Day 30 of #30DaysOfCode with Educative Challenge: Continuous Subarray Sum (Medium) Core Idea: Detect a subarray whose sum is a multiple of a given number The Pattern: Running total with remainder tracking Why It Works Here: Keep a running total while walking through the array and record the first position where each remainder (after dividing by the given number) appears. When the same remainder shows up again at least two positions later, it means the numbers between those two positions add up to a value that fits the requirement. Production Lesson: This pattern is useful whenever repeated “states” over a stream need to be detected quickly. Log analysis, rolling counters, or anomaly detection can often be reduced to tracking how cumulative values repeat over time instead of re-scanning all ranges. Edge Worth Remembering: Storing an initial remainder state at a virtual index before the array starts allows subarrays beginning at the first element to be handled cleanly. #Educative #Python #SoftwareEngineering #ContinuousLearning #ProblemSolving
Detecting Subarray Sums with Educative's 30 Days of Code Challenge
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🚀 DSA Practice – Daily Progress 📌 Problem: Sort an array of 0s, 1s, and 2s (without using built-in sort) Today I solved a classic DSA problem using the Dutch National Flag algorithm. 🔍 Approach: Used three pointers (low, mid, high) Segregated 0s, 1s, and 2s in a single pass Achieved in-place sorting with constant extra space ⏱ Complexity: Time: O(n) Space: O(1) 📖 Key Takeaway: Simple pointer-based greedy strategies can lead to highly efficient solutions when constraints are well understood. #DSA #ProblemSolving #Python #Algorithms #CodingPractice #GeeksforGeeks #LearningJourney
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🧠 LEETCODE CONSISTENCY SERIES 🚀 Day 1️⃣ of 365 Days 🔁 📘 Topic: Array + Hashing 🧩 Problem: LeetCode #1 – Two Sum ⏱ Time Taken: ~30 mins 💡 Key Learning: .Brute force approach works but has O(n²) time complexity .Nested loops are simple but not scalable for large inputs .Problem guarantees exactly one solution, which simplifies logic ❌ Mistake I made: .Started with brute force without thinking about time optimization 📌 Improvement for tomorrow: .Solve the same problem using HashMap (O(n)) approach .Focus on recognizing patterns for optimization early 🚀 Consistency > Complexity #DSA #LeetCode #ProblemSolving #LearningInPublic #Python #CodingJourney #Consistency
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Daily leetCode Question Day -7 Solved LeetCode 1630 – Arithmetic Subarrays ✅ 🔹 Approach: For each query, I extracted the subarray, sorted it, and verified whether the difference between consecutive elements remains constant — a direct and reliable way to check if it can form an arithmetic sequence. 🔹 Time Complexity: Each query: O(k log k) (due to sorting) Overall: O(q × k log k) 🔹 Space Complexity: O(k) for the temporary subarray This solution prioritizes clarity and correctness, making it easy to reason about while still meeting performance constraints. #LeetCode #DSA #Algorithms #ProblemSolving #Python #CodingJourney #TimeComplexity #SpaceComplexity
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🚀 𝟲𝟬 𝗗𝗮𝘆𝘀 𝗼𝗳 𝗖𝗼𝗱𝗶𝗻𝗴 | 𝗗𝗦𝗔 𝘅 𝗥𝗲𝗮𝗹 𝗪𝗼𝗿𝗹𝗱 𝗣𝗿𝗼𝗷𝗲𝗰𝘁𝘀 #Day34 | 𝗖𝗼𝘂𝗿𝘀𝗲 𝗣𝗿𝗲𝗿𝗲𝗾𝘂𝗶𝘀𝗶𝘁𝗲 𝗣𝗹𝗮𝗻𝗻𝗲𝗿 Built a Course Prerequisite Planner using Topological Sort (Kahn’s Algorithm) to resolve course dependencies and generate a valid completion order. Focused on: • Directed graphs • Indegree-based processing • Cycle detection • Understanding how dependency resolution works in real systems 📌 𝗖𝗼𝗱𝗲 𝗮𝗻𝗱 𝗱𝗼𝗰𝘂𝗺𝗲𝗻𝘁𝗮𝘁𝗶𝗼𝗻: https://lnkd.in/gxzGJ4nB Feedback and suggestions are welcome. #DSA #Graphs #TopologicalSort #Dependencies #Python #60DaysOfCoding #LearningInPublic #SoftwareEngineering
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🧠 LEETCODE CONSISTENCY SERIES 🚀 Day 3️⃣ of 365 Days 🔁 📘 Topic: String / Sliding Window 🧩 Problem: LeetCode #3 – Longest Substring Without Repeating Characters ⏱ Time Taken: 35 mins 💡 Key Learning: - Sliding Window helps reduce brute-force O(n²) solutions to O(n) - Using a set or hashmap avoids repeated substring scanning ❌ Mistake I made: - Initially used string operations (`in`, `index`) which increased time complexity 📌 Improvement for tomorrow: - Practice more sliding window problems with hashmaps Consistency > Complexity 🚀 #DSA #LeetCode #SlidingWindow #Python #ProblemSolving #LearningInPublic
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LeetCode Journey 🚀 | Solved Problem #9 – Palindrome Number ✅ Just solved LeetCode Problem 9 – Palindrome Number ✅ 📌 Key learnings from this problem: Understood how to check palindromes without converting to string Learned the importance of edge cases (negative numbers, numbers ending with 0) Practiced reversing only half of the number for better optimization This problem helped me improve my logical thinking and confidence in handling conditions step by step. Small progress, but moving forward consistently 💪 #LeetCode #DSA #ProblemSolving #LearningJourney #Consistency #CareerRestart #Python https://lnkd.in/gK7nVYjV
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LeetCode Problem 54: "Spiral Matrix" : Given an m x n matrix, return all elements of the matrix in spiral order. The logic is quite simple- Define the grid boundaries, store traversed elements in a list and shrink boundaries as you traverse through the grid spirally. The time complexity will be O(n*m) where n and m are no. of rows and columns in grid (traverse each element) #dsa #matrix #leetcode #optimalsolution #python #competitiveprogramming #algorithms #datastructures #problemsolving #dailycoding #codingchallenge #math #logic #datastructures
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🚀 Day 6 of LeetCode | Integer to Roman Today’s challenge was Integer to Roman — converting a given integer into its Roman numeral representation. Approach: Use predefined integer values and their Roman symbols Apply a greedy strategy to subtract the largest possible value Handle subtractive cases like IV, IX, XL, XC Build the result step by step until the number becomes zero ⏱ Time Complexity: O(1) 📦 Space Complexity: O(1) #LeetCode#DSA #ProblemSolving #Python #CodingJourney
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Day 99: Find Minimum in Rotated Sorted Array 📉The penultimate day! We’re using Binary Search to find the "inflection point" in a rotated array in $O(\log n)$. The trick is comparing nums[mid] to the left boundary: if nums[mid] is greater, the minimum must be to the right; otherwise, it's to the left or is the mid itself. We also include a shortcut—if the current window is already sorted (nums[l] < nums[r]), the leftmost element is our winner! 🏆#Day99 #BinarySearch #LeetCode #Algorithms #Python #100DaysOfCode #DSA #CodingChallenge
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