Two Sum Problem Solution with HashMap

DSA Journey Continues 🚀 Today I solved the “Product of Array Except Self” problem on LeetCode. This problem really helped me understand how to optimize without using division and extra space. Problem Goal: For each index, return the product of all elements except the current one. Step-by-Step Approach : 1. Left Product Pass Traverse from left to right Store product of all elements before current index Keep updating a variable left 2. Right Product Pass Traverse from right to left Multiply with product of all elements after current index Maintain another variable right Core Idea: First pass → store left products Second pass → multiply with right products Final array gives the result Why This is Efficient: No division used Time Complexity: O(n) Space optimized (no extra arrays, only output array) What I Learned Today: Prefix and suffix product concepts How to solve problems in two passes Writing optimized solutions without extra space Understanding patterns like this makes problem solving much easier over time. #Day4 #DSAJourney #Java #LeetCode #ProblemSolving #Coding #Learning #Consistency #100DaysOfCode 10000 Coders Pathulothu Navinder

🚀 Day 6 of LeetCode Problem Solving Solved today’s problem — LeetCode #1480: Running Sum of 1d Array 💻🔥 ✅ Approach: Prefix Sum (Running Sum) ⚡ Time Complexity: O(n) 📊 Space Complexity: O(n) The task was to compute the running sum of an array where each element represents the sum of all previous elements including itself. 👉 Example: Input: [1,2,3,4] Output: [1,3,6,10] 💡 Key Idea: Keep adding elements as you traverse the array and store the cumulative sum. 👉 Core Logic: Initialize sum = 0 Traverse array Add current element → sum += nums[i] Store in result array 💡 Key Learning: Simple problems help build strong fundamentals — mastering basics like prefix sum is very important for advanced problems. Grateful to my mentor Pulkit Aggarwal — your guidance is helping me strengthen my fundamentals every day 🙌 Consistency is the key — small steps lead to big results 🚀 #Day6 #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode

🚀 Day 8 of 30 Days Coding Challenge Today I solved a problem on LeetCode: Minimum Distance Between Three Equal Elements. 🔍 Approach: Initially implemented a brute-force solution using three nested loops (O(n³)). It worked for smaller constraints but highlighted the importance of optimizing. 💡 Key Learning: Instead of checking all possible triplets, grouping indices of identical elements and analyzing consecutive occurrences leads to a much more efficient solution. ⚡ Optimization Insight: Reduced time complexity from O(n³) → O(n) Leveraged HashMap to store indices and process only relevant combinations. 📈 Takeaway: This problem reinforced an important concept: 👉 When dealing with repeated elements, think in terms of grouping and patterns rather than brute force. Consistency is key — learning something new every day and improving step by step. #Day8 #30DaysOfCode #CodingChallenge #LeetCode #Java #DataStructures #Algorithms #ProblemSolving #CodingJourney #SoftwareDevelopment #LearnToCode #TechGrowth

Day 55 of 100 Days of LeetCode 💻 Today I solved Longest Consecutive Sequence — and honestly, this one taught me more about coding discipline than algorithms. At first, my approach was correct: Used HashSet for O(1) lookup Applied the “start of sequence” logic But I still got TLE. The reason? A tiny mistake: if(!set.contains(num-1)); That single ; made my condition useless and turned my O(n) solution into O(n²). 💡 Lesson learned: Don’t just think your logic is right → verify what your code actually does Small syntax mistakes can completely break optimal solutions Debugging is just as important as problem-solving Finally fixed it and got Accepted ✅ Slowly improving not just in DSA, but in writing cleaner and more careful code. #100DaysOfLeetCode #DSA #Java #CodingJourney #Learning

🚀 Day 7 of LeetCode Problem Solving Solved today’s problem — LeetCode #414: Third Maximum Number 💻🔥 ✅ Approach: Tracking Top 3 Maximums ⚡ Time Complexity: O(n) 📊 Space Complexity: O(1) The challenge was to find the third distinct maximum number in an array. If it doesn’t exist, return the maximum number. 👉 Instead of sorting (which takes extra time), I tracked the top 3 distinct maximum values in a single pass. 💡 Key Idea: Maintain three variables (max, smax, tmax) to store the first, second, and third maximum values while traversing the array. 👉 Core Logic: Skip duplicates Update max, second max, third max accordingly If third max doesn’t exist → return max 💡 Key Learning: Optimizing from sorting to a single-pass solution can significantly improve efficiency. Grateful to my mentor Pulkit Aggarwal — your guidance is helping me think more optimally 🙌 Consistency is the key — improving step by step 🚀 #Day7 #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode

🚀 Solved LeetCode Problem #47 – Permutations II Today I tackled a classic backtracking problem that adds an interesting twist—handling duplicate elements while generating permutations. 🔍 Problem Insight: Given an array that may contain duplicates, the goal is to generate all unique permutations without repetition. 💡 Approach Used: I used a Backtracking + Sorting strategy: First, sort the array to bring duplicates together Use a visited[] array to track used elements Apply a smart condition to skip duplicate choices during recursion This ensures that we only generate distinct permutations efficiently. 🧠 Key Learning: Handling duplicates in recursive problems is crucial Sorting helps simplify duplicate detection Backtracking becomes powerful when combined with pruning conditions 📈 Complexity: Time: O(n!) Space: O(n) ✨ This problem strengthened my understanding of: Backtracking techniques Recursion control and pruning Writing optimized solutions for combinatorial problems Consistency in solving such problems is helping me build stronger problem-solving skills every day! 💪 #LeetCode #DSA #Backtracking #Algorithms #Coding #ProblemSolving #Java #TechJourney

Day-3 of My Coding Journey Today I worked on the LeetCode problem: 👉 Find Numbers with Even Number of Digits What I learned: How to count digits in a number using String.valueOf(num).length() How to check if a number has even digits using % 2 == 0 ❌ My Mistake: I wrote: String.value(num).length % 2 == 0 ✔ Issues: Used value() instead of valueOf() Forgot () in length ✅ Fix: String.valueOf(num).length() % 2 == 0 After correcting this, my solution got Accepted! ✨ Takeaway: Small syntax mistakes can cause errors, but fixing them improves attention to detail and strengthens fundamentals. #Day3 #LeetCode #Java #CodingJourney #Learning 10000 Coders

🚀 Day 50 of My Coding Journey Today’s problem was simple but interesting: 👉 Replace all 0’s in a number with 5 Example: Input: 1004 Output: 1554 💡 Approach: Traverse each digit of the number Whenever a digit is 0, replace it with 5 Reconstruct the number 🧠 Key Learning: Even easy problems improve thinking about edge cases and number manipulation. Consistency matters more than difficulty. 🔥 50 days completed — not stopping here. Next target: Day 100 💯 #Day50 #CodingJourney #Java #ProblemSolving #Consistency #Learning My way ===== class Solution {   int convertfive(int num) {     int l=0,sum=0;     String s=""+num,s1="",s2="";     for(int i=0;i<s.length();i++)     {       if(s.charAt(i)=='0')       {         s1=s1+'5';        }       else       {         s1=s1+s.charAt(i);       }     }     for(int i=0;i<s1.length();i++)     {      s2=""+s1.charAt(i);       l=Integer.parseInt(s2);      sum=(sum*10)+l;     }     return sum;   } }

🚀 Day 13 / 50 – Wild Coding Kickoff 🔥 Today’s problem looked like it wanted a complicated solution… but it ended up teaching me something much simpler. I read the question and instantly thought: “Okay, substring search… this might get tricky.” 👀 Loops, comparisons, edge cases… my mind started going there. But then I stopped and asked: 👉 “Is there a simpler way?” That’s when I used this: class Solution { public int strStr(String haystack, String needle) { int index = haystack.indexOf(needle); return index; } } 💡 What does indexOf() actually do? It searches for the first occurrence of needle inside haystack Returns: ✅ The starting index if found ❌ -1 if the substring is not present 🔥 Example haystack = "sadbutsad" needle = "sad" 👉 Output: 0 (because "sad" starts at index 0) 🧠 What I learned today Sometimes we try to prove we know everything… But real skill is knowing when to keep things simple. Using built-in methods effectively is also a skill. #Day13 #50DaysOfCode #WildCodingKickoff #LeetCode #Java #CodingJourney #Consistency #KeepItSimple #ProblemSolving

🚀 Day 13/100 – LeetCode Journey 📌 Problem: Rotate Array (LeetCode 189) Today’s problem looked simple at first… but it taught me something deeper about how arrays and math work together. 🔹 Task: Rotate an array to the right by k steps. Example: [1,2,3,4,5,6,7], k = 3 → [5,6,7,1,2,3,4] --- 💡 Key Learning: Instead of shifting elements again and again (which is inefficient), I learned a smarter way using: 👉 "(i + k) % n" This single formula helped me: - Move elements to the correct position - Avoid index out-of-bounds - Understand how modulo (%) creates a circular effect --- 🧠 Biggest takeaway: At first, I struggled with: 👉 Why "3 % 7 = 3" 👉 Why we don’t use decimals in modulo Now it finally makes sense: ✔ Modulo gives the remainder after full division ✔ It helps wrap values within array limits ✔ Arrays can be treated like a circle 🔄 --- 💻 Approach Used: - Extra array to store rotated values - Place each element using "(i + k) % n" - Copy back to original array --- ✨ What I realized: Sometimes the problem isn’t coding… It’s understanding the math behind it. --- #Day13 #100DaysOfCode #LeetCode #Java #DSA #CodingJourney #ProblemSolving