#Pythoncoding 𝗤𝘂𝗲𝘀𝘁𝗶𝗼𝗻: 𝗪𝗿𝗶𝘁𝗲 𝗮 𝗣𝘆𝘁𝗵𝗼𝗻 𝗳𝘂𝗻𝗰𝘁𝗶𝗼𝗻 𝘁𝗼 𝗺𝗮𝗻𝘂𝗮𝗹𝗹𝘆 𝗿𝗲𝘃𝗲𝗿𝘀𝗲 𝗮 𝗴𝗶𝘃𝗲𝗻 𝗹𝗶𝘀𝘁 𝘄𝗶𝘁𝗵𝗼𝘂𝘁 𝘂𝘀𝗶𝗻𝗴 𝘁𝗵𝗲 𝗯𝘂𝗶𝗹𝘁-𝗶𝗻 𝗿𝗲𝘃𝗲𝗿𝘀𝗲() 𝗺𝗲𝘁𝗵𝗼𝗱 𝗼𝗿 𝘀𝗹𝗶𝗰𝗶𝗻𝗴 ([::-𝟭]). 𝗖𝗼𝗻𝘀𝘁𝗿𝗮𝗶𝗻𝘁𝘀: -Do not use reverse() or [::-1]. -The function should work for both numbers and strings. -The input list may contain duplicates and should maintain their order in reverse. -The function should return a new reversed list (not modify the original list). #Python #PythonDeveloper #Coding #100DaysOfCode #LearnPython #DataScience #MachineLearning
Reverse List Function in Python
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CPython Unveils Chained Assignment Mechanics in Python Bytecode Execution 📌 Chained assignments in Python don’t create separate objects-they share the same underlying instance, a bytecode-level quirk that can silently break your code. A deep dive into CPython’s execution reveals how a = b = [] uses COPY and STORE_FAST to assign the same list to multiple variables, exposing a common pitfall for developers working with mutable data. 🔗 Read more: https://lnkd.in/dUfiAJ44 #Cpython #Pythonbytecode #Chainedassignment #Interpreterexecution #Variablebinding
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LeetCode #102 – Binary Tree Level Order Traversal | Python Implementation I implemented an iterative BFS approach using a queue to collect nodes level by level. Core Insight: The len(q) snapshot before the inner loop prevents mixing levels. New children added during iteration don't affect the current level's processing count, ensuring clean level separation. Time: O(n) | Space: O(w) where w = maximum tree width #LeetCode #DataStructures #Python #BinaryTree #BFS #LevelOrderTraversal #CodingInterview #SoftwareEngineering
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I recently implemented the K-Medoids clustering algorithm from scratch in Python and visualized the clusters using graphs to understand how the algorithm groups data points. #MachineLearning #PythonProjects https://lnkd.in/gczybNGr
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Day 12/100 – #100DaysOfCode 🚀 Solved LeetCode #217 – Contains Duplicate (Python). Today I worked on an array problem to determine whether any value appears at least twice in the array. Approach: 1) Sort the array first. 2) Traverse through the array from index 0 to n-2. 3) Compare each element with the next element. 4) If two adjacent elements are equal, a duplicate exists. 5) If no duplicates are found after traversal, return False. Time Complexity: O(n log n) due to sorting Space Complexity: O(1) Understanding how sorting can simplify duplicate detection in arrays 💪 #LeetCode #Python #DSA #Arrays #ProblemSolving #100DaysOfCode
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Day 10/100 – #100DaysOfCode 🚀 Solved LeetCode #66 – Plus One (Python). Today I worked on an array manipulation problem that simulates adding one to a large integer represented as a list of digits. Approach: 1) Traverse the array from the last digit to the first. 2) If the current digit + 1 is not equal to 10, simply increment it and return the array. 3) If the digit becomes 10, set it to 0 and carry over to the next digit. 4) If the carry reaches the first digit, add 1 at the beginning of the array. Time Complexity: O(n) Space Complexity: O(1) Understanding how carry propagation works in arrays 💪 #LeetCode #Python #DSA #Arrays #ProblemSolving #100DaysOfCode
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ove all zeros to the end of an array while preserving the relative order of non-zero elements — implemented in-place using the two-pointer technique. ✔ Single pass ✔ O(n) time complexity ✔ O(1) space complexity ✔ Stable order maintained A clean and efficient array manipulation problem focused on logic optimization and space efficiency. #geekstreak60 #npci #Python #DataStructures #ProblemSolving
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Solved the Minimum Window Substring problem using the Sliding Window technique. Goal: Find the smallest substring in s that contains all characters of p (including duplicates). Key Idea • Maintain character frequency • Expand window to include required characters • Shrink window to get the minimum valid substring Time Complexity: O(n) #geekstreak60 #npci #algorithms #datastructures #python #slidingwindow #codingpractice #problemSolving #100DaysOfCode
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Day 7/100 – #100DaysOfCode 🚀 Solved LeetCode #27 – Remove Element (Python). Today I practiced the Two Pointer technique to remove all occurrences of a given value in-place and return the new length of the array. Approach: 1) Initialize a pointer k to track the position of valid elements. 2) Traverse the array using index i. 3) If nums[i] is not equal to the given value, assign it to nums[k]. 4) Increment k to expand the valid portion of the array. 5) Return k as the new length. Time Complexity: O(n) Space Complexity: O(1) – In-place solution Learning how in-place array manipulation works step by step 💪 #LeetCode #Python #DSA #Arrays #TwoPointers #ProblemSolving #100DaysOfCode
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feb-22-26 LeetCode 762 – Prime Number of Set Bits in Binary Representation Solved today’s problem! ✅ The task is to count numbers in a given range [left, right] whose number of set bits (1s in binary form) is a prime number. ⚡ Complexity: Time: O(n log n) Space: O(1) A simple yet powerful problem combining binary operations and number theory 🔥 #LeetCode #DailyCoding #BitManipulation #Python #ProblemSolving #CodingJourney
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Logistic Regression comes alive when applied to actual datasets — and we’ve demonstrated how it’s used to solve classification problems with practical Python implementations. 💻 Start Exploring [Part 2: Practical] 𝐋𝐨𝐠𝐢𝐬𝐭𝐢𝐜 𝐑𝐞𝐠𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐀𝐥𝐠𝐨𝐫𝐢𝐭𝐡𝐦 – 𝐀𝐩𝐩𝐥𝐢𝐜𝐚𝐭𝐢𝐨𝐧𝐬 𝐮𝐬𝐢𝐧𝐠 𝐏𝐲𝐭𝐡𝐨𝐧. 🔗 https://lnkd.in/gk6gaB9f
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::-1 gives reverse string using extented slicing operation