Early Humans to Java: The Timeless Logic of Efficiency

Day 6/100 – LeetCode Challenge 🚀 Problem: #189 Rotate Array   Difficulty: Medium   Language: Java   Approach: Array Reversal Technique   Time Complexity: O(n)   Space Complexity: O(1) 🔍 Key Insight: Instead of shifting elements one by one, the array can be rotated efficiently using a three-step reversal strategy. Steps: 1️⃣ Reverse the entire array   2️⃣ Reverse the first k elements   3️⃣ Reverse the remaining elements  This achieves the required rotation in-place with constant extra space. 🧠 Solution Brief: Calculated k % n to handle cases where k is greater than array length.   Reversed the entire array first.   Then reversed the first k elements and the remaining n-k elements.   This sequence correctly rotates the array to the right. 📌 What I Learned: Understanding patterns like array reversal can simplify problems that initially seem complex.   Optimizing from brute force shifting to an in-place O(n) solution improves efficiency. #LeetCode #Day6 #100DaysOfCode #Java #DSA #Arrays #RotateArray #ProblemSolving #CodingJourney

Here I am again! Today I practiced building a simple multiplication table using loops in Java. This time, I explored a more robust way to validate user input. Instead of using Scanner.hasNextInt() like before, I used try-catch to handle invalid inputs. This approach allows the program to safely handle errors and continue running, even when the user enters unexpected values. While Scanner.hasNextInt() can handle validation, using try-catch provides a simpler and more robust solution. You can check the full project and source code here: https://lnkd.in/dRzfk_3D Check out the logic below! 👇 #Java #Algorithms #SoftwareDevelopment #Coding

🚀 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 18/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1528. Shuffle String Used a direct indexing approach by placing each character at its correct position using the given indices array. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) A simple yet important problem to strengthen understanding of arrays and index mapping. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀 Day 20/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 350. Intersection of Two Arrays II Used sorting + two-pointer technique to efficiently find common elements appearing in both arrays while maintaining correct frequency. ⏱️ Time Complexity: O(n log n + m log m) 📦 Space Complexity: O(min(n, m)) (for storing the intersection) Strengthening understanding of array traversal and two-pointer pattern through consistent practice. 💪 Consistency keeps the progress moving 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

I almost overcomplicated this problem… until I realized it was just prefix sum. 👉 Equal Sum Grid Partition (LeetCode 3546) At first, I thought: “Try all possible cuts.” But that quickly gets messy. Then the real insight hit: If the total sum is even, we just need to find a prefix (row-wise or column-wise) equal to total / 2. That’s it. No brute force. Just clarity. 💡 Lesson: The problem is not about splitting the grid — it’s about finding the right prefix. Time: O(m × n) Space: O(n) Sharing the clean Java solution on GitHub 👇 #leetcode #dsa #java #coding #developers #softwareengineer #codewithishwar

🚀 Day 13 of #LeetCode Challenge 🔢 Problem: 1758 – Minimum Changes To Make Alternating Binary String Today’s problem was about converting a binary string into an alternating pattern with minimum changes. 💡 Key Insight: An alternating string can only start with either ‘0’ or ‘1’. So instead of trying many possibilities, I compared both patterns and counted mismatches — then returned the minimum. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ This problem improved my thinking in: Pattern observation Optimization approach Writing clean and efficient Java code Consistency > Motivation 💪 #Day13 #Java #DSA #LeetCode #CodingJourney #FutureFullStackDeveloper

🚀 Day 31/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 69. Sqrt(x) Used the Binary Search approach to find the integer square root without using built-in power functions. The search space is reduced by checking mid * mid against x. ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) Strengthening problem-solving skills with binary search patterns on answer space. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

LeetCode Problem || Check if Binary String Has at Most One Segment of Ones(1784)🚀 we need to check: The string should have only one continuous block of '1's. After a '0' appears, '1' should never appear again. ✨ Insight: Instead of manually counting segments using loops, we can simply check if "01" exists in the string. 📌 Time Complexity: O(n) Practicing problems like these helps improve pattern recognition and problem-solving efficiency. #LeetCode #DSA #Java #CodingPractice #ProblemSolving

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