Day 37: Smallest Pair with Different Frequencies in Java

💡 Day 38 of LeetCode Problem Solved! 🔧 🌟643. Maximum Average Subarray I🌟 Task : • You are given an integer array nums consisting of n elements, and an integer k. • Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value. • Any answer with a calculation error less than 10-5 will be accepted.   Example 1: Input: nums = [1,12,-5,-6,50,3], k = 4 Output: 12.75000 Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75 Example 2: Input: nums = [5], k = 1 Output: 5.00000 #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

🚀 DSA Consistency – Day 54 Solved “Check if Binary String Has at Most One Segment of Ones” on LeetCode. The task is to determine whether a binary string contains at most one contiguous block of 1s. Example: 1100111 → ❌ False (two segments of 1s) 111000 → ✅ True (only one segment) 🔹 Approach 1: Traverse the String Idea: While traversing the string, once we encounter a 0, no 1 should appear afterward. If 1 appears again, it means a second segment of ones has started. 🔹 Approach 2: String Pattern Check A binary string with only one segment of 1s should never contain "01" followed by another 1. So, we simply check for the pattern "01". Both the approaches have same complexities: Time Complexity: O(n) Space Complexity: O(1) #DSA #LeetCode #Java #ProblemSolving #CodingJourney #Consistency #100DaysOfCode

🚀 Day #82/100 – 𝐑𝐞𝐦𝐨𝐯𝐞 𝐄𝐥𝐞𝐦𝐞𝐧𝐭 (LeetCode) Today’s problem was Remove Element — a simple yet important question to understand in-place array manipulation. 🔍 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: We don’t need extra space! We can solve this using a two-pointer approach efficiently.  𝐖𝐡𝐲 𝐢𝐭 𝐰𝐨𝐫𝐤𝐬? We overwrite unwanted elements and keep only the valid ones at the beginning of the array. ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Initialize k = 0 Traverse array: If element ≠ val → place it at index k Increment k Return k (new length) ⏱️ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(n) 📦 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: O(1) #Day82 #100DaysOfCode #Java #DSA #LeetCode #TwoPointers #CodingJourney

🚀 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

Day: 72/365 📌 LeetCode POTD: Count Submatrices With Equal Frequency of X and Y Medium Key takeaways/Learnings from this problem: 1. The key idea is converting the problem into numbers (like +1 and -1) so equal frequency turns into a sum = 0 problem. 2. 2D prefix sums really help here—you can quickly get submatrix sums without recalculating every time. 3. It’s a nice extension of 1D subarray problems to 2D, so knowing those basics pays off big time. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

🚀 Day 38 of #100DaysOfCode 🌱 Topic: Linked List / Recursion ✅ Problem Solved: LeetCode 24 – Swap Nodes in Pairs 🛠 Approach: Used a recursive approach to swap every two adjacent nodes in the linked list. If the list has 0 or 1 node, return it directly (base case). Store the second node (head.next) as a temporary node. Recursively swap the remaining list starting from temp.next. Adjust pointers so the second node becomes the new head of the pair. Connect the swapped pair with the recursively processed list. This swaps nodes without modifying their values, only changing pointers. #100DaysOfCode #Day38 #DSA #LinkedList #Recursion #LeetCode #Java #ProblemSolving #CodingJourney #Consistency

Day 57 — LeetCode Progress (Java) Problem: Jump Game Required: Given an array where each element represents the maximum jump length at that position, determine if you can reach the last index. Idea: Track the farthest index reachable at every step. If at any point your current index exceeds this range, you’re stuck. Approach: Maintain a variable maxReach to store the farthest reachable index. Traverse the array: If current index > maxReach, return false (cannot proceed). Update maxReach = max(maxReach, i + nums[i]). If traversal completes, return true. Time Complexity: O(n) Space Complexity: O(1) #LeetCode #DSA #Java #Greedy #Arrays #Algorithms #CodingJourney #100DaysOfCode

Day 9/100 – LeetCode Challenge Problem: 3Sum Today’s challenge was to find all unique triplets in an array whose sum equals 0. Approach: Sort the array to efficiently apply the two-pointer technique. Fix one element nums[i] and use two pointers (left and right) to find the remaining two numbers. If the sum equals 0, store the triplet. Skip duplicate elements to ensure only unique triplets are included. Key Idea: Sorting + Two Pointers helps reduce the brute-force O(n³) approach to O(n²). Concepts Practiced: Sorting Two-pointer technique Duplicate handling Array traversal optimization #100DaysOfCode #LeetCode #DSA #Java #TwoPointers #ProblemSolving #CodingJourney

Solved "Container With Most Water" on LeetCode today using the Two-Pointer Technique. 🚀 The key insight is that the area formed by two lines depends on the shorter height and the distance between them. Starting with pointers at both ends of the array, we compute the area and move the pointer at the smaller height inward to potentially find a taller boundary and maximize the area. This approach efficiently reduces the problem from O(n²) brute force to O(n) time with O(1) space. Problems like this are a great reminder that the right observation can drastically optimize a solution. 💡 #LeetCode #DSA #TwoPointers #Java #ProblemSolving

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