Niaz Chowdhury, PhD’s Post

🐍 Day 92 — Linear Regression (Concept) Day 92 of #python365ai 📈 Linear regression models relationships between variables. Equation: y = mx + c 📌 Why this matters: It’s one of the simplest and most important ML models. 📘 Practice task: Think of predicting salary based on experience. #python365ai #LinearRegression #MachineLearning #Python

🐍 Day 93 — Linear Regression (Implementation) Day 93 of #python365ai 🧑💻 Example: from sklearn.linear_model import LinearRegression model = LinearRegression() model.fit(X, y) 📌 Why this matters: This is your first real ML model. 📘 Practice task: Fit a simple regression model. #python365ai #MLModel #Python #DataScience

🐍 Day 95 — Model Evaluation (Mean Squared Error) Day 95 of #python365ai 📏 Evaluate models using metrics like MSE. Example: from sklearn.metrics import mean_squared_error 📌 Why this matters: We need to measure how good a model is. 📘 Practice task: Compute error for predictions. #python365ai #ModelEvaluation #ML #Python

Day 37 / #120DaysOfCode – LeetCode Challenge ✅ Problem Solved: • Search a 2D Matrix 💻 Language: Python 📚 Key Learnings: • Applied Binary Search on a 2D matrix • Learned how to treat matrix as a flattened sorted array • Practiced converting 1D index → 2D index (row, col) • Improved understanding of search space reduction • Strengthened logarithmic time complexity (O(log n)) thinking Better logic → Faster execution 🚀 🔗 LeetCode Profile: https://lnkd.in/gbeMKcv5 #LeetCode #Python #DSA #BinarySearch #Algorithms #CodingJourney #Consistency #120DaysOfCode

Search Insert Position: Binary Search with Left Pointer as Insertion Index Standard binary search returns -1 when not found. This variant returns insertion index instead. Key insight: when loop terminates without finding target, left pointer naturally points to where target should be inserted to maintain sorted order. Left Pointer Property: Binary search terminates when l > r. At this point, left has crossed past where target would fit. This makes left the correct insertion index without extra logic. Time: O(log n) | Space: O(1) #BinarySearch #InsertPosition #PointerProperties #SortedArrays #Python #AlgorithmDesign #SoftwareEngineering

Day 248 of #365DaysOfCode Solved a problem involving mirror pairs and index distance minimization using a hashmap-based approach. For each element, computed its digit-reversed counterpart and tracked indices to identify valid mirror pairs. The minimum distance was updated by comparing current positions with previously stored indices. The solution runs in O(n · d) time, where d is the number of digits, and efficiently leverages hashing for constant-time lookups. Continuing to build intuition around number manipulation and indexing strategies. #365DaysOfCode #Day248 #DSA #LeetCode #Python #Algorithms #HashMap #ProblemSolving #Consistency

Anagram Validation: Frequency Counting Beats Sorting Sorting both strings enables O(n log n) comparison. HashMaps reduce this to O(n) by counting character frequencies — anagrams have identical distributions. Early length check eliminates mismatches instantly. Frequency Pattern: Character counting appears in: group anagrams, substring problems, permutation validation. .get(key, 0) avoids KeyError exceptions cleanly. Time: O(n) | Space: O(1) — max 26 chars #HashMap #FrequencyCounting #Anagrams #StringAlgorithms #Python #AlgorithmOptimization #SoftwareEngineering

🐍 Day 89 — Features and Labels Day 89 of #python365ai 📌 Features (X) → input variables Labels (y) → output Example: X = [size, rooms] y = price 📌 Why this matters: Clear distinction is essential for building ML models. 📘 Practice task: Identify features and labels in a dataset. #python365ai #Features #MachineLearning #Python

💡 Solved Group Anagrams today! 🧠 Key idea: Instead of comparing strings, I used a frequency array (size 26) to create a unique key for each word. ⚠️ Catch in constraints: All characters are lowercase English letters, which allows us to use a fixed-size array of 26 for efficient hashing. 🚀 Result: Achieved O(n · k) time complexity. Great example of converting a comparison problem into a hashing problem! #LeetCode #DSA #Algorithms #Python #CodingInterview

Group Anagrams: Frequency Signature Beats Sorting for Keying Sorting each string as key costs O(k log k) per string. Character frequency array is O(k) and creates same signature for anagrams. Fixed 26-element array converted to tuple serves as hashable HashMap key — cleaner, faster grouping. Frequency as Key: Character counts uniquely identify anagram groups. Tuple conversion enables using array as dict key (lists aren't hashable). This pattern applies to document clustering, duplicate detection. Time: O(n × k) vs O(n × k log k) with sorting | Space: O(n × k) #FrequencySignature #HashMap #Anagrams #KeyOptimization #Python #AlgorithmDesign #SoftwareEngineering