Solved LeetCode 125 Valid Palindrome in Python

Day 50/100 – #100DaysOfCode 🚀 Solved LeetCode #28 – Find the Index of the First Occurrence in a String (Python). Today I practiced string matching using a brute-force approach to find the first occurrence of a substring. Approach: 1) Traverse the main string (haystack). 2) For each index, try to match the substring (needle). 3) Compare characters one by one. 4) If all characters match, return the starting index. 5) If mismatch occurs, break and move to the next index. 6) If no match is found, return -1. Time Complexity: O(n × m) Space Complexity: O(1) Understanding basic string matching techniques step by step 💪 #LeetCode #Python #DSA #Strings #ProblemSolving #100DaysOfCode

Day 33/100 – #100DaysOfCode 🚀 Solved LeetCode #1480 – Running Sum of 1d Array (Python). Today I practiced prefix sum logic to compute the running sum of an array. Approach: 1) Initialize an empty list to store the running sum. 2) Maintain a variable sum = 0. 3) Traverse the array and keep adding each element to sum. 4) Append the updated sum to the result list. 5) Return the final running sum array. Time Complexity: O(n) Space Complexity: O(n) Understanding prefix sums helps solve many array problems efficiently 💪 #LeetCode #Python #DSA #Arrays #PrefixSum #ProblemSolving #100DaysOfCode

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

Day 46/100 – #100DaysOfCode 🚀 Solved LeetCode #2733 – Neither Minimum nor Maximum (Python). Today I practiced sorting and simple array logic to find any element that is neither the minimum nor the maximum in the array. Approach: 1) Sort the array. 2) Check the length of the array. 3) If the size is less than or equal to 2, return -1 (no valid element exists). 4) Otherwise, return the second element as it will be neither min nor max. Time Complexity: O(n log n) Space Complexity: O(1) Understanding how sorting simplifies boundary-based problems 💪 #LeetCode #Python #DSA #Arrays #Sorting #ProblemSolving #100DaysOfCode

Day 55/100 – #100DaysOfCode 🚀 Solved LeetCode #205 – Isomorphic Strings (Python). Today I practiced hashmap (dictionary) usage to check whether two strings follow the same pattern. Approach: 1) Create two hashmaps to store character mappings in both directions. 2) Traverse both strings together using zip(). 3) Check if the current mapping is consistent in both maps. 4) If any mismatch is found, return False. 5) Otherwise, update the mappings and continue. 6) If all mappings are valid, return True. Time Complexity: O(n) Space Complexity: O(n) Understanding how bidirectional mapping ensures consistency 💪 #LeetCode #Python #DSA #HashMap #Strings #ProblemSolving #100DaysOfCode

🚀 Day 72 of #100DaysOfCode Today’s problem: Words Within Two Edits of Dictionary. At first, it looked like a simple string comparison… but checking all possibilities made it tricky. Had to focus on counting differences instead of overcomplicating transformations. 💡 What helped me: Compare character by character → count mismatches Stop early if differences exceed the limit. It’s more about controlling comparisons than trying every change. Slowly getting better at string problems. #DSA #LeetCode #CodingJourney #Python #Strings

Day 39/100 – #100DaysOfCode 🚀 Solved LeetCode #2215 – Find the Difference of Two Arrays (Python). Today I practiced set operations to efficiently find distinct elements between two arrays. Approach: 1) Convert both arrays into sets to remove duplicates. 2) Find elements present in nums1 but not in nums2 using set difference. 3) Find elements present in nums2 but not in nums1. 4) Convert both results back to lists. 5) Return the final list of differences. Time Complexity: O(n + m) Space Complexity: O(n + m) Understanding how set operations simplify comparison problems 💪 #LeetCode #Python #DSA #Sets #Arrays #ProblemSolving #100DaysOfCode

Two sum: Using dictionary (Optimized Solution) Instead of checking every pair: For each number, calculate what number is required. Formula to calculate required : required = target - current_number  Time complexity → O(n) #Python #LogicBuilding #ProblemSolvingSkills #Beginners

Python Clarity Series – Episode 24 Topic: Late Binding in Loops (Functions) ⚠️ Advanced pitfall: Late binding in loops funcs = [] for i in range(3): funcs.append(lambda: i) for f in funcs: print(f()) Output: 2 2 2 ❗ 👉 Expected: 0 1 2 👉 Got: same value 💡 Reason: Lambda captures variable, not value. 💡 Fix: funcs.append(lambda i=i: i) 💡 Rule: Default arguments capture current value. This is a classic interview trap. #PythonAdvanced #CodingPitfalls #DeveloperLevel #python #clarity

Here's a Python dictionary merge challenge from @dontmisstmr — can you get it right without running the code? dict1 = {"name": "Alice", "age": 25} dict2 = {"city": "Noida", "age": 26} merged = dict1 | dict2 print(merged) The | operator was introduced in Python 3.9. When both dicts share a key, which value wins? Drop your answer in the comments! #Python #SoftwareDevelopment #CodingChallenge #ProgrammingTips #TechCommunity