 🎯 Optimized the approach by narrowing down the search space efficiently with each guess — classic divide & conquer in action. 💡 Key Takeaway: When the search space is sorted and bounded, Binary Search is your best friend. ⚡ Runtime: 0 ms (Beats 100%) 📊 Efficient and clean implementation Consistency > Intensity. Showing up every day 💪 #Day20 #100DaysOfCode #LeetCode #DSA #BinarySearch #CodingJourney #Java #ProblemSolving

#Day360 of #1001DaysOfCode 📘 LeetCode Daily Challenge Problem: Search in a Binary Search Tree (LeetCode 700) 💡 Approach: Used the properties of a Binary Search Tree to efficiently search for a value. At each step: If the value is smaller, move to the left subtree If larger, move to the right subtree This reduces the search space at every step. ⏱ Time Complexity: O(h) 🧠 Space Complexity: O(h) Continuing daily consistency in problem solving 🚀 #DSA #Java #LeetCode #BinaryTree #ProblemSolving #Coding

🚀 𝗗𝗮𝘆 𝟭𝟬 𝗼𝗳 𝟭𝟬𝟬: 𝗦𝗼𝗹𝘃𝗶𝗻𝗴 LeetCode! Today I focused on applying the 𝗣𝗿𝗲𝗳𝗶𝘅 𝗦𝘂𝗺 concept to identify balanced positions in an array. 📌 𝗣𝗿𝗼𝗯𝗹𝗲𝗺𝘀 𝗦𝗼𝗹𝘃𝗲𝗱 𝗧𝗼𝗱𝗮𝘆 (𝗗𝗮𝘆 𝟭𝟬) 🔹 𝗙𝗶𝗻𝗱 𝘁𝗵𝗲 𝗠𝗶𝗱𝗱𝗹𝗲 𝗜𝗻𝗱𝗲𝘅 𝗶𝗻 𝗔𝗿𝗿𝗮𝘆 ✅ Goal: Find an index where the sum of elements on the left is equal to the sum on the right. If multiple exist, return the leftmost one. Approach: Calculated the total sum of the array, then traversed it while maintaining a running left sum. At each index, compared left sum with "(total sum - left sum - current element)" to check if it satisfies the condition, achieving O(n) time complexity. #100DaysOfCode #DSA #LeetCode #Java #ProblemSolving #CodingJourney #PrefixSum #Arrays