Minimum Difference Between K Elements: Simplifying with Sorting

🚀 DSA Progress Update Solved: Single Element in Sorted Array (Binary Search) Initially struggled with: Handling edge cases Index out of bounds Deciding search direction without a target After multiple attempts, I realized: 👉 Binary Search is not always about finding a target — sometimes it’s about identifying a pattern. 💡 Key Insight: Before the single element → pairs follow an even index pattern After that → the pattern breaks This problem helped me improve: - Binary Search intuition - Edge case handling - Thinking beyond brute force Still learning and improving step by step 💪 #DSA #Java #CodingJourney #Learning

🚀 DSA Preparation 💪 Solved a basic yet important Array + Two Pointer problem. Focused on rearranging elements by grouping even and odd numbers efficiently. Great for improving in-place operations and partitioning logic 🔥 🧠 Problem 🔎 Sort Array By Parity Given an integer array nums, move all even integers to the beginning followed by all odd integers. 👉 Return any array that satisfies this condition. Example Input: nums = [3,1,2,4] Output: [2,4,3,1] Input: nums = [0] Output: [0] ⚡ Key Learning 📌 Use two pointers / partition approach 📌 Efficiently rearrange elements in one pass 📌 Time Complexity: O(n) → traverse the array once 📌 Space Complexity: O(1) → in-place rearrangement (no extra space) Improving DSA with strong fundamentals 🚀 #DSA #LeetCode #Arrays #TwoPointers #Java #ProblemSolving #Consistency

🔥 DSA Challenge – Day 136/360 🚀 📌 Topic: Bit Manipulation 🧩 Problem: Number of 1 Bits (Hamming Weight) Problem Statement: Given an integer, return the number of set bits (1’s) present in its binary representation. 🔍 Example: Input: 6 Output: 2 Explanation: Binary of 6 → 110 → 2 set bits 💡 Optimized Approach (Brian Kernighan’s Algorithm): Instead of checking each bit one by one, we use a smart trick: n & (n - 1) removes the rightmost set bit in each iteration This reduces the number of operations to the number of set bits only ⚡ Time Complexity: O(k) (k = number of set bits) ⚡ Space Complexity: O(1) 🚀 Key Takeaway: Using bit manipulation can drastically optimize performance compared to brute force approaches. Always look for patterns in binary operations! #DSA #Coding #Java #BitManipulation #136DaysOfCode #LeetCode #ProblemSolving #SoftwareEngineering

📘 DSA Journey — Day 35 Today’s focus: Binary Search with index patterns. Problem solved: • Single Element in a Sorted Array (LeetCode 540) Concepts used: • Binary Search • Index parity (even/odd pattern) • Search space reduction Key takeaway: The array is sorted and every element appears twice except one. A key observation: Before the single element, pairs start at even indices After the single element, this pattern breaks. Using binary search: • If mid is even and nums[mid] == nums[mid + 1], the single element lies on the right side • Else, it lies on the left side (including mid) By leveraging this pattern, we can find the answer in O(log n) time and O(1) space. Continuing to strengthen binary search intuition and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney

📊 DSA Progress Update – Week 5 Over the past few days, I’ve been practicing array problems and improving my approach step by step. What I learned: • Starting with a brute force solution helps in understanding the problem clearly • Optimization becomes easier once the pattern is identified • Key techniques : - Two Pointers - Sliding Window - Prefix Sum - Kadane’s Algorithm Something new: Started learning Binary Search and understanding its basic approach. Plan: Continue practicing and get more comfortable with these concepts. Taking it one step at a time. #DSA #Java #LeetCode #Consistency #CodingJourney

📘 DSA Journey — Day 36 Today’s focus: Binary Search for boundaries. Problem solved: • Find First and Last Position of Element in Sorted Array (LeetCode 34) Concepts used: • Binary Search • Lower bound & upper bound • Searching boundaries efficiently Key takeaway: The goal is to find the first and last occurrence of a target in a sorted array. Instead of scanning linearly, we perform two binary searches: • First search → find the leftmost (first) occurrence • Second search → find the rightmost (last) occurrence Key idea: When we find the target: • For left boundary → continue searching in the left half • For right boundary → continue searching in the right half This ensures we capture the full range in O(log n) time. Continuing to strengthen binary search patterns and problem-solving consistency. #DSA #Java #LeetCode #CodingJourney

Day 43 #SDE backtracking and constraint-based problems. Solved: • Restore IP Addresses • N-Queens Key Learning: • “Restore IP Addresses” involves backtracking with constraints, where we build valid segments (0–255) while handling edge cases like leading zeros. • “N-Queens” is a classic backtracking problem, where we place queens while ensuring no conflicts in rows, columns, and diagonals — strengthening constraint checking and pruning. #LeetCode #DSA #Backtracking #Recursion #Algorithms #Java #SoftwareEngineering

🔥 DSA Challenge – Day 135/360 🚀 📌 Topic: Bit Manipulation 🧩 Problem: Check if a Number is Power of 2 Problem Statement: Given an integer n, determine whether it is a power of 2 or not. 🔍 Example: Input: n = 4 Output: true Input: n = 6 Output: false 💡 Approach: Optimized (Bit Manipulation) 1️⃣ Step 1 – If n <= 0, return false (powers of 2 are always positive) 2️⃣ Step 2 – Use bit trick: n & (n - 1) removes the lowest set bit 3️⃣ Step 3 – If result becomes 0, then only one set bit was present ✔ This avoids looping and works in constant time ⏱ Complexity: Time: O(1) Space: O(1) 📚 Key Learning: A number is a power of 2 if it has exactly one set bit in binary representation. #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #135DaysOfCode #BitManipulation #LeetCode

🚀 Day 15 of My DSA Journey Today’s problem: Array Rank Transform 💡 Problem Insight: Given an array, replace each element with its rank when the array is sorted. Rank starts from 1 Equal elements → same rank Ranks must be continuous (no gaps) 🧠 Approach: 🔹 Clone and sort the array 🔹 Use a HashMap to assign ranks to unique elements 🔹 Traverse original array and replace values using the map ⚡ Key Learning: 👉 This is a classic example of Coordinate Compression 👉 Helps reduce large values into a smaller ranked range 💻 Code Logic: Sort copy of array Assign rank only if element not already mapped Replace original values using the map 📊 Result: ✅ 43 / 43 test cases passed ⏱ Runtime: 31 ms 📈 Beat: 58.58% 💾 Memory: 74.74 MB (85.14% better than others) 💭 Takeaway: Simple problems can test clean thinking + handling duplicates properly. Hashing + sorting is a powerful combo 🔥 🔥 Consistency Check: Day 15 Complete Building discipline one problem at a time. #DSA #CodingJourney #LeetCode #Java #ProblemSolving #Consistency #100DaysOfCode

😎 Day 39 of Solving dsa -->longest consecutive sequence. 🚀 Approach First, add all elements of the input array nums into a hash set for efficient O(1) lookups. Initialize a variable longest_streak to 0. Iterate through the array nums again. For each number num: Check if num - 1 exists in the hash set. If it does, num is part of a longer sequence that will be processed later, so we skip it. If num - 1 does not exist in the hash set, num is the start of a consecutive sequence. Start a nested loop from num, incrementing by one, and check if each subsequent number exists in the set. Count the length of this sequence. Update longest_streak with the maximum length found so far. Return longest_streak. 📈 Complexity Time complexity: O(n) — Although there is a nested loop, each element in the hash set is visited at most twice: once during insertion and once during a streak check. Space complexity: O(n) — We store all n elements in the hash set. #programming #solving #java #dsa #100 days challenge #logic #leetcode #striversheeet #tuf #daily challenge