Prefix Sum LeetCode Solution in Python

Day 49 of my #100DaysOfCode challenge 🚀 Today I worked on a Python program to find the Equilibrium Index of an array. An equilibrium index is an index where the sum of elements on the left is equal to the sum of elements on the right. What the program does: • Takes an array as input • Finds an index where left sum = right sum • Returns the index if found • Returns -1 if no such index exists How the logic works: • Calculate the total sum of the array • Initialize left_sum = 0 • Traverse the array using enumerate() • For each element: – Right sum = total_sum - left_sum - current element – If left_sum == right_sum, return the index • Add the current element to left_sum • If no equilibrium index is found, return -1 Example: Input: [-7, 1, 5, 2, -4, 3, 0] Output: 3(Left sum = Right sum = 1) Another example: Input: [1, 2, 3] Output: -1 (No equilibrium index) Another example: Input: [1, 0, -1] Output: 1 Why this approach is efficient: – Uses prefix sum concept – Avoids nested loops – Time Complexity: O(n) Key learnings from Day 49: – Understanding prefix sums – Optimizing from brute force to O(n) – Working with running totals – Strengthening array problem-solving #100DaysOfCode #Day49 #Python #PythonProgramming #Arrays #PrefixSum #Algorithms #DataStructures #ProblemSolving #CodingPractice #InterviewPrep #LearnByDoing #ProgrammingJourney #DeveloperGrowth #BTech #CSE #AIandML #VITBhopal #TechJourney

💻 Solved a great problem today: Count Increasing Subarrays 🚀 Given an array, the task was to count all strictly increasing subarrays of size ≥ 2. 🔍 Approach I used: Broke the array into continuous increasing segments For each segment of length k, calculated subarrays using the formula: 👉 k(k-1)/2 Summed all results to get the final answer ⚡ Time Complexity: O(n) 📦 Space Complexity: O(n) 💡 This problem helped me understand how breaking a problem into patterns can simplify complex logic. Here’s my Python solution 👇 class Solution: def countIncreasing(self, arr): list2 = [] list2.append(arr[0]) list3 = [] for i in range(1, len(arr)): if arr[i-1] < arr[i] and arr[i] > 2: list2.append(arr[i]) else: list3.append(list2) list2 = [] list2.append(arr[i]) list3.append(list2) ans = 0 for i in list3: if len(i) > 1: ans += (len(i) * (len(i) - 1)) // 2 return ans 🔥 Always learning, always improving. #python #dsa #coding #problemSolving #programming #developers #learning

🚀 DSA Day — Merge Two Sorted Arrays (LeetCode 88) Today I worked on a classic problem that looks simple but teaches an important optimization technique 👇 🔹 Problem Statement You are given two sorted arrays and need to merge them into one sorted array in-place. Example: nums1 = [1, 7, 8, 0, 0, 0], m = 3 nums2 = [2, 5, 6], n = 3 ✅ Output: [1, 2, 5, 6, 7, 8] 🔸 Approach 1: Brute Force ✔ Copy nums2 into nums1 ✔ Sort the entire array ⏱ Time Complexity: O((m+n) log(m+n)) 👉 Simple but not efficient 🔸 Approach 2: Optimized (Two Pointers) 💡 Key Idea: Start filling from the end ✔ Compare last elements of both arrays ✔ Place the larger one at the end ✔ Move pointers accordingly ⏱ Time Complexity: O(m+n) 📦 Space Complexity: O(1) 🔥 Why this works? Because nums1 already has extra space at the end — we use it smartly without shifting elements. 💻 Code Snippet (Optimized) https://lnkd.in/g9Z7yf3d def merge(nums1, m, nums2, n): i = m - 1 j = n - 1 k = m + n - 1 while i >= 0 and j >= 0: if nums1[i] > nums2[j]: nums1[k] = nums1[i] i -= 1 else: nums1[k] = nums2[j] j -= 1 k -= 1 while j >= 0: nums1[k] = nums2[j] j -= 1 k -= 1 🎯 Key Takeaway 👉 Always think from the end when dealing with in-place array problems. #DSA #Python #CodingInterview #LeetCode #ProblemSolving #SoftwareEngineering #LearnInPublic

Day 43 of my #100DaysOfCode challenge 🚀 Today I worked on a Python program to find the Longest Common Prefix (LCP) among a list of strings. This is a popular problem that helps strengthen string manipulation and comparison logic. What the program does: • Takes a list of strings as input • Finds the common starting characters shared by all strings • Returns the longest common prefix • Handles edge cases like empty lists and single strings How the logic works: • If the list is empty, return an empty string • Sort the list of strings • Compare only the first and last strings (they will have the maximum difference) • Iterate character by character • Add matching characters to the prefix • Stop when characters don’t match • Return the final prefix Example: Input: ["flower", "flow", "flight"] Output: "fl" Another example: Input: ["dog", "racecar", "car"] Output: "" (No common prefix) Another example: Input: ["apple", "apricot", "april"] Output: "ap" Why this approach works well: – Sorting reduces comparisons to just two strings – Efficient and easy to implement – Time Complexity: O(n log n + m) Key learnings from Day 43: – String comparison techniques – Using sorting to simplify problems – Handling edge cases effectively – Writing optimized and clean logic #100DaysOfCode #Day43 #Python #PythonProgramming #Strings #Algorithms #ProblemSolving #CodingPractice #DataStructures #InterviewPrep #LearnByDoing #DeveloperGrowth #ProgrammingJourney #ComputerScience #BTech #CSE #AIandML #VITBhopal #TechJourney

My Day 11 of 90 Days Growth Challenge AMDOR ANALYTICS Today, we’ll focus on for Loop – What is a for loop? A for loop in Python is used to iterate over a sequence (such as a list, tuple, string, or dictionary) or iterable objects. It allows you to execute a block of code repeatedly for each item in that sequence.   What is the Basic Syntax The standard structure of a for loop is as follows:   Python for item in sequence: # Code to execute for each item print(item)   Explaining each of the components in the for loop: ·        for: The keyword that starts the loop. ·        Item: A temporary variable that takes the value of the current element in the sequence during each iteration. ·        In: Links the variable to the sequence. ·        Sequence: The object being iterated over (e.g., a list or range). ·        Indentation: Python uses indentation to define the block of code inside the loop. Tomorrow, we delve into while loop #Techjourney #90daysgrowthchallenge #consistency #growth #aiengineering #Amdoranalytics

Day 64 of LeetCode Grind ⚡🔥 Two classic Medium problems today — both are interview must-knows and each teaches a pattern you'll reuse constantly. 1️⃣ Top K Frequent Elements (347) <<< return [x for x, _ in Counter(nums).most_common(k)] >>> * Counter.most_common(k) uses a heap internally to find top-k in O(n log k) — better than sorting which is O(n log n). Want true O(n)? Use Bucket Sort: > Create n+1 buckets indexed by frequency. > Elements with frequency f go into bucket[f]. > Scan buckets from highest frequency down, collect until you have k elements. > Since frequency can't exceed n, bucket index is bounded → pure O(n)! 2️⃣ Product of Array Except Self (238) > No division allowed. O(n). O(1) extra space. Classic prefix + suffix product pattern: * nums = [1, 2, 3, 4] * prefix = [1, 1, 2, 6] ← everything LEFT of i * suffix = [24, 12, 4, 1] ← everything RIGHT of i * result = [24, 12, 8, 6] ← multiply both * Two passes, one output array. The trick is reusing the result array itself — store prefix in the first pass, multiply suffix in-place on the second pass. ✨ Reflection: Both problems follow the same meta-pattern: precompute something so each query is O(1). Whether it's frequency counts or prefix products, the "precompute → combine" mindset is fundamental to efficient algorithm design. #LeetCode #Day64 #Python #PrefixProduct #BucketSort #HashMap #Arrays #Medium #100DaysOfCode #DSA

✅ Day 87 of 100 Days LeetCode Challenge Problem: 🔹 #2840 – Check if Strings Can be Made Equal With Operations II 🔗 https://lnkd.in/gY73RBb5 Learning Journey: 🔹 Today’s problem extended the previous one, allowing swaps between indices where the difference is even. 🔹 I observed that this again partitions the string into two independent groups: • Even indices (0, 2, 4, …) • Odd indices (1, 3, 5, …) 🔹 I extracted characters from even and odd positions separately for both strings. 🔹 Then, I sorted these groups and compared them between s1 and s2. 🔹 If both even-index groups and odd-index groups match, the strings can be made equal. Concepts Used: 🔹 String Manipulation 🔹 Index Grouping (Parity-based) 🔹 Sorting 🔹 Greedy Observation Key Insight: 🔹 Since swaps are allowed only between indices with even distance, characters can only move within their parity group. 🔹 Therefore, the problem reduces to checking if both parity groups have identical character distributions. Complexity: 🔹 Time: O(n log n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

Most FastAPI codebases look clean at first glance. Until you try to change something. I’ve noticed a pattern — a lot of complexity doesn’t come from the problem itself, but from where the logic lives. When routes start handling more than just request/response, things get harder to reason about. Lately, I’ve been keeping one constraint: Routes should stay thin. They handle the HTTP layer. All business logic moves to services. It’s a small shift, but it changes a lot: 1) Clearer separation of concerns 2) Easier testing 3) Fewer side effects when making changes Also started appreciating dependency injection more. Not as a framework feature, but as a way to keep things decoupled and predictable. Nothing groundbreaking here. But in a time where a lot of code is being generated faster than it’s being designed, maintainability comes down to how consistently we apply these basics — not whether we know them. Curious how others approach structuring FastAPI projects at scale. #FastAPI #BackendDevelopment #CleanCode #SoftwareEngineering #Python

🚀 Another LeetCode Problem Solved: Palindrome Number! 🔗 Check out my solution: https://lnkd.in/dwDMqXXn 💡 Problem Overview Given an integer x, determine whether it is a palindrome — meaning it reads the same forward and backward. (LeetCode) Examples: ✔ 121 → Palindrome ❌ -121 → Not a palindrome ❌ 10 → Not a palindrome 🧠 My Approach (Digit Reversal) Instead of converting the number to a string, I used a mathematical approach: Extract digits using % 10 Reverse the number step by step Compare reversed number with original ⚙️ Key Learnings ✔ Strong understanding of number manipulation ✔ Importance of handling edge cases (negative numbers, trailing zeros) (leet-solution.com) ✔ Practicing clean and efficient logic ⏱️ Complexity • Time Complexity: O(log n) • Space Complexity: O(1) 🔥 Why this problem matters Even though it’s an “easy” problem, it builds: Logical thinking Problem-solving fundamentals Confidence for bigger challenges #LeetCode #DSA #Python #CodingJourney #ProblemSolving #100DaysOfCode

LeetCode POTD 💫: Trying to be consistent with POTDs 🫠 Description: You are given two strings, str1 and str2, of lengths n and m, respectively. A string word of length n + m - 1 is defined to be generated by str1 and str2 if it satisfies the following conditions for each index 0 <= i <= n - 1: -> If str1[i] == 'T', the substring of word with size m starting at index i is equal to str2, i.e., word[i..(i + m - 1)] == str2. -> If str1[i] == 'F', the substring of word with size m starting at index i is not equal to str2, i.e., word[i..(i + m - 1)] != str2. Return the lexicographically smallest possible string that can be generated by str1 and str2. If no string can be generated, return an empty string "". Here's my solution: https://lnkd.in/g6B2-GWz #Python #DSA #Leetcode #DailyChallenge