Maximizing Minimum Power Supply with LeetCode Challenge

🚀 Day 40 of My #LeetCode Challenge 🚀 📘 Problem: Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold (LeetCode #1343) 💡 Approach: Implemented the Sliding Window technique to efficiently calculate the sum of each subarray of size k. First, compute the sum of the initial k elements. Then, slide the window forward — subtract the element leaving and add the new one entering. Check if the average of the current window meets or exceeds the given threshold. This method optimizes the time complexity to O(n) by avoiding repetitive summation. ⏱️ Concepts Used: Sliding Window | Arrays | Optimization | Time Complexity 🔥 Consistency and logic go hand in hand — every day adds a new layer to my problem-solving skills! #Day40 #LeetCode #100DaysOfCode #CodingChallenge #Java #DSA #ProblemSolving #CodingJourney #TechLearning #CodeEveryday #SoftwareEngineering #ProgrammerLife #CodingCommunity #Developer #JavaProgramming #LearnToCode #TechGoals #CodeNewbie #Algorithm #DataStructures #CodingPractice

🚀 Day 131 of #150DaysOfCode – LeetCode Challenge 🚀 Today’s problem was 3346. Maximum Frequency of an Element After Performing Operations I. Problem in short: We are given an array of integers and a number k. Each element can be increased or decreased by at most k. The goal is to determine the maximum possible frequency of any element after performing these operations. Key Insight: Each element can reach a value in [num - k, num + k]. Sorting the array and using a sliding window approach or interval overlap counting helps to efficiently find the maximum frequency. ✅ Learning takeaway: Sliding window and sorting strategies are very powerful for problems involving ranges and frequency maximization. #Java #LeetCode #DSA #Algorithms #CodingChallenge #ProblemSolving #150DaysOfCode #Array #SlidingWindow #Greedy #Intervals #DivideAndConquer

🚀 Day 54 of my LeetCode Journey 🚀 Today, I worked on Search in Rotated Sorted Array II. 🔍 Key takeaway: This problem looks similar to searching in a rotated sorted array, but the twist is duplicates. Duplicates make it difficult to identify which half of the array is sorted. Instead of giving up — we adapt the binary search! ✅ Logic used: If nums[mid] == nums[right]: we can't decide the sorted half → shrink the search space (right--). Else, check which half is sorted and move the pointers accordingly. 🧠 Concept learned: Sometimes, when data is ambiguous (like duplicates), instead of forcing a direction — reduce uncertainty first. 💻 Skills improved: • Binary Search • Handling ambiguous cases • Optimizing decision-making logic Excited to continue improving one problem at a time. Consistency > Speed 🔥 👨💻 "Every problem teaches something — even if it looks like just another binary search." #100DaysOfLeetCode #Day54 #codingjourney #java #problemsolving #binarysearch #leetcode #learningeveryday

🚀 Leveling Up My Problem-Solving Skills | #LeetCodeJourney Recently solved some interesting LeetCode problems that tested my thinking in arrays, hashmaps, and frequency patterns: 💡 1️⃣ Reshape the Matrix Learned how to transform a 2D matrix into a new dimension while maintaining element order. Focused on data traversal and logical constraints. 🍬 2️⃣ Distribute Candies Practiced the use of Set to efficiently count unique types and apply mathematical constraints (n / 2). Reinforced understanding of hashing and uniqueness concepts. 📊 3️⃣ Longest Harmonious Subsequence Explored real-world frequency analysis with HashMap. Built logic to find pairs with a difference of exactly 1 — a beautiful balance between pattern detection and optimization. These problems helped me sharpen: ✅ Logical reasoning ✅ Data structure handling (List, Set, Map) ✅ Problem-solving approach using clean and optimized Java code ✨ You can check out my LeetCode profile and explore my progress here: 🔗 leetcode.com/u/amans_01/ I believe consistent problem-solving is the best way to grow as a developer — one problem at a time 💻💪 #Java #LeetCode #ProblemSolving #CodingJourney #SoftwareEngineering #DataStructures #Algorithms #CareerGrowth #ChandigarhUniversity

🔥 LeetCode Day 27 Challenge Problem Attempted: Search in Rotated Sorted Array II 💡 Approach & Progress: Today, I explored how to search for a target element in a rotated sorted array that contains duplicates. This version is trickier than the original because duplicates can make it hard to determine which half of the array is sorted. I learned that the key is to handle edge cases carefully — especially when the elements at low, mid, and high are equal. In such cases, it becomes impossible to decide which side is sorted, so the best approach is to shrink the search space by moving both pointers inward. After managing duplicates, I applied the standard binary search logic by identifying which half of the array is sorted and checking if the target lies within that range. 🧠 Key Learnings: Understood how duplicates affect binary search in rotated arrays. Learned to identify and manage ambiguous cases where sorting order is unclear. Improved logical reasoning for range-based conditions in binary search. Strengthened confidence in solving complex search variations efficiently. #LeetCode #Day27 #BinarySearch #DSA #ProblemSolving #CodingChallenge #Java #Algorithm #100DaysOfCode #LeetCodeJourney #CodeEveryday #TechLearning #ProgrammerLife

🚀 Day 8 of #100DaysOfLeetCode Challenge 😎 📌Today's problem: Palindrome Number (LeetCode #9) 💡 🔹 Concept: Check whether a given integer reads the same backward and forward — without converting it to a string! 🔹 Logic Used: Instead of reversing the entire number, I reversed only half of it to improve efficiency. This avoids unnecessary computation and eliminates integer overflow risks. 🔹 Key Learnings: ✅ Optimized logic using mathematical manipulation ✅ Improved understanding of number reversal techniques ✅ Importance of edge case handling (negative numbers, trailing zeros) ⚙️ Result: 💻 Runtime: 4 ms (Beats 100%) 💾 Memory: 44.84 MB 😎Small optimizations make a big difference in performance! 💪 📌Problem: https://lnkd.in/ef6AC2j6 🔥Each day is a step closer to writing cleaner and more optimized code. ✨ Let’s keep pushing forward and refining our problem-solving skills! 💻🔥 #LeetCode #Day8 #100DaysOfCode #ProblemSolving #Java #CodingChallenge #PalindromeNumber #DataStructures #Algorithms #LearningEveryday

🚀 Day 48 of My LeetCode Journey 🚀 💡 Problem: Valid Perfect Square 🧩Topic: Binary Search Approach: To determine if a number is a perfect square without using sqrt(), I used binary search. By checking the mid-point’s square in each iteration, we efficiently narrow down whether the number lies between two squares. ⚙️ Logic Summary: Set low = 1, high = num While low <= high: Find mid = (low + high) / 2 Compare mid * mid with num Adjust range accordingly Return true if perfect square found 🧠 Time Complexity: O(log n) 🧾 Space Complexity: O(1) ✨ Key Takeaway: Binary search isn’t just for arrays — it’s a powerful tool for solving numeric range problems too. #Day48 #LeetCode #100DaysOfCode #Java #CodingJourney #BinarySearch #ProblemSolving

🚀 Day 51 of My LeetCode Journey 🚀 💡 Problem: Binary Search 🧠 Concept: Binary Search is one of the most fundamental algorithms — it efficiently searches a sorted array by repeatedly dividing the search space in half. If the target value is equal to the middle element, return the index. If the target is smaller, search the left half; if larger, search the right half. ⚙️ Complexity: ⏱ Time: O(log n) 💾 Space: O(1) ✨ Learning: Binary search is not just an algorithm — it’s a mindset to “divide and conquer.” Mastering it helps in many advanced problems like searching in rotated arrays, finding peaks, and solving range queries efficiently. #100DaysOfCode #LeetCode #CodingJourney #Java #DSA #BinarySearch #ProblemSolving

🔥 LeetCode Day 28 Challenge Problem Attempted: Find Minimum in Rotated Sorted Array 💡 Approach & Progress: Today, I focused on identifying the minimum element in a rotated sorted array — a classic binary search variation that relies on recognizing the inflection (pivot) point. The key observation is that in a rotated array, all elements to the left of the pivot are greater than the elements to the right of the pivot. Using binary search, I compared the mid element with the rightmost element to decide which side to continue searching: If nums[mid] > nums[right], the minimum lies to the right. Otherwise, it lies on the left (including mid). This approach efficiently reduces the search space and finds the minimum in O(log n) time without needing to sort or traverse the array fully. 🧠 Key Learnings: Learned how to locate the pivot or minimum element using binary search logic. Understood how comparisons between mid and right help identify sorted vs rotated segments. Reinforced the principle that binary search can be used beyond simple value matching — for conditions and structure detection. Improved understanding of edge cases like unrotated arrays or minimal rotations. #LeetCode #Day28 #BinarySearch #DSA #ProblemSolving #CodingChallenge #Java #Algorithm #100DaysOfCode #LeetCodeJourney #CodeEveryday #TechLearning #ProgrammerLife

NeetCode 29 | LeetCode #153 | Find Minimum in Rotated Sorted Array Used binary search to find the rotation pivot. Compared nums[mid] with nums[right] to shrink search space efficiently. Time complexity: O(log n) | Space: O(1) #LeetCode #NeetCode #Java #Algorithms #DataStructures #Optimization #Coding