LeetCode 66: Array Manipulation and Incrementing Numbers

LeetCode Problem || Prime Number of Set Bits in Binary Representation(762)🚀 Solved an interesting bit manipulation problem today! Problem : Given two integers left and right, count how many numbers in that range have a prime number of set bits (1s) in their binary representation. This problem shows how: Binary representation connects with number theory Built-in methods like bitCount() simplify bit problems Constraint analysis helps optimize thinking Small problems like this build strong fundamentals in bit manipulation. #LeetCode #Java #BitManipulation #PrimeNumbers #ProblemSolving #CodingJourney #100DaysOfCode

Day 27/100 – LeetCode Challenge 🚀 Problem: Sort List Approach: Applied Merge Sort on the linked list Used slow and fast pointers to find the middle Recursively sorted both halves Merged the sorted halves Time Complexity: O(n log n) Space Complexity: O(log n) (recursion stack) Key takeaway: Merge sort is the most efficient sorting technique for linked lists, as it avoids random-access operations required by other algorithms. #LeetCode #100DaysOfCode #DSA #Java #LinkedList #MergeSort #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

🚀 Day 23/100 – LeetCode Challenge Today’s problem: Partitioning Into Minimum Number of Deci-Binary Numbers 🔹 Key Insight: The minimum number of deci-binary numbers required is equal to the maximum digit present in the string. 🔹 Approach: Traverse through each character in the string Convert it to integer (ch - '0') Track the maximum digit Return the maximum value 🔹 Time Complexity: O(n) 🔹 Space Complexity: O(1) ✨ Simple logic, but powerful observation! Instead of constructing numbers, we just analyze the digits. Consistency > Motivation 💪 #Day23 #100DaysOfCode #LeetCode #Java #ProblemSolving #CodingJourney

Day 31/100 - LEETCODE Challenge ✅ Problem : Find First and Last Position of Element in Sorted Array Solved the Search for First and Last Position of Element in Sorted Array problem using an optimized Binary Search approach in Java. Instead of using a linear scan, I implemented two separate binary searches to efficiently find the first and last occurrence of the target element in O(log n) time complexity. This solution improves performance for large datasets and achieved 0 ms runtime (100% beats) on LeetCode. Problems like this help strengthen understanding of binary search variations and edge case handling in sorted arrays. #100DaysOfCode #java #Coding #SoftwareDeveloper

🚀 Day 523 of #750DaysOfCode 🚀 📌 Problem Solved: Check if Binary String Has at Most One Segment of Ones (LeetCode 1784) 🔎 Problem Summary: Given a binary string s (without leading zeros), we need to check whether the string contains at most one contiguous segment of 1s. If the 1s appear in more than one separate segment, we return false. (LeetCode) 💡 Key Insight: Since the string starts with 1, the only valid structure is: 111...000... If we ever encounter the pattern "01", it means a 1 appeared again after a 0, which creates multiple segments of ones. (AlgoMonster) ⚙️ Approach: Traverse the string or simply check if "01" exists. If "01" is found → more than one segment → return false. ⏱ Complexity: Time: O(n) Space: O(1) 📚 Takeaway: Sometimes string problems become much easier when we identify patterns instead of counting segments. Consistency > Motivation. On to Day 524 tomorrow. 💪 #LeetCode #Java #DataStructures #Algorithms #CodingChallenge #Consistency #ProblemSolving #100DaysOfCode #750DaysOfCode

Day 12/100 – LeetCode Challenge Problem Solved: Permutations Today’s problem was about generating all possible permutations of a given array of distinct integers. This is a classic backtracking problem where the objective is to build permutations step by step while ensuring each element is used exactly once in every arrangement. I implemented a recursive solution supported by a boolean array to track which elements were already included in the current permutation. At every recursive call, I select an unused element, add it to the current list, mark it as used, and continue exploring deeper. Once a permutation reaches the required length, it is added to the result set. Then comes the most important part — backtracking. I remove the last element and reset its state so other combinations can be explored. Time Complexity: O(n × n!) Space Complexity: O(n) excluding the output list #100DaysOfLeetCode #Java #Backtracking #Recursion #Algorithms #ProblemSolving #Consistency

Day 10 | LeetCode Daily Solved LeetCode Problem #166 – Fraction to Recurring Decimal Concept: • HashMap to detect repeating cycles • Long division simulation Key Learnings: • Repeating decimals occur when remainders repeat • Careful handling of sign and integer overflow is important Building consistency with daily problem-solving and learning to handle tricky edge cases. Submission link: https://lnkd.in/gKQweNyY #LeetCode #DSA #HashMap #Java #ProblemSolving

🚀 Day 82 of #100DaysOfCode Solved LeetCode Problem #1382 – Balance a Binary Search Tree ✅ This problem is a great reminder that sometimes rebuilding is better than fixing. By leveraging the sorted nature of BSTs, converting it to a balanced tree becomes clean and intuitive. Key Takeaways: -> Inorder traversal gives a sorted sequence in BST -> Divide & conquer helps rebuild a height-balanced tree -> Choosing the middle element ensures balance -> Clean recursion beats complex rotations here Language: Java -> Runtime: 2 ms (Beats 97.82%) ⚡ -> Memory: 48.44 MB Consistency compounds. Trees today, forests tomorrow. 🌳💻🔥 #LeetCode #Java #BinarySearchTree #Recursion #DivideAndConquer #ProblemSolving #100DaysOfCode

Day 36/100 – LeetCode Challenge 🚀 Problem: 3Sum Closest Approach: Sorted the array Fixed one element and applied the two-pointer technique Tracked the closest sum by comparing absolute differences Returned immediately if an exact match was found Time Complexity: O(n²) Space Complexity: O(1) Key takeaway: Many optimization problems are variations of classic patterns. Understanding 3Sum deeply makes solving 3Sum Closest straightforward. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving #InterviewPrep