Solved Base 7 Conversion in Java with Efficient Algorithm

Day 85 of #100DaysOfCode Solved Range Sum Query - Immutable (NumArray) in Java ➕ Approach Today's problem was a classic that demonstrates the power of pre-computation: finding the sum of a range in an array many times. The optimal solution is the Prefix Sum technique. Pre-computation: In the constructor, I built a prefixSum array where prefixSum[i] holds the sum of all elements from index 0 up to index $i-1$ in the original array. This takes $O(N)$ time. $O(1)$ Query: The magic happens in the sumRange(left, right) method. The sum of any range $[left, right]$ is found instantly by calculating prefixSum[right + 1] - prefixSum[left]. The cost of a single $O(N)$ setup is outweighed by the ability to perform every subsequent query in $O(1)$ time! #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #Arrays #PrefixSum #Optimization #ProblemSolving

🧠 Daily LeetCode Grind — Java Edition Today’s challenge: ✅ Container With Most Water (#11 - Medium) 📌 Goal: Given an integer array height[], find two lines that, together with the x-axis, form a container that holds the most water. 📌 Approach: 🔹 Use the Two-Pointer Technique to optimize the search for the maximum area. 🔹 Start with two pointers at both ends of the array and calculate the area at each step. 🔹 Move the pointer pointing to the smaller height inward to maximize potential area. 🔹 Continue until both pointers meet. 🧩 Test Case: Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 💡 Key Takeaways: 🔹 Learned to balance area optimization using width × height. 🔹 Reinforced the power of the two-pointer approach for reducing time complexity. 🔹 Strengthened problem-solving speed through mathematical reasoning. 💻 Language: Java 🧠 Complexity: O(n) — single pass using two pointers. #LeetCode #Java #CodingPractice #ProblemSolving #TwoPointer #Algorithm #DeveloperLife #CybernautEdTech #AcceptedSolution

🔢 Today I worked on a classic matrix problem in Java — Diagonal Sum. The goal was simple: Calculate the sum of both the primary and secondary diagonals of a square matrix. Initially, I used a nested loop approach with O(n²) complexity. But then I optimized it to a clean O(n) solution by directly accessing diagonal indexes. While optimizing, I made an interesting mistake: ❌ I wrote if (i != matrix[i][matrix.length - 1 - i]) Here I accidentally compared an index with an element value. ✔ Correct approach: if (i != matrix.length - 1 - i) This ensures the center element in odd-sized matrices isn’t counted twice. 🧠 Key Learning: Indexes and values may look similar, but mixing them can break logic silently. Optimization is not just about speed — it’s about accuracy. #Java #leetcode #Coding #DSA #ProblemSolving #LearningInPublic #SoftwareEngineering

(Largest Perimeter Triangle — Java 🔺) Hey everyone 👋 Today I explored how to find the largest perimeter triangle from a given array of side lengths using Java. While solving this, I understood how a simple mathematical condition can help in optimizing the entire logic 🔥 Here’s the breakdown 👇 🧠 Concept: To form a valid triangle from 3 sides, the sum of any two sides must be greater than the third one. → a + b > c → a + c > b → b + c > a 💡 Approach 1: Sort the array in descending order and check triplets from the start. If any 3 consecutive sides satisfy the triangle condition, their sum gives the largest perimeter. ⚡ Approach 2 (Optimized): Instead of reversing the array, just sort it in ascending order and iterate from the end — gives the same result but with cleaner logic and less work. 🧩 Time Complexity: O(N log N) 💾 Space Complexity: O(1) This problem helped me understand how sorting order and traversal direction can make an algorithm more elegant and efficient 🚀 📁 Code is available on my GitHub repo: https://lnkd.in/ekjD9s22 #Java #DSA #ProblemSolving #Arrays #CodingJourney #Developers #LearningByDoing #FullStackDeveloper #MCA #GitHub

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

🚀 Day 55 of #100DaysOfCode Challenge Problem: LeetCode #219 – Contains Duplicate II Language: Java ☕ Today I solved an interesting array problem that checks whether a duplicate element exists within a given distance k in the array. 💡 Logic: Use a HashMap to remember the last index of each number. If the same number appears again, check if the difference between indices ≤ k. If yes → return true, else keep checking. 💻 Code: import java.util.HashMap; public class Solution { public boolean containsNearbyDuplicate(int[] nums, int k) { HashMap<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { if (map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) { return true; } map.put(nums[i], i); } return false; } } 🧠 Example: Input: [1,2,3,1], k = 3 Output: true ✅ (same number 1 appears within distance 3) ⚙️ Key Concepts: HashMap for quick lookup Difference check using indices Time Complexity → O(n) Space Complexity → O(n) 💬 Every day’s problem teaches me something new — today it was about using maps smartly to track elements efficiently. On to the next challenge! 💪 #Day55 #LeetCode #Java #100DaysOfCode #CodingJourney #ProblemSolving #HashMap #DSA

Day 90 of #100DaysOfCode Solved Squares of a Sorted Array in Java 🔠 Approach The task was to take an array of integers sorted in non-decreasing order, square each number, and then return the result array also sorted in non-decreasing order. Brute-Force Method The solution implemented here is a straightforward two-step brute-force approach: Squaring: I iterated through the input array nums and replaced each element with its square (i.e., nums[i] * nums[i]). This handles both positive and negative numbers correctly. Sorting: After squaring all elements, I used Java's built-in Arrays.sort(nums) method to sort the entire array. While correct, this approach has a time complexity dominated by the sorting step, which is O(NlogN), where N is the number of elements. The runtime of 10 ms shows that a more efficient, two-pointer approach (which can solve this in O(N) time) is generally preferred for optimal performance. #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #Array #Sorting #ProblemSolving

Day 89 of #100DaysOfCode Solved Find Numbers with Even Number of Digits in Java 🔢 Approach The goal of this problem was to count how many integers in a given array have an even number of digits. I implemented two methods: findNumbers(int[] nums): This is the main function that iterates through every number in the input array nums. isEvenOrOdd(int n): This helper function takes an integer and determines if its digit count is even or odd. Inside the helper function, I used a while loop to count the digits: I initialized a count to 0. The loop continues as long as $n > 0$. In each iteration, I increment count and then perform integer division by 10 (n = n / 10) to remove the least significant digit. Finally, I return true if the total count of digits is even (count \% 2 == 0). This efficient, digit-by-digit checking approach resulted in a strong performance, beating 98.90% of other submissions. #Java #100DaysOfCode #LeetCode #CodingChallenge #Algorithms #Array #ProblemSolving #Optimization

📌 Day 16/100 - Reverse String (LeetCode 344) 🔹 Problem: Reverse a given string in-place — meaning you must modify the original array of characters without using extra space. 🔹 Approach: Used the two-pointer technique — one starting at the beginning and one at the end of the array. Swap characters at both pointers, then move them closer until they meet. Efficient, clean, and runs in linear time without additional memory allocation. 🔹 Key Learnings: In-place algorithms optimize space complexity significantly. The two-pointer pattern is a versatile tool for many array and string problems. Understanding mutable vs immutable structures in Java is crucial for memory efficiency. Sometimes, the simplest logic beats the most complex one. 🧠 “True efficiency lies in simplicity, not complexity.” #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA #CodingJourney #TwoPointers