Today’s random problem was 1743. Restore the Array From Adjacent Pairs. The question asks us to reconstruct an original array given only the adjacent pairs. We know the array has no duplicates and every adjacent pair is unique. Key Insight: 1. The endpoints (first and last elements) will only have one neighbor. 2. All middle elements will have exactly two neighbors. 3. To restore the array, we just need to find one of those "endpoints" and follow the trail! Problem Is Tricky because of Traversal: 1. Finding the Start: We build an graph and look for a node with a degree of 1. If we started in the middle, we’d have to go in two directions, which makes the logic hard. 2. Avoiding Infinite Loops: Since the graph is undirected, when moving from node A to node B, node B will list A as a neighbor. We must pass a prev reference to ensure we don't bounce back and forth between the same two numbers. My Approach: I used a DFS starting from an endpoint. By keeping track of the prev node, we can traverse the entire path linearly until we hit the other endpoint. Complexity: Time: O(n) Space: O(n) #LeetCode #CodingChallenge #SoftwareEngineering #PythonProgramming #DataStructures #Algorithms #GraphTheory #DFS #ProblemSolving #TechInterviewPrep #Python #Java
Restore Array from Adjacent Pairs with DFS
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🔍 Problem Summary: Given an array and multiple queries, for each query index, we need to find the minimum circular distance to another index having the same value. If no such element exists, return -1. 💡 Key Idea: Since the array is circular, distance can be calculated in two ways: forward and backward. For each value, store all its indices. For every query, check the nearest occurrence of the same value and compute the minimum circular distance. ⚡ Example Insight: For value 1 at index 0, nearest same value is at index 2 → distance = 2 If an element appears only once → result = -1 🧠 What I Learned: Efficient use of hash maps for indexing Handling circular traversal logic Optimizing search using preprocessing #DataStructures #ProblemSolving #CodingPractice #Python #Java #SoftwareEngineering #LearningJourney
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🚀 Day 39/60 — LeetCode Discipline Problem Solved: Sqrt(x) Difficulty: Easy Today’s challenge was to compute the square root of a number without using built-in functions. Instead of brute force, I used Binary Search — a classic, elegant approach that narrows down the answer efficiently. 💡 Key Learnings: • Binary Search application beyond arrays • Handling edge cases (x < 2) • Avoiding overflow using conditions carefully • Finding floor value of square root • Optimized thinking over brute force ⚡ Performance: Runtime: 4 ms Like walking in a foggy path, I didn’t see the answer directly… But step by step— cutting the search space in half— the truth revealed itself. That’s the beauty of algorithms. #LeetCode #60DaysOfCode #DSA #BinarySearch #ProblemSolving #CodingJourney #Python #Consistency #TechGrowth
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Shipped a small end-to-end backend pipeline this month: • Ingested noisy vendor-style location events (dedupe + retries + basic health stats) • Normalized positions with a simple quality policy (accuracy / jitter / floor sanity) • Derived events: zone enter/exit + asset proximity • Sketched the relational model + example SQL for “where was X between T0–T1?” Nice reminder that most of the work isn’t algorithms—it’s clear boundaries, idempotency, and making the data trustworthy before you build features on top. #backend #python #systemdesign #dataengineering
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Vector databases are great, but they aren't always the right tool for complex document intelligence. 🧠📉 If you are tired of context fragmentation and untraceable LLM hallucinations, it is time to look at Vectorless RAG with Page Index. By swapping out mathematical embeddings for a reasoning-based, hierarchical document tree, you can achieve upwards of 98% accuracy on complex Q&A tasks with perfect citation traceability. I wrote a complete guide on how this architecture works, including a full Python code implementation. Read it here: https://lnkd.in/gRuXiSxK #ArtificialIntelligence #RAG #PythonDeveloper #MachineLearning #AIEngineering
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LeetCode Problem 1448. Count Good Nodes in Binary Tree: "Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X. Return the number of good nodes in the binary tree." Approach: make use of pre-order traversal i.e., visit root->left->right child, in each visit, keep track of the max element upto that point by maintaining a stack, top of the stk should represent the max element upto that point in the path, pop() elements from stack when both the children (left,right) are explored. Time Complexity: O(n) Space Complexity: O(n) b/c of maintaining a stack #Python #LeetCode #DSA #DataStructures #Algorithms #Stack #Recursion #DFS #BinaryTree #ProblemSolving
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Handling Time Data: Logic over Strings. Today I worked on a common challenge: comparing time values like '7:15' and '10:30' stored in a list.The Problem: Standard string comparison can be unreliable (e.g., '7:15' vs '10:30'), and using float numbers leads to mathematical inaccuracies.The Solution: I converted all time entries into a single unit — Total Minutes from the start of the day (hours * 60 + minutes).This transformation turns time-strings into simple integers, creating a robust and scalable logic for sorting and filtering. A solid foundation is everything, whether it's infrastructure or code. 🛡️🦾#Python #Coding #ProblemSolving #SoftwareEngineering #Backend #Summerson
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🚀 Solved Another Sliding Window Problem on LeetCode! Today’s problem: Maximum Number of Vowels in a Substring of Given Length (LeetCode #1456) 💡 Problem Summary: Given a string s and an integer k, find the maximum number of vowels in any substring of length k. ❌ Brute Force Approach: Generate all substrings of size k Count vowels in each substring Time Complexity: O(n × k) → not efficient ✅ Optimized Approach: Sliding Window Instead of recalculating everything: Count vowels in the first window Slide the window forward: Add next character Remove previous character Track the maximum count 👉 Core Idea: count = count + new_char - old_char 💻 Clean Code: def maxVowels(s, k): vowels = set("aeiou") count = 0 for i in range(k): if s[i] in vowels: count += 1 max_count = count for i in range(k, len(s)): if s[i] in vowels: count += 1 if s[i - k] in vowels: count -= 1 max_count = max(max_count, count) return max_count ⚡ Complexity: Time: O(n) Space: O(1) 🧠 Key Takeaway: Sliding Window is not just a technique — it’s a mindset. You reuse previous computation instead of recalculating everything. 🔥 This pattern applies to: Strings & Arrays Substring / Subarray problems Real-world streaming data #DSA #LeetCode #Coding #SlidingWindow #InterviewPreparation #Python #ProblemSolving
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Day 14/100 – Data Structures & Algorithms Today, I worked on the problem “First Unique Character in a String.” Overview The task is to identify the first non-repeating character in a string and return its index. If no such character exists, the result is -1. Approach I used a two-pass strategy: • First pass to store character frequencies using a hashmap • Second pass to identify the first character with a frequency of one Complexity • Time Complexity: O(n) • Space Complexity: O(1) Key Takeaway This problem reinforces how effective hashmaps are for frequency-based problems and how a simple two-pass approach can lead to optimal solutions. Staying consistent and building problem-solving intuition step by step. #Day14 #100DaysOfCode #DSA #Python #LeetCode #ProblemSolving #SoftwareEngineering
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Topic 5/100 🚀 🧠 Topic 5 — Iterators Ever wondered how a for loop actually works behind the scenes? 🤔 This is the concept powering it. 👉 What is it? Iterators are objects that allow you to traverse through data step-by-step using __iter__() and __next__() methods. 👉 Use Case: Used in real-world applications for: Custom data pipelines Streaming data Building your own iterable objects 👉 Why it’s Helpful: Gives full control over iteration Enables custom looping logic Foundation for generators 💻 Example: class Counter: def __init__(self, max): self.max = max self.current = 0 def __iter__(self): return self def __next__(self): if self.current < self.max: self.current += 1 return self.current raise StopIteration for num in Counter(3): print(num) 🧠 What’s happening here? We created a custom object that behaves like a loop by controlling how values are returned one by one. ⚡ Pro Tip: If you understand iterators, you’ll unlock how Python handles loops internally. 💬 Follow this series for more Topics #Python #BackendDevelopment #100TopicOfCode #SoftwareEngineering #LearnInPublic
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Topic 4/100 🚀 🧠 Topic 4 — Generators Processing large data and running out of memory? 😵 This concept solves that problem. 👉 What is it? Generators are functions that return values one at a time using yield instead of returning everything at once. 👉 Use Case: Used in real-world applications for: Reading large files Data streaming Handling APIs with pagination 👉 Why it’s Helpful: Saves memory Improves performance Enables lazy evaluation (data generated only when needed) 💻 Example: def count_up_to(n): for i in range(n): yield i for num in count_up_to(5): print(num) 🧠 What’s happening here? Instead of storing all values in memory, the function generates them one by one when needed. ⚡ Pro Tip: If you're working with large datasets, always think “generator” before using lists. 💬 Follow this series for more Topics #Python #BackendDevelopment #100TopicOfCode #SoftwareEngineering #LearnInPublic
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