LeetCode Challenge: Minimize Alternating Binary String Changes

🚀 Day 4 of 100 Days LeetCode Challenge. Problem: Special Positions in a Binary Matrix Today’s problem focused on matrix traversal + counting logic—simple concept, but requires careful observation. 💡 Key Insight: A position (i, j) is special if: mat[i][j] == 1 All other elements in the same row and column are 0 🔍 Efficient Approach: Count number of 1’s in each row Count number of 1’s in each column A position is special only if: Row count = 1 Column count = 1 👉 This avoids unnecessary repeated checks and improves efficiency. 🔥 What I Learned Today: Preprocessing (row & column counts) simplifies problems Avoid brute force → think in terms of frequency/counting Clean logic > complex code 📈 Challenge Progress: Day 4/100 ✅ Staying consistent! LeetCode, Matrix Problem, Arrays, Counting Technique, DSA Practice, Coding Challenge, Problem Solving, Algorithm Thinking, Optimization #100DaysOfCode #LeetCode #DSA #CodingChallenge #Matrix #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 Day 29 of 100 Days LeetCode Challenge Problem: Check if Strings Can be Made Equal With Operations I Day 29 is a short but pattern + constraint observation problem 🔥 💡 Key Insight: We can only swap characters where: 👉 j - i = 2 For a string of length 4, possible swaps: Index 0 ↔ 2 Index 1 ↔ 3 👉 That means: Positions (0,2) form one group Positions (1,3) form another group 🔍 Core Approach: 1️⃣ Group Characters Extract characters from: Even indices → [0, 2] Odd indices → [1, 3] 2️⃣ Compare Groups Sort both groups for s1 and s2 If both corresponding groups match → ✅ true Else → ❌ false 💡 Why This Works: You can rearrange freely within each group But cannot mix between groups 🔥 What I Learned Today: Limited operations define independent groups Think in terms of index grouping Small problems still test logical clarity 📈 Challenge Progress: Day 29/100 ✅ One day to 30! LeetCode, Strings, Swapping, Greedy, Pattern Recognition, Arrays, DSA Practice, Coding Challenge, Problem Solving #100DaysOfCode #LeetCode #DSA #CodingChallenge #Strings #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 Day 6 of 100 Days LeetCode Challenge Problem: Check if Binary String Has at Most One Segment of Ones Today’s problem is all about pattern observation in strings—simple, but easy to overthink. 💡 Key Insight: The string should contain only one continuous block of '1's. 👉 That means: Once a 0 appears after a 1, There should be no more '1's later 🔍 Simplest Trick: Just check if the pattern "01" appears more than once OR even better → check if "10" appears followed by another "1" 💡 Cleaner approach: Traverse the string Count transitions from 1 → 0 If you ever see 1 again after that → ❌ Invalid 🔥 What I Learned Today: Many problems are just pattern validation Clean logic beats complex conditions Always try to reduce the problem to a simple rule 📈 Challenge Progress: Day 6/100 ✅ Consistency building strong! LeetCode, Strings, Pattern Recognition, Greedy, DSA Practice, Coding Challenge, Problem Solving, Algorithm Thinking, Programming #100DaysOfCode #LeetCode #DSA #CodingChallenge #Strings #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 Day 14 of 100 Days LeetCode Challenge Problem: The K-th Lexicographical String of All Happy Strings of Length n Today’s problem is a perfect blend of Backtracking + Lexicographical Ordering 🔥 💡 Key Insight: A happy string: Uses only 'a', 'b', 'c' No two adjacent characters are the same 👉 Instead of generating all strings blindly, we: Build valid strings using backtracking Maintain lexicographical order naturally 🔍 Core Approach: 1️⃣ Backtracking (DFS) Start building string character by character At each step: Choose from 'a', 'b', 'c' Skip if same as previous character 2️⃣ Lexicographical Order Always try characters in order: 'a' → 'b' → 'c' 3️⃣ Stop Early Once we reach the k-th string, stop recursion 👉 If total strings < k → return "" 🔥 What I Learned Today: Backtracking is powerful for constraint-based generation Ordering decisions early helps avoid extra sorting Early stopping = major optimization 📈 Challenge Progress: Day 14/100 ✅ 2 weeks strong! LeetCode, Backtracking, Recursion, Lexicographical Order, Strings, DFS, DSA Practice, Coding Challenge, Problem Solving #100DaysOfCode #LeetCode #DSA #CodingChallenge #Backtracking #Recursion #Strings #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

DAY->31 🚀 DSA Progress Update – Reverse Linked List Solved LeetCode 206: Reverse Linked List and explored multiple approaches to understand the efficiency behind each solution. 🔹 Approach 1: Using Stack Push all nodes into a stack and rebuild the list in reverse order ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) (extra space for stack) 🔹 Approach 2: Iterative (Optimal Approach – as shown in my code) Reverse pointers in-place using three variables ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) (no extra space used) 💡 The key learning here is that while both approaches take the same time, the iterative approach is more efficient because it avoids extra memory usage and works directly on the linked list. This problem helped me strengthen my understanding of pointer manipulation and optimization in linked lists. More DSA problems coming soon 🔥 #LeetCode #DSA #LinkedList #CodingJourney #ProblemSolving #100DaysOfCode

🚀 Day 19 of 100 Days LeetCode Challenge Problem: Count Submatrices With Equal Frequency of X and Y Today’s problem was a solid combination of 2D Prefix Sum + Hashing concept 🔥 💡 Key Insight: We need submatrices that: Include the top-left cell (0,0) Have equal number of 'X' and 'Y' Contain at least one 'X' 👉 Trick: Convert the grid into values: 'X' → +1 'Y' → -1 '.' → 0 🔍 Core Approach: 1️⃣ 2D Prefix Sum Transformation Build prefix sum based on above conversion 2️⃣ Check Each Submatrix (0,0 → i,j) If sum == 0 → equal number of X and Y ✅ 3️⃣ Extra Condition: Ensure at least one 'X' exists 👉 Track X count separately using another prefix matrix 🔥 What I Learned Today: Transforming problems into numbers simplifies logic Prefix sums + condition checks = powerful combo Always handle extra constraints carefully 📈 Challenge Progress: Day 19/100 ✅ Almost hitting Day 20! LeetCode, Prefix Sum, Matrix, Hashing, Problem Transformation, Algorithms, DSA Practice, Coding Challenge, Problem Solving #100DaysOfCode #LeetCode #DSA #CodingChallenge #PrefixSum #Matrix #Hashing #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 Day 7 of 100 Days LeetCode Challenge Problem: Minimum Number of Flips to Make the Binary String Alternating Today’s problem leveled up the difficulty with a mix of sliding window + string rotation + greedy logic. 💡 Key Insight: We can rotate the string (Type-1 operation), which means we should consider all possible rotations of the string. 👉 Trick: Double the string → s + s to simulate rotations Compare every window of size n with: "010101..." "101010..." 🔍 Approach: Use sliding window on the doubled string Count mismatches for both patterns Minimum flips across all windows = final answer 🔥 What I Learned Today: Rotation problems → think of string doubling Sliding window helps optimize repeated checks Combining multiple concepts = real DSA growth 📈 Challenge Progress: Day 7/100 ✅ One full week completed! LeetCode, Sliding Window, Strings, Greedy Algorithm, Pattern Matching, DSA Practice, Coding Challenge, Problem Solving, Optimization #100DaysOfCode #LeetCode #DSA #CodingChallenge #SlidingWindow #Strings #GreedyAlgorithm #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep

🚀 DSA Day 48 – LeetCode Problem 435: Non-overlapping Intervals Today’s problem was a great example of applying a Greedy Algorithm effectively 🧠 🔍 Problem Insight: We are given a set of intervals, and the goal is to remove the minimum number of intervals so that the rest do not overlap. 💡 Key Idea: Sort intervals based on their end time Always pick the interval that finishes earliest If the next interval overlaps, we remove it Otherwise, we include it and move forward ⚡ Why this works? Choosing the interval with the earliest end time leaves more room for upcoming intervals — a classic greedy strategy for optimal selection. 🧠 What I Learned: How greedy algorithms simplify complex problems Importance of sorting in interval-based questions Making optimal local choices to achieve global optimum 💻 Time Complexity: O(n log n) (due to sorting) 📦 Space Complexity: O(1) (excluding input) Another day, another concept mastered 💪 — consistency is compounding! #DSA #LeetCode #GreedyAlgorithm #CodingJourney #100DaysOfCode #ProblemSolving #TechGrowth

Day 2: Leetcode – 704. Binary Search Today’s problem looked straightforward, but it exposed how small mistakes in control flow can completely break the logic. The Key Insight: Binary search isn’t just about comparing values — it’s about maintaining the correct search space and letting the loop run until it’s fully exhausted. Returning too early can silently ruin the entire algorithm. My Mistake: I placed return -1 inside the loop, which caused the function to exit after just one iteration if the target wasn’t found immediately. The code compiled fine but the logic was wrong — a good reminder that passing compilation doesn’t mean correctness. My Approach: Initialize low = 0 and high = n-1 Calculate mid each iteration Compare nums[mid] with target Adjust search space accordingly Only return -1 after the loop ends Takeaway: Binary search is simple in theory but demands discipline in implementation. One misplaced return or bracket can turn a correct idea into a faulty solution. Day 2 Completed #LeetCode #DSA #BinarySearch #Learning #CodingJourney