Solved LeetCode Problem #852 with Java and Binary Search

🚩 Problem: 78. Subsets 🔥 Day 56 of #100DaysOfLeetCode 🔍 Problem Summary: Given an integer array nums, return all possible subsets (the power set). The solution set must not contain duplicates, and the order does not matter. 🧠 Intuition: This is one of the simplest and cleanest backtracking problems. At each index, you have only two choices: Include the current element Exclude the current element This naturally builds the entire power set. Alternatively, you can also build subsets by expanding a list from left to right. ✅ Backtracking Approach: Start with an empty subset Recursively explore adding each number After each choice, backtrack and remove it Add all generated combinations into the result Super clean and beginner-friendly. ⚙️ Performance: ⏱️ Runtime: 1 ms 🚀 💪 Beats: 97% of Java solutions 💾 Memory: 43 MB ⚡ (Beats ~90% of users) 📊 Complexity: Time Complexity: O(n × 2ᶰ) Space Complexity: O(n) (recursion + temp list) ✨ Key Takeaway: This problem teaches the classic recursive structure used for: Subsets Combinations Nested choices Decision-tree exploration It's the perfect stepping stone to more advanced backtracking problems. Link:[https://lnkd.in/gfQV_8w4] #100DaysOfLeetCode #Day56 #Problem78 #Subsets #Backtracking #Recursion #Algorithms #DSA #Java #ProblemSolving #LeetCode #CodingChallenge #CodingCommunity #InterviewPreparation #CrackingTheCodingInterview #SoftwareEngineering #DataStructures #ArjunInfoSolution #DeveloperJourney #LearnToCode #TechCareers #CareerGrowth #CodeNewbie #ZeroToHero #CodingIsFun #JavaDeveloper #GameDeveloper #Unity #AI #MachineLearning

🚀 Day 52 of #100DaysOfCode 🚀 Today I solved LeetCode Problem #389 – Find the Difference 🧩 This problem is a great example of using ASCII value manipulation to solve what seems like a string comparison challenge. 💡 Key Learnings: Strengthened understanding of character encoding (ASCII values) and how they can simplify logic. Learned how summing character values can help detect differences efficiently without extra data structures. Reinforced the value of O(n) solutions for string-based problems. 💻 Language: Java ⚡ Runtime: 1 ms — Beats 99.32% 🚀 📉 Memory: 41.9 MB — Beats 65.84% 🧠 Approach: 1️⃣ Convert both strings to character arrays. 2️⃣ Compute the sum of ASCII values of both. 3️⃣ The difference between sums gives the ASCII value of the extra character in t. Simple, elegant, and efficient! ⚙️ Sometimes, a clever use of ASCII arithmetic can replace complex logic — efficiency lies in simplicity. #100DaysOfCode #LeetCode #Java #ProblemSolving #CodingChallenge #Algorithms #DataStructures #SoftwareDevelopment #LearningEveryday #CleanCode

🚀 Day 60 of #100DaysOfCode 🚀 Today, I solved LeetCode Problem #137 – Single Number II 🧩 📘 Problem Statement: Given an integer array where every element appears three times except for one that appears exactly once, find that single element and return it. The challenge: solve it with O(n) time and O(1) space complexity. 💻 Language: Java ⚡ Runtime: 0 ms — Beats 100.00% 📉 Memory: 45.54 MB — Beats 56.30% 🧠 Concept Used: 🔹 Bit Manipulation — A powerful yet tricky technique that leverages binary operations to track occurrences of bits efficiently. 🔹 Instead of using extra space or hash maps, we track bits that appear once and twice using two integer variables: ✅ once → bits that appeared once ✅ twice → bits that appeared twice The trick lies in updating them using XOR (^) and NOT (~) to "cancel out" bits appearing three times. 🧩 Approach Summary: 1️⃣ Initialize two variables once = 0 and twice = 0. 2️⃣ For every number in the array: - Update once and twice based on how many times a bit has appeared. 3️⃣ After processing all numbers, once holds the value of the single element. ✅ Complexity: Time: O(n) Space: O(1) ✨ Takeaway: This problem teaches how bitwise logic can replace traditional data structures, achieving the same result with constant space — a crucial optimization in system-level programming and interviews. #100DaysOfCode #LeetCode #Java #BitManipulation #ProblemSolving #CleanCode #DataStructures #Algorithms #TechLearning #CodingChallenge #SoftwareEngineering #InterviewPreparation

🚀 Day-75 of #100DaysCodeOfChallenge 💡 LeetCode Problem: 1526. Minimum Number of Increments on Subarrays to Form a Target Array (Hard) 🧠 Concepts Practiced: Greedy Algorithms | Array Manipulation | Difference Computation Today’s challenge was a Hard-level problem, focusing on finding the minimum number of operations required to transform an initial array of zeros into a given target array — using only subarray increments. The key insight here was realizing that each increase in value from one element to the next represents a required new operation on LeetCode. Instead of simulating every subarray operation, we can simply sum up the positive differences between consecutive elements — a great example of mathematical optimization through observation 💭 🔹 Intuition: Only when a number increases compared to the previous one, a new increment operation is needed. ⚙️ Language: Java ⚡ Runtime: 3 ms (Beats 100%) 💾 Memory: 56.82 MB (Beats 61.41%) ✅ Result: 129 / 129 test cases passed — Accepted! 🎯 Each “hard” problem solved adds one more layer of confidence — and reminds me that persistence pays off 💪 #100DaysOfCode #LeetCode #Java #ProblemSolving #Algorithms #CodingChallenge #GreedyAlgorithm #LearningJourney #TechMindset #KeepCoding #SoftwareEngineering #DeveloperLife #LogicBuilding

#Day_27 Today’s problem was an interesting twist on binary search — “Find Peak Element” 🔍 Given an array of numbers, the task is to find an index i such that nums[i] is greater than its neighbors — basically, a peak element. At first glance, it seems like a simple linear scan could solve it in O(n). But I wanted to do better — and that’s where binary search comes in 💡 🧠 Approach We observe that: If nums[mid] > nums[mid + 1], it means we are on a descending slope, so the peak lies on the left side (including mid). Otherwise, we’re on an ascending slope, so the peak lies on the right side (after mid). Using this logic, we can shrink our search space by half in every iteration — a classic divide and conquer move 🔥 ⏱ Complexity Time: O(log n) Space: O(1) This is a great example of how binary search isn’t just for sorted arrays — it can also be applied to problems involving patterns or directional decisions. 💬 Every day of this challenge reinforces that problem-solving isn’t just about writing code — it’s about recognizing patterns, making logical deductions, and applying optimized thinking. #CodingJourney #LeetCode #BinarySearch #ProblemSolving #100DaysOfCode #Java #LearnByDoing

🚀 Day 146 / 180 of #180DaysOfCode ✅ Today’s Highlight: Revisited LeetCode 3234 — “Count the Number of Substrings With Dominant Ones.” 🧩 Problem Summary: Given a binary string s, the task is to count how many of its substrings have dominant ones. A substring is considered dominant if: number of 1s ≥ (number of 0s)² This creates an interesting balance check between zeros and ones, pushing you to think about substring ranges, prefix behavior, and efficient counting techniques. 💡 Core Takeaways: Great exercise in string analysis and prefix-based reasoning Highlights how mathematical conditions influence algorithm design Reinforces handling frequency relationships inside a sliding or expanding window Efficient counting becomes essential due to potential O(n²) substrings 💻 Tech Stack: Java ⏱ Runtime: 115 ms (Beats 84.11%) 💾 Memory: 47 MB 🧠 Learnings: Strengthened understanding of substring behavior under unique constraints Refined thinking around optimizing brute-force patterns A good reminder of how theoretical conditions (like squaring zeros) change practical implementation choices 📈 Progress: Today’s problem significantly improved my intuition around constraint-based substring evaluation—valuable for more advanced algorithmic challenges ahead. #LeetCode #Java #Algorithms #SubstringProblems #DSA #ProblemSolving #CodingJourney #SoftwareEngineering #180DaysOfCode #LogicBuilding #LearningEveryday

✅Day 42 : Leetcode 154 - Find Minimum in Rotated Sorted Array-2 #60DayOfLeetcodeChallenge 🧩 Problem Statement You are given an array nums that is sorted in ascending order and then rotated between 1 and n times. The array may contain duplicates. Your task is to find and return the minimum element in the rotated sorted array. You must minimize the number of overall operations as much as possible. 💡 My Approach I used a modified binary search technique to handle both rotation and duplicates. Initialize two pointers — low = 0 and high = n - 1. Calculate the middle index mid = low + (high - low) / 2. Update the answer as ans = min(ans, nums[mid]). If nums[low] == nums[mid] && nums[mid] == nums[high], move both low++ and high-- to skip duplicates. If the left half is sorted (nums[low] <= nums[mid]), update the answer and move to the right half (low = mid + 1). Otherwise, move to the left half (high = mid - 1). Continue until low > high. This efficiently finds the minimum even when duplicates exist. ⏱️ Time Complexity Worst Case: O(n) — when many duplicates exist. Average Case: O(log n) — behaves like binary search when duplicates are few. #BinarySearch #LeetCode #RotatedArray #Java #DSA #ProblemSolving #CodingPractice #LeetCodeHard

🌟 Day 26 of #100DaysOfCode 🌟 🔍 Remove Duplicate Letters — Lexicographically Smallest Unique String 🔹 What I Solved Today, I tackled the “Remove Duplicate Letters” problem — a fascinating challenge that combines stack operations, greedy logic, and lexicographical ordering. It’s a perfect test of both analytical and implementation skills. 📝 Problem Statement Given a string s, remove duplicate letters so that every letter appears once and only once. You must ensure the resulting string is the smallest in lexicographical order among all possible results. ✅ Example 1: Input: s = "bcabc" Output: "abc" ✅ Example 2: Input: s = "cbacdcbc" Output: "acdb" Constraints: 1 ≤ s.length ≤ 10⁴ s consists of lowercase English letters 🧠 Concepts Used Stack Greedy Algorithm HashMap / Frequency Counting Character Visitation Tracking ⚙️ Approach Count the frequency of each character using a map. Use a stack to build the result string. Iterate through each character: Decrease its frequency (as it’s now visited). If the character is already in the stack, skip it. Otherwise, while the current character is smaller than the top of the stack and the top of the stack will appear later again, pop it out to maintain lexicographical order. Push the current character into the stack. Convert the stack to the final result string. 🚀 What I Learned This problem beautifully demonstrates how greedy thinking and data structures like stacks can work together to maintain order and constraints. It deepened my understanding of how lexicographical optimization problems can be efficiently solved using character frequency and conditional popping. #100DaysOfCode #ProblemSolving #Java #DataStructures #Algorithms #CodingJourney #GeeksforGeeks #KeepLearning

🚩 Problem: 153. Find Minimum in Rotated Sorted Array 🔥 Day 59 of #100DaysOfLeetCode 🔍 Problem Summary: You’re given a rotated sorted array of unique integers. Your task: find the minimum element in O(log n) time. Example of rotated array: [4,5,6,7,0,1,2] → minimum = 0 🧠 Intuition: Because the array is originally sorted, rotation keeps two sorted halves. We can use binary search: Check if the mid element is part of the right sorted half Or the left sorted half Narrow down the search to where the rotation break happens The minimum is exactly at the rotation point Key observation: If nums[mid] > nums[right] → min is in the right half Else → min is in the left half Clean, elegant, and strictly log-time.⚙️ Performance: ⏱️ Runtime: 0 ms 🚀 💪 Beats: 100% of Java solutions 💾 Memory: 42 MB ⚡ (Beats 95%+ users) 📊 Complexity: Time Complexity: O(log n) Space Complexity: O(1) ✨ Key Takeaway: This is the perfect example of using binary search on conditions, not just sorted arrays. Mastering this pattern helps solve advanced problems on rotated arrays and search intervals. Link:[https://lnkd.in/ggdZwAKJ] #100DaysOfLeetCode #Day59 #Problem153 #FindMinimumInRotatedArray #BinarySearch #Algorithms #DSA #Java #CodingChallenge #ProblemSolving #LeetCode #InterviewPreparation #CrackingTheCodingInterview #BinarySearchPattern #CodingCommunity #SoftwareEngineering #DataStructures #DeveloperJourney #ArjunInfoSolution #TechCareers #CareerGrowth #Programming #ZeroToHero #LearnToCode #CodeNewbie #CodingIsFun #JavaDeveloper #GameDeveloper #Unity #AI #MachineLearning