Binary Search Solution for Koko Eating Bananas

Want to check whether a #Python #Pandas series contains another string? Use .str.contains: df['x'].str.contains('a') This returns a boolean series, whose index matches that of df. Keep only those rows containing 'a': df.loc[ pd.col('x').str.contains('a') ] # Pandas 3 syntax

You can't use a list as a #Python dict key: mylist = ['a', 'b'] d = {mylist: 10} # TypeError! Why not? Dicts run "hash" on a key to choose a pair's storage location. To avoid pairs getting lost, mutable builtins (list, set, dict) cannot be used as dict keys.

Day 48/100 – #100DaysOfCode 🚀 Solved LeetCode #13 – Roman to Integer (Python). Today I practiced string processing and mapping logic to convert Roman numerals into integers. Approach: 1) Create a dictionary to map Roman symbols to their integer values. 2) Traverse the string from left to right. 3) If the current value is less than the next value, subtract it. 4) Otherwise, add it to the total. 5) Return the final result. Time Complexity: O(n) Space Complexity: O(1) Understanding pattern-based problems in strings 💪 #LeetCode #Python #DSA #Strings #ProblemSolving #100DaysOfCode

🐍 Day 103 — Decision Trees (Implementation) Day 103 of #python365ai 🧑💻 Example: from sklearn.tree import DecisionTreeClassifier model = DecisionTreeClassifier() model.fit(X, y) 📌 Why this matters: Decision Trees handle both classification and regression tasks. 📘 Practice task: Train a simple decision tree model. #python365ai #DecisionTree #MachineLearning #Python

Most implementations of the State pattern in Python look very “clean”. Lots of small classes. A base interface. One class per state. But if you’ve ever worked with one in a real project, you know the downside: transitions are scattered, behaviour is hard to see in one place, and adding new states often means touching multiple files. In today’s video, I rebuild the State pattern in a very different way. Instead of relying on inheritance, I make the state machine explicit as data and use decorators to define transitions. The result is a small, reusable engine where the entire flow becomes visible at a glance. If you’re interested in writing Python that’s easier to reason about and extend, this is a pattern worth understanding. 👉 Watch here: https://lnkd.in/e9Y3xGNF. #python #softwaredesign #designpatterns #statemachine #cleancode

Most implementations of the State pattern in Python look very “clean”. Lots of small classes. A base interface. One class per state. But if you’ve ever worked with one in a real project, you know the downside: transitions are scattered, behaviour is hard to see in one place, and adding new states often means touching multiple files. In today’s video, I rebuild the State pattern in a very different way. Instead of relying on inheritance, I make the state machine explicit as data and use decorators to define transitions. The result is a small, reusable engine where the entire flow becomes visible at a glance. If you’re interested in writing Python that’s easier to reason about and extend, this is a pattern worth understanding. 👉 Watch here: https://lnkd.in/eg22yEHR. #python #softwaredesign #designpatterns #statemachine #cleancode

Day 60/100 — #100DaysOfCodingChallenge 60 days in… consistency is slowly turning into a habit now. 🔹 Python (DSA) Solved Search a 2D Matrix — used binary search by treating the matrix like a flattened sorted array. It was a nice reminder of how powerful binary search can be when applied smartly. 🔹 SQL Did some light practice to keep concepts fresh and maintain the streak. #Python #SQL #DSA #LeetCode #Day60 #100DaysOfCode #LearningInPublic #Consistency

Day 42/100 – #100DaysOfCode 🚀 Solved LeetCode #2574 – Left and Right Sum Differences (Python). Today I practiced prefix sum logic to calculate the absolute difference between left and right sums for each index. Approach: 1) Calculate the total sum of the array. 2) Initialize leftSum = 0. 3) Traverse the array. 4) For each index, compute rightSum = total - leftSum - nums[i]. 5) Calculate the absolute difference and append it to the result. 6) Update leftSum by adding nums[i]. Time Complexity: O(n) Space Complexity: O(n) Understanding prefix sum helps solve problems efficiently 💪 #LeetCode #Python #DSA #Arrays #PrefixSum #ProblemSolving #100DaysOfCode