LeetCode Problem 1545: Binary String S Construction

LeetCode Problem 1582: "Special Positions in a Binary Matrix": Given an m x n binary matrix mat, return the number of special positions in mat. A position (i, j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed). Approach: Initialize two arrays, one stores the number of 1s in each row of the matrix and other stores the number of 1s in each column, by traversing the matrix for filling each of the arrays. Now re-traverse the matrix and if mat[i][j]==1, check for the number of 1s in corresponding row and column, if number of 1s in that row and column is equal to one, we get our required position, calculate the total number of such positions. #Python #LeetCode #Matrices #ProblemSolving #CompetitiveProgramming #DataStructures #Algorithms #Arrays #Coding

LeetCode 23: Merge k sorted list: You are given an array of k linked-lists lists, each linked-list is sorted in ascending order. Merge all the linked-lists into one sorted linked-list and return it. Approach: Simple and straightforward, iterate through each of the linked-list in the given list, push each node's value into a priority queue using heapq module, create a new linked-list using the elements of priority queue. Time complexity: O(n logn) where n is the total number of values. Space complexity: O(n) #Python #LeetCode #DSA #CompetitiveProgramming #DataStructures #Algorithms #ProblemSolving

Day 62 of 365 Days of code Qn1) insert into a binary search🌳 Approach: recursion def insert(root,val): if root==None: set root=Treenode(val) return root elif root.val<val: root.right=insert(root.right,val) else: root.left=insert(root.left,val) return root Sleep well during exams bud, a piece of advice :"). #365daysOfCode #NeetCode  #leetcode  #DSA  #python #LeetCode #ProblemSolving  #Algorithms #365dayschallenge

Reducing 3Sum from O(n³) to O(n²): The Power of Sorting + Two Pointers Most developers first approach 3Sum with nested loops, hitting O(n³) complexity. The breakthrough: sort the array first, then reduce it to n iterations of the Two Sum problem. For each element as the fixed anchor, use two pointers on the remaining sorted portion to find pairs that complete the triplet. Sorting unlocks directional movement (move left if sum too high, right if too low), collapsing a dimension of complexity. The Critical Detail: Duplicate handling isn't just about correctness — it's about recognizing that sorted order lets you skip duplicates in O(1) rather than using a HashSet for deduplication. The while nums[l] == nums[l-1] skip after finding a valid triplet prevents processing identical combinations. Time: O(n²) | Space: O(1) excluding output #AlgorithmOptimization #TwoPointers #Sorting #ComplexityReduction #Python #CodingInterview #SoftwareEngineering

✅ Day 64 of 100 Days LeetCode Challenge Problem: 🔹 #1784 – Check if Binary String Has at Most One Segment of Ones 🔗 https://lnkd.in/gpJYed9J Learning Journey: 🔹 Today’s problem focused on determining whether a binary string contains only one contiguous segment of '1's. 🔹 Since the string has no leading zeros, the first character is always '1'. 🔹 If a '1' appears again after a '0', it indicates the start of a second segment. 🔹 A single pass through the string is enough to detect this pattern. Concepts Used: 🔹 String Traversal 🔹 Pattern Detection 🔹 Conditional Logic Key Insight: 🔹 The pattern "01" followed by another '1' reveals multiple segments of ones. 🔹 Detecting this transition allows us to solve the problem efficiently. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) 💬 What’s your favorite trick for solving binary string problems? #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

Merge Intervals (LeetCode 56) - Medium Key Learnings: Sorting: By sorting intervals based on their start times, we can process them in a single pass (O(N)). Overlap Logic: Comparing the current_start with the previous_end time is the secret sauce to identifying overlaps. Space-Time Tradeoff: Sorting takes O(N log N) time, and we use O(N) space to store the results. #CodingJourney #LeetCode #Blind75 #SDEPreparation #SoftwareEngineering #Python #DataStructures

🚀 Day 11 of Consistency | #75DaysLeetCodeChallenge 🧠 Today’s Problem : Two Sum II – Input Array Is Sorted (#167) 💡 Key Learning: This problem highlights the power of the two-pointer technique on a sorted array, helping reduce time complexity efficiently compared to brute force. ⚡ Approach: Use two pointers → left (l) at start & right (r) at end → If sum == target → return indices If sum < target → move l++ If sum > target → move r-- 🧠 Why this works: Takes advantage of sorted array Reduces complexity → O(n) No extra space required → O(1) 🔥 Result : ✔️ Runtime: 0 ms (Beats 100%) 📈 Mastering patterns like two pointers is key to cracking medium & hard problems. Consistency is compounding. Keep going. 💪 #Day11 #LeetCode #DSA #CodingJourney #100DaysOfCode #Python #TwoPointers #Consistency

LeetCode Problem 1758: "Minimum changes to make alternating binary string": You are given a string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa. The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not. Return the minimum number of operations needed to make s alternating. Approach: Here are two possibilities, first do not make change in the 1st character and apply rules following it, second consider step of changing the very first character and apply rules following it. The below implementation in Python works on this principle only, simple and straightforward. #Python #LeetCode #Strings #OptimalSolution #CompetitiveProgramming #DataStructures #Algorithms #ProblemSolving

🚀 Day 16 – DSA Daily Series Today’s Problem: Two Sum II – Input Array Is Sorted (LeetCode 167) Today’s problem was simple yet really interesting — especially because of how efficiently it can be solved using the two-pointer technique. 🧠 Problem Given a sorted array, find two numbers such that they add up to a target value and return their indices (1-based). Example: Input: [2,7,11,15], target = 9 Output: [1,2] 💡 Approach Instead of using extra space like a hashmap, I used: • Two pointers — one at the start and one at the end • If sum is too large → move right pointer left • If sum is too small → move left pointer right • Stop when target is found Clean and efficient ⚡ ⏱ Complexity Time Complexity: O(n) Space Complexity: O(1) 🔎 Key Learning When the array is already sorted, always think of two pointers before anything else — it saves both time and space. Solved it today and it felt really smooth to implement! 💯 Continuing to stay consistent and improve step by step 🚀 #DSA #LeetCode #Python #TwoPointers #CodingJourney #ProblemSolving

🚀 #100DaysOfCode – Day 52 📌 Problem: Sort Colors (LeetCode 75) 💡 Problem Idea: You are given an array containing only 0s, 1s, and 2s, representing three colors. The task is to sort the array in-place so that all 0s come first, then 1s, and then 2s. ⚡ Key Insight: Instead of using a sorting algorithm, we can solve this in one pass using three pointers. We maintain three regions: Left pointer → position for next 0 Right pointer → position for next 2 Current pointer (i) → traverses the array 🎯 Approach (Dutch National Flag Algorithm): If element is 0 → swap with left, move both left and i If element is 1 → just move i If element is 2 → swap with right, decrease right 📈 Complexity: Time Complexity: O(n) Space Complexity: O(1) (in-place sorting) ✅ Efficient because the array is processed only once. 💭 Learning: This problem is a great example of how pointer techniques can optimize sorting problems without using built-in sort functions. #DSA #LeetCode #Python #CodingJourney #ProblemSolving #Algorithms #100DaysOfCode #LinkedInLearning