Solved LeetCode Problem 28: Find First Occurrence of Substring

🔍 Day 66 of #100DaysOfCode 🔍 🔹 Problem: Binary Search – LeetCode ✨ Approach: Implemented an iterative binary search to efficiently locate the target element within a sorted array. By halving the search range each time — adjusting low and high around the mid-point — the algorithm achieves blazing-fast lookups! ⚡ 📊 Complexity Analysis: Time Complexity: O(log n) — array size halves with each iteration Space Complexity: O(1) — constant extra space ✅ Runtime: 0 ms (Beats 100.00%) ✅ Memory: 46.02 MB 🔑 Key Insight: Binary Search is proof that efficiency isn’t about doing more — it’s about eliminating what’s unnecessary. 🚀 #LeetCode #100DaysOfCode #ProblemSolving #DSA #AlgorithmDesign #BinarySearch #LogicBuilding #Efficiency #ProgrammingChallenge #CodeJourney #CodingDaily

🧩 LeetCode Challenge – Day 72 ✅ Dived into a string decoding problem today — a perfect blend of stacks, recursion, and pattern tracking. 🔗 LeetCode 394 – Decode String This challenge revolves around decoding expressions like 3[a2[c]], requiring careful handling of nested patterns and character reconstruction. It’s a great exercise in managing multiple layers of logic — keeping track of counts, substrings, and the overall decoded output step by step. 💡 Key Takeaways: • Stack-based parsing builds precision in handling nested structures. • Breaking problems into layers simplifies even the most complex logic. • Attention to detail is everything — one misplaced index can change the entire output. #Day72 #LeetCodeChallenge #100DaysOfCode #DSA #CodingJourney #ProblemSolving #Stack #Strings #Parsing #Recursion

🌟 Day 93 of My LeetCode Journey — Problem 686: Repeated String Match 💡 Problem Insight: Today’s problem explored how many times string A must be repeated so that string B becomes a substring of it. It’s a fun mix of string repetition and pattern matching — testing both logic and edge-case awareness. 🧠 Concept Highlight: The core idea is to keep repeating A until it’s at least as long as B, then check if B exists within it (or one more repetition). This problem reinforces concepts of string concatenation, searching, and boundary conditions in text processing. 💪 Key Takeaway: Sometimes, solutions aren’t about complexity but about knowing when to stop repeating — in both coding and persistence! ✨ Daily Reflection: Simple problems like these highlight the beauty of algorithmic reasoning — transforming trial-and-error into structured logic. #Day93 #LeetCode #100DaysOfCode #StringMatching #ProblemSolving #DSA #CodingJourney #LearnByDoing #SoftwareEngineering #Persistence

🚀 Day 37 of #100DaysOfCode Challenge 🚀 🧠 Problem: 🧩 LeetCode 282 — Expression Add Operators (Hard) This problem was all about exploring backtracking with arithmetic logic. We’re given a string of digits and need to insert +, -, or * between them to reach a target value. ⚙️ Approach Used backtracking to: -Pick every possible substring as the next number. -Try adding each operator before it (+, -, *). -Track three things: -current value → total so far -previous operand → to handle multiplication correctly -expression → the string we’re building -Also skipped invalid paths like numbers with leading zeros. 📘 Learning -Understood how to evaluate expressions on the fly without using eval(). -Learned to manage operator precedence (* before +/-) using recursion and previous operand adjustment. -Realized that some problems can’t be optimized further — they test how well you can prune and organize recursion. #100DaysOfCode #DSA #CodingChallenge #StriversSheet #ProblemSolving #TakeUForward  #Recursion #CodingInterview #ProgrammingTips #LeetCode  #Backtracking 

💡 Day 78 of #100DaysOfCode 💡 🔹 Problem: Find Closest Number to Zero – LeetCode ✨ Approach: Traversed through the array while keeping track of the number closest to zero using absolute difference comparison. Handled ties by preferring the positive number — because even in code, positivity wins 😉 📊 Complexity Analysis: ⏱ Time Complexity: O(n) — single pass through the array 💾 Space Complexity: O(1) — no extra space used ✅ Runtime: 3 ms (Beats 44.80%) ✅ Memory: 46.87 MB 🔑 Key Insight: Even the smallest differences matter — especially when you’re finding what’s closest to zero! ⚡ #LeetCode #100DaysOfCode #DSA #ProblemSolving #CodingChallenge #JavaProgramming #AlgorithmDesign #CodeJourney #StayPositive

There are a bunch of hard problems on LeetCode that use basic graph traversal, which is pretty straightforward. However, they might be a bit more complex because we need to create, or identify how to create, a graph representation. The post: https://lnkd.in/dtQCR26x The problem: https://lnkd.in/dEfwjQPN #LeetCode #DSATips #DSA #ThinkDSA #ProblemSolving

🚀 LeetCode POTD — 3228. Maximum Number of Operations to Move Ones to the End 🎯 Difficulty: Medium | Topics: Greedy, String Manipulation, Two-Pointer Logic 🔗 Solution Link: https://lnkd.in/gAAxvfFV Today’s #LeetCode Problem of the Day was an interesting greedy + two-pointer problem. We’re given a binary string, and the task is to repeatedly move '1' over blocks of '0' until it reaches either another '1' or the end — and count the maximum operations possible. 🧠 My Approach: The key observation is that every '0' segment contributes operations equal to the number of '1's seen before it. I used a two-pointer technique: l → counts how many '1's we’ve encountered so far r → scans the string Whenever we encounter a '0' block, we add l to the result Every time we see a '1', we increment l This handles all possible valid moves optimally without actually simulating the operations. 💡 Core Idea: 👉 Each '1' can “jump” over every '0' segment that appears after it — the count of such jumps is the total number of operations. 📈 Complexity: Time: O(n) Space: O(1) 💬 Takeaway: Greedy algorithms often work beautifully when you track just the right variables. This problem shows how understanding movement relationships in a sequence can simplify what looks like a complicated operation. #LeetCode #ProblemOfTheDay #StringManipulation #GreedyAlgorithms #TwoPointers #DSA #CodingChallenge #SoftwareEngineering #CodingJourney #TechCommunity #100DaysOfCode

#Day67 of my LeetCode Journey 🚀   Starting with Recursion problems !  🧩 Question #8 – String to Integer (atoi) (Medium)  Brute Force Approach:   1) Convert the string manually step-by-step remove whitespaces, handle signs, read digits, and clamp values.   2) Check for invalid inputs and handle edge cases like overflow/underflow explicitly.   3) Use iterative parsing to process characters sequentially.   - Time Complexity: O(N)   - Space Complexity: O(1)  Optimal (Recursive) Approach:   1) Use recursion to simplify the parsing process each helper function handles one part (like skipping spaces, checking sign, removing zeros, or reading digits).   2) Extract digits recursively until a non-numeric character appears.   3) Convert digits to an integer and clamp it within the 32-bit signed range.   - Time Complexity: O(N)   - Space Complexity: O(N) (due to recursive calls)  ✨ That’s it for Day 67. The journey continues — see you on Day 68. Happy coding! 🚀 #Day67 #LeetCodeJourney #LeetCode #Recursion #String #Atoi #Python #DSA #CodingPrep

🔹 Day 70 of #100DaysOfLeetCodeChallenge 🔹 🚀 Problem: Combination Sum III 🔑 Topic: Backtracking 🧠 Approach: We need to find all combinations of k distinct numbers (1–9) that sum to n. Here’s how I solved it 👇 Use backtracking to explore all combinations starting from 1 to 9. Add the number and move forward recursively, reducing both k (count) and n (target). Stop when k == 0 and n == 0 → valid combination found! Backtrack by removing the last number and continue exploring. ⏳ Time Complexity: O(2⁹) ≈ O(512) 💾 Space Complexity: O(k) (recursion + temp list) 📌 Example: Input: k = 3, n = 9 Output: [[1,2,6],[1,3,5],[2,3,4]] ✅ 🎯 Takeaway: This problem teaches the power of controlled recursion — when both depth and sum conditions guide the search efficiently. 💡 #LeetCode #DSA #Backtracking #ProblemSolving #CodingChallenge #100DaysOfLeetCodeChallenge 🚀

I recently tackled LeetCode 2654 – Minimum Number of Operations to Make All Array Elements Equal to 1. Most solutions simulate every operation. The approach that worked for me came from stepping back and thinking mathematically: 1️⃣ If there’s already a 1 in the array, it can be spread to all other elements efficiently. 2️⃣ If no 1 exists, find the shortest subarray whose GCD is 1 – creating a 1 from that subarray is the minimal first step. 3️⃣ Once a 1 exists, the rest follows in a straightforward way. It’s a reminder that sometimes understanding the structure of a problem beats brute-force coding. And here’s a little validation: ✅ Runtime: 1ms, beats 100% of submissions ✅ Memory: 56.1MB, beats 100% of submissions For anyone doing coding challenges or preparing for interviews, focus on insights like these – they often save more time than writing extra lines of code. #ProblemSolving #Algorithms #CodingMindset #LeetCode #SoftwareEngineering