LeetCode 1: Two sum 2️⃣➕7️⃣=9️⃣ Suppose you have a list of numbers and you need to find two numbers that add up to a target value. You want to know their positions in the list. How does the code work? - We use a dictionary to remember the numbers we’ve seen so far and their positions. - As we look at each number, we check if there’s another number (the complement) that would add up to the target. - If we’ve seen that complement before, we return the positions. - If not, we store the current number and its index for future checks. Complexity: - Time Complexity: O(n) — Only one pass through the array. - Space Complexity: O(n) — Extra space for the dictionary to store numbers and their indices. Check out the problem here: https://lnkd.in/gppmcPDB Keep going, keep revising, and keep building confidence! 💪🔥 #DSA #Coding #ProblemSolving #Learning

LeetCode 219: Contains duplicate II ⁇ Suppose you have a list of numbers, and you want to check if any number appears at least twice close together—within k positions apart. How does the code work? - We use a dictionary to track the last index where each number appeared. - As we go through the list: - If the current number has been seen before, - Check the gap between indices. - If the gap (i−j)is less than or equal to k, we found a nearby duplicate! - Otherwise, update the index to the current position. - If the number is new, just store this index. - If we finish the list with no matches, return False. Complexity: - Time Complexity: O(n) — Scan through all numbers in a single pass. - Space Complexity: O(n) — Need to remember each distinct number’s last position. Check out the problem here: https://lnkd.in/gUtKpsHf Keep going, keep revising, and keep building confidence! 💪🔥 #DSA #Coding #ProblemSolving #Learning

LeetCode 136: Single number 🔂 Approach 1: Hash Map - We use a dictionary to count occurrences of each number. - Go through the list and update counts. - Finally, look for the number with a count of 1—that’s our answer! Complexity: - Time Complexity: O(n) — Check all items. - Space Complexity: O(n) — Dictionary to track counts. Approach 2: XOR (Bit Manipulation) - Initialize a variable ans = 0. - XOR every number in the list with ans. - Pairs cancel out (because a ^ a = 0 and 0 ^ b = b), so only the single number remains at the end. Complexity: - Time Complexity: O(n) — Just one pass through all numbers. - Space Complexity: O(1) — No extras, just ans. The XOR trick is super efficient and works like magic due to properties of binary operations! Check out the problem here: https://lnkd.in/giUbz53v Keep going, keep revising, and keep building confidence! 💪🔥 #DSA #Coding #ProblemSolving #Learning

✨ Generating Valid Parentheses Using DFS Problem Link:https://lnkd.in/e97DASha Today I worked on a classic backtracking problem generating all valid parentheses combinations for a given n. The interesting part is that every step has rules: 1.You can add '(' only if you still have remaining open brackets to use 2.You can add ')' only if it does not break the balance 3.The final string must be exactly 2 * n characters and perfectly balanced ✅ This problem beautifully shows how constraints guide creativity. The solution uses DFS (Depth First Search) to explore all possible sequences, but only keeps those that remain valid at every step like making the right choices early leads to the right result later. This is one of those problems that strengthens both logic and patience in coding. Small steps → careful decisions → perfect outcome. 🚀 #DSA #Backtracking #Recursion #CPP #CodingJourney #LeetCode #SoftwareEngineering #Algorithm #ProblemSolving #DeveloperMindset #KeepCoding #LearningEveryday

💻 Day 34 of #100DaysOfLeetCode Today’s Challenge: 83. Remove Duplicates from Sorted List 🔹 Problem: Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well. Key Insight: Because the list is already sorted, duplicates will always be adjacent—this allows us to remove them in a single pass using pointer manipulation. 🔍 Approach: Traverse the list using a pointer. Compare current node with the next node. If values are equal → skip the next node by linking to the next of next. Otherwise → move forward. 🕒 Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ Key Takeaway: This problem reinforces the importance of pointer manipulation and understanding how memory references work in linked lists. A simple check can eliminate duplicates efficiently without using extra space. Link:[https://lnkd.in/gsjVxHXM] #100DaysOfLeetCode #Day34 #LeetCode #ProblemSolving #DSA #Algorithms #LinkedList #CodingChallenge #CodeNewbie #InterviewPreparation #SoftwareEngineering #Programming #CodingCommunity #TechCareers #CareerGrowth #ArjunInfoSolution

𝐏𝐫𝐞𝐟𝐢𝐱 𝐒𝐮𝐦 — 𝐓𝐡𝐞 𝐏𝐚𝐭𝐭𝐞𝐫𝐧 𝐓𝐡𝐚𝐭 𝐓𝐮𝐫𝐧𝐬 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 𝐈𝐧𝐭𝐨 𝐂𝐥𝐚𝐫𝐢𝐭𝐲 when you’re solving DSA, some problems feel heavy because we repeatedly calculate the same thing again and again. Prefix Sum fixes that. It simply says: 👉 “Calculate once. Reuse forever.” You create an extra array that stores cumulative values, and suddenly: Range queries become O(1) Subarray problems become predictable Target-sum problems feel easier Frequency-based logic becomes clean Prefix Sum is not just a trick — it’s a mindset shift. You stop doing brute force. You start thinking smarter. Problems where Prefix Sum helps: Range Sum Queries Subarrays equal to K Subarray count / longest subarray types Even–odd frequency problems Difference arrays Kadane variations Many DP optimizations Keep learning these patterns — they are the real shortcuts in DSA. One pattern a day. One improvement a day. Keep going. for more let's Connect : Anil Rathod #dsa #leetcode #codingjourney #problemsolving #prefixsum #codingcommunity #programming #dailycoding

LeetCode #33 - Search in Rotated Sorted Array This is the classic variation of Binary Search, where the sorted array is rotated at some pivot point. Even tough it looks unsorted at first glance, one half of the array is always sorted and that's the key insight to solving this problem efficiently. Problem Summery: We are given an array that was initially sorted but then rotated at an unknown index. We need to find the index of the target element in O(log n) time. It it doesn't exist, return -1. Approach: Use Modified Binary Search At each step: - Determine which side (left or right) is sorted - Check if the target lies within that sorted half - Narrow the search range accordingly This ensures that every iteration halves the search space - keeping it logarithmic in complexity. Key Insights: - One of the two halves is always sorted in a rotated sorted array - Comparing nums[start], nums[mid], and nums[end] helps identify the sorted side - Efficiently narrows down to the correct segment without linear scanning Complexity Analysis: - Time Complexity: O(log n) - Space Complexity: O(1) Example: Input : nums = [4,5,6,7,0,1,2], target = 6 Output: 2 The algorithm quickly identifies that 6 lies in the sorted left half and returns its index. #LeetCode #DSA #CPP #C++ #ProblemSolving #BinarySearch #Coding #InterviewPreparation #TechLearning #Coding #Programming #DataStructure

🔹 DSA Daily | Count Trailing Zeroes Today, I solved the **Count Zeroes** problem — a simple yet insightful challenge that tests **array traversal and efficiency**. 💡  🧩 Problem Statement:   Given a **binary sorted array** (containing only 0s and 1s), count the number of **zeroes** efficiently.   Since the array is sorted, all zeroes appear together either at the start or end. ⚙️ Approach:   1️⃣ Start iterating from the **end of the array**.   2️⃣ Count zeroes until the first non-zero element is found.   3️⃣ Break the loop immediately when a non-zero is encountered to save time.  ⏱️ Time Complexity: O(n) — traverse from the end only   💾 Space Complexity: O(1) — constant extra space  This problem reinforces how **small logic tweaks** can make code more **efficient and readable**. 🚀  #DSA #CPlusPlus #Coding #ProblemSolving #Arrays #BinaryArray #CodeEveryday #GeeksforGeeks #LeetCode #ProgrammingJourney #DataStructures #Algorithms #InterviewPrep #CodingCommunity #LogicBuilding #EfficientCode #LearnToCode #TechJourney

LeetCode 217: Contains duplicate ⁇ Suppose you have a list of numbers, and you want to know if any number appears more than once. It’s like making sure everyone in class has a unique roll number! Approach 1: Set - Go through each number. - If it’s not in your set, add it. - If you ever see a number already in your set, you’ve got a duplicate! Approach 2: Hash Table (Counter) - Use a Counter to count how often each number appears in the list. - If any number appears more than once, return True (there’s a duplicate). Complexity: - Time Complexity: O(n) — One pass through all numbers. - Space Complexity: O(n) — Storing seen numbers or counts. Both approaches are fast and effective for catching duplicates quickly! Let’s check our numbers and make sure everyone’s unique! Check out the problem here: https://lnkd.in/g-JgRTEn Keep going, keep revising, and keep building confidence! 💪🔥 #DSA #Coding #ProblemSolving #Learning

🔹 DSA Daily | Check for Balanced Parentheses (C++) Balanced parentheses problems are a classic way to test your understanding of **stacks** and **logical thinking**. Today, I solved one such problem — checking if a string of parentheses, braces, and brackets is balanced. 💡 Problem Statement: Given a string containing '(', ')', '{', '}', '[' and ']', determine if the string is **balanced**. A string is balanced if every opening bracket has a corresponding closing bracket in the correct order. Approach: I used a **stack** to keep track of opening brackets.   - Push every opening bracket onto the stack.   - For closing brackets, check if the top of the stack has the matching opening bracket.   - If it does, pop it; otherwise, the string is unbalanced.   - At the end, if the stack is empty, the string is balanced. Time Complexity: O(n) — traversing the string once   Space Complexity: O(n) — for the stack  This problem highlights how **stacks make bracket-matching elegant and efficient**. 🚀 #DSA #CPlusPlus #Coding #ProblemSolving #Stack #BalancedParentheses #CodeEveryday #GeeksforGeeks #LeetCode #ProgrammingJourney #DataStructures #Algorithms #InterviewPrep #CodingCommunity #LogicBuilding #EfficientCode #LearnToCode #TechJourney