Solved LeetCode Problem 3228: Moving Ones to the End

🗓 Day 14 / 100 – #100DaysOfLeetCode 📌 Problem 1437: Check If All 1’s Are at Least K Places Away The goal was to determine whether every pair of consecutive 1s in the array is separated by at least k zeros. 🧠 My Approach: Traversed the array while tracking the index of the previous 1. On encountering a new 1, checked the distance from the last one. If the gap was smaller than k, the condition failed instantly. This single-pass logic keeps the solution efficient and clean. 💡 Key Learning: This problem reinforced the importance of index tracking and using simple arithmetic comparisons to validate structural constraints. It’s a great reminder that not all problems require heavy algorithms — sometimes, clarity and careful traversal are enough for an optimal solution. Small steps forward build strong reasoning over time 🚀 #100DaysOfLeetCode #LeetCodeChallenge #Python #ProblemSolving #LogicBuilding #Arrays #Algorithms #DataStructures #DSA #CompetitiveProgramming #CodingJourney #SoftwareEngineering #LearningInPublic #DeveloperJourney #TechStudent #CareerGrowth #CodeEveryday #CodingCommunity #KeepLearning #Programming #TechCareer

🗓 Day 4 / 100 – #100DaysOfLeetCode 📌 Problem 728: Self Dividing Numbers The task was to find all numbers in a given range that are self-dividing — i.e., numbers that are divisible by each of their digits and contain no zero. 🧠 My Approach: Iterated through the range of numbers. Checked each digit of a number to ensure: It’s non-zero. The original number is divisible by that digit. Collected all numbers that satisfied both conditions. ⏱ Time Complexity: O(n × d) — where d is the number of digits in each number. 💾 Space Complexity: O(1) 💡 Key Learning: This problem strengthened my understanding of digit-wise iteration and conditional logic — simple concepts that are essential for handling number-based and math-intensive problems efficiently. Small problems like these build the habit of writing clean, readable, and logical code — one step closer each day 🚀 #100DaysOfLeetCode #LeetCodeChallenge #Python #ProblemSolving #LogicBuilding #Programming #DataStructures #Algorithms #DSA #CodingJourney #CodingChallenge #CodeEveryday #LearningInPublic #CompetitiveProgramming #DeveloperJourney #TechStudent #CareerGrowth #CodingCommunity #KeepLearning

🚀 Day 54 of #100DaysOfCode Solved LeetCode Problem 2011. Final Value of Variable After Performing Operations ✅ This problem tests simple yet essential programming logic — understanding how pre/post increment and decrement operations affect variable states. Given a list of operations like ["--X", "X++", "++X"], the goal is to compute the final value of X after applying all updates sequentially. 💡 Key Insight: Each operation (++X, X++) increases the value by 1, while (--X, X--) decreases it by 1. The implementation can be efficiently handled in O(n) time by iterating through the operations once. ⚙️ Result: Runtime: 0 ms ⚡ Beats 100% of Python submissions Memory Usage: 17.76 MB (Beats 60.70%) Another step forward in improving my algorithmic problem-solving and code optimization skills 💪 #LeetCode #Python #100DaysOfCode #CodingJourney #ProblemSolving #DailyPractice #TechLearning #MythylyCodes

🚀 LeetCode Daily Challenge 📘 Problem: Implement Stack using Queue (LeetCode #225) Today, I implemented a stack (LIFO) using a single queue (FIFO) — a classic problem that strengthens understanding of data structure manipulation. 💪 Key learning points: Efficiently rotating elements in a queue to simulate stack behavior Mastering core operations: push, pop, top, and empty Reinforcing how data structure constraints lead to creative solutions #LeetCode #DataStructures #Python #DSA #Learning #Programming #TechJourney #Queue #Stack

🚀 Day 4 – LeetCode Problem Solving Journey 💻 Today’s challenge was “Merge Sorted Array” (LeetCode #88) — a classic array manipulation problem that helps improve understanding of two-pointer techniques and in-place merging. Problem: You are given two sorted arrays, nums1 and nums2, and the task is to merge them into a single sorted array, stored inside nums1. Example: Input: nums1 = [1,2,3,0,0,0], m = 3 nums2 = [2,5,6], n = 3 Output: [1,2,2,3,5,6] 🧠 Concept Used: Two-pointer approach starting from the end Comparison and in-place insertion Time Complexity: O(m + n) Space Complexity: O(1) Learning how to handle in-place data modification efficiently is a key step in mastering DSA! #LeetCode #Day4 #ProblemSolving #DSA #CodingJourney #100DaysOfCode #Python #Programming #DevelopersCommunity #SoftwareEngineering #LearningInPublic #CareerGrowth

🚀 Day 87/100 of the #100DaysOfCode Challenge! #Day87 of #100Dayscodingchallenge:Today's focus was on mastering control flow and precision with nested loops in Python. I tackled a series of classic pattern problems that are fantastic for understanding logical structuring and boundary conditions. The Challenge: Build hollow geometric patterns using only for loops and conditional statements. Patterns Solved: ✅ Hollow Pyramid: A test of symmetry and managing increasing spaces. ✅ Hollow Right-Angled Triangle: Focusing on the first and last elements of each row. ✅ Inverted Hollow Pyramid: The reverse logic of its upright counterpart. ✅ Hollow Diamond: Combining the logic of both the standard and inverted pyramids for a complex, symmetrical shape. These exercises are more than just printing stars; they're about sharpening problem-solving skills, breaking down complex problems into manageable steps, and writing efficient, clean code. It's a powerful reminder that solid fundamentals in loops and conditionals are the building blocks for more advanced algorithms. Check out the code snippet in the comments! 👇 #Nxtwave #ccbp #intensive #Python #Programming #CodingChallenge #Algorithms #ProblemSolving #SoftwareDevelopment #LearnToCode #CodeNewbie #Developer #ProgrammingPatterns

✅ Day 53 – GFG 160 Days of Problem Solving Challenge 💡 Problem: Find Pair with Sum Closest to Target on GeeksforGeeks At first, I considered checking all pairs using two nested loops — but that approach would be O(n²) and inefficient for large arrays. 🧠 Optimized Approach (Two-Pointer Technique): Sort the array. Initialize two pointers — left at the start and right at the end. Calculate the pair sum and track the difference from the target. Update the closest pair whenever a smaller difference is found. Move pointers intelligently based on whether the sum is less than or greater than the target. This logic brings down the time complexity to O(n log n) due to sorting and O(n) for the two-pointer traversal. 🔥 Result: Solved 1115/1115 test cases successfully with 100% accuracy in just 0.55 sec! 🚀 The two-pointer method is not only efficient for exact sums but also powerful for problems involving closest sums — a great optimization over brute force! Onwards to Day 54! 💪 #DSA #GeeksforGeeks #ProblemSolving #Python #TwoPointer #Optimization #CodingChallenge #160DaysOfCode #Day53

From equations on paper to code: Define state space models instantly with python and c4dynamics 🔥 c4dynamics brings the legacy of state space modeling into the Python era: The state vector isn’t just math. It’s programmable: s = state(x=180, v=0) print(s.X) # [180, 0] One framework. One philosophy. The same pursuit that started centuries ago: understanding how systems move, react, and adapt. State space programming is the future of dynamical systems. What’s stopping you from turning your models into code?

c4dynamics brings the legacy of state space modeling into the Python era: The state vector isn’t just math. It’s programmable: s = state(x=180, v=0) print(s.X) # [180, 0] One framework. One philosophy.

✅Day 83 of #100DaysOfLeetCode 1.📌Problem: Find the Pivot Integer Given a positive integer n, find the pivot integer 𝑥 such that the sum of all elements between 1 and 𝑥 inclusively equals the sum of all elements between 𝑥 and n inclusively. Return the pivot integer 𝑥. If no such integer exists, return -1. 2.🟢 Difficulty: Easy 3.📍Topic: Arithmetic & Basic Reasoning 4.🎯 Goal: Identify the integer x where the sum of numbers before and after (including) 𝑥z are equal for a given range.image.jpg 5.🧠Key idea: Approach 1: Find the total sum of numbers from 1 to n using 𝑛(𝑛+1). Incrementally compute the prefix sum from 1 up to 𝑖 and compare it with the suffix sum from 𝑖 to n. If prefix sum equals suffix sum for any i, that 𝑖 is the pivot integer; otherwise, return -1. #LeetCode #100DaysOfCode #CodingChallenge #Programming #DataStructures #Algorithms #CodingJourney #LearningJourney #GrowthMindset #ProblemSolving #DeveloperCommunity #Python #Java #SoftwareEngineering #TechCareersbest