LeetCode Daily – #1784 Check if Binary String Has at Most One Segment of Ones [Easy] The Insight: Since the string always starts with '1' (no leading zeros), the only way a second segment of ones can appear is if a '0' comes BEFORE another '1' — i.e., the substring "01" exists in the string. So the entire problem reduces to one check: → Does "01" appear in the string? If yes → false. If no → true. Time: O(N) | Space: O(1) This is one of those problems where stripping away the noise leads to an almost trivially elegant solution. The key was recognizing the constraint — no leading zeros — which guarantees the string always opens with a '1', making "01" the only possible signal of a broken segment. Sometimes the best code is the code you don't write. 🚀 #LeetCode #CodingChallenge #CPlusPlus #ProblemSolving #Programming #SoftwareEngineering #POTD #DSA
Check if Binary String Has at Most One Segment of Ones
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42 of #100DaysOfCode Solved LeetCode 930 – Binary Subarrays With Sum 💡 Today’s problem was a great example of how powerful Prefix Sum + HashMap can be when dealing with subarrays. 🔹 Approach Used: Instead of checking all subarrays (which would be inefficient), I used a running prefix sum and stored its frequency in a hashmap. This reduces the time complexity to O(n) — much more efficient than brute force! ⚡ 🔹 What I Learned: Prefix sum helps convert subarray problems into lookup problems HashMap optimizes repeated computations Always think: “Can I store previous results to save time?” #LeetCode #DSA #CodingJourney #Programming #Cpp #SoftwareEngineering #ProblemSolving #LearnToCode
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LeetCode Practice: Today I solved Problem no.328(Odd Even Linked List) Statement- We are given the head of a singly linked list , group all the node with odd indices together followed by the node of even indices , and return the reordered list. Approach(Brute Force) Key Idea: I used ana auxiliary array to store the node values 1-Traverse the list by using temp->next->next starting from head to collect the odd indexed nodes 2-Then temp=head->next and traverse it using temp->next->next to collect the even indexed nodes 3-This ensures that all the odd index were stored first followed by even indexed nodes 4- Finally convert the array into the linked list. Time Complexity: O(N) Space Complexity: O(N) (due to the extra array) #LeetCode #DataStructures #LinkedList #CodingPractice #DSA #Programming #ProblemSolving
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Found a great tool: Dozzle Dozzle is a self-hosted application for monitoring and logging Docker containers in real-time. It provides shell access, multi-host support, alerts, and SQL-based log analysis capabilities. Check it out: https://lnkd.in/dBnjBJ7p #DevTools #Programming #WebDev
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🚀 Day 182 of #200DaysOfCoding 📌 Problem: Check if Binary String Has at Most One Segment of Ones (LeetCode 1784) Today’s challenge was about checking whether a binary string contains at most one continuous segment of '1's. 🧠 Problem Understanding We are given a binary string s that starts with 1 and contains only 0 and 1. We need to verify that all 1s appear in one contiguous block. ✔ Valid Examples 111000 110 1000 ❌ Invalid Example 1001 → Here, the segment of 1s breaks and appears again later. 💡 Key Idea If we ever encounter 0 followed by 1, it means a new segment of 1s has started, which violates the condition. ⏱ Complexity Time Complexity: O(n) Space Complexity: O(1) Consistent practice like this is helping me strengthen my problem-solving and data structure skills every day. #Day182 #200DaysOfCoding #LeetCode #DSA #Programming #CodingJourney #ProblemSolving
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When you miss a base case🧠 What is the best way to check it, what do you guys think??💬 When working with recursion, the most important step is often the simplest one: 💻Define your base case first. Before thinking about the recursive calls, pause and ask, like what’s the worst or simplest condition where the function should stop? Most of the time, that’s your base case. A small reminder from experience: 👉 Start with the base case 👉 Think about edge conditions first 👉 Then build the recursive flow Strong foundations make complex problems manageable. You can also check out✨, and learn CPP from the basics💻: https://lnkd.in/gdPC4D6f #Recursion #Programming #CPP #DSA #problemSolving #leetcode #basecase #lpu
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🚀 Just solved LeetCode 3643 - Reverse Submatrix Today I worked on a really interesting matrix manipulation problem where the goal was to reverse a k×k submatrix efficiently. 💡 Key Insight: Instead of using extra space, I applied a two-pointer approach to swap rows from top to bottom — making the solution clean and optimal. ⚡ What I learned: • In-place matrix transformations • How to optimize nested loops • Importance of visualizing row/column operations 🧠 Complexity: Time → O(k²) Space → O(1) 📌 Small problems like this build strong fundamentals for bigger DSA challenges! 💻 Code + Explanation PPT available on my GitHub 👇 #LeetCode #DSA #CodingJourney #CPlusPlus #ProblemSolving #Tech #Programming #100DaysOfCode
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Revisiting fundamentals with a structured approach. Built a prime number checker in C++ using a boolean flag pattern to enhance readability and control flow. Emphasizing: • Logical clarity • Maintainable structure • Efficient iteration Strong fundamentals → Better engineering decisions. #DeveloperMindset #Optimization #FAANGPreparation #Cplusplus #CPP #Programming #SoftwareEngineering #Coding #DataStructure #Algorithms #DSA #ProblemSolving #LogicBuilding #CleanCode #CodeQuality #Developer #Tech
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LeetCode Parctice: Problem no.217 (Contains Duplicate) Statement- A array is given , we have to return true if any value appears at least twice in the array, and return false if every element is distinct. Approach- Brute Force Key Idea: - I used sets data structure for this problem as set only store unique elements. -While traversing the array, I check whether the current element already exists in the set. If the element is already present → it means a duplicate exists, so return true. Otherwise, insert the element into the set and continue checking. Time Complexity = O(n) Space Complexity = O(n) #DSA #CodingPractice #Cpp #Programming #ProblemSolving
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Leetcode POTD : Find Unique Binary Strings... Given an array that contains n unique binary strings, each of length n, and some possible strings are missing. Target: find a binary string of length n that is not present in the array. Thought process... Initially I thought of generating all possible binary strings of length n. Since the constraint was small, it passed all the test cases. But then I felt generating all 2^n strings just to find one missing is not really optimal, so I started thinking about a better approach 🤔 If we observe the examples carefully, we notice something interesting. There are n strings and each string has length n. Ohh yes, here is the trick 👀 Since the ansmust also be a string of length n, we can construct a new one ourselves. All the given strings are already unique, so if we flip the bit at index i from the i-th string (0 -> 1 or 1 -> 0), the newly formed string will differ from every string at least at one position. That guarantees the string will be unique and missing from the array ✨ Using this idea, the missing binary string can be found in O(n) time. #leetcode #potd #binary #strings #programming #problem_solving #consistency
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