Minimum Changes To Make Alternating Binary String

🚀 LeetCode Problem Solved: Maximum Number of Vowels in a Substring of Given Length Today I solved LeetCode 1456 – Maximum Number of Vowels in a Substring of Given Length using the Sliding Window Technique. 🔹 Problem Statement: Given a string s and an integer k, find the maximum number of vowels in any substring of length k. 🔹 Approach: Instead of checking every substring separately, I used the Sliding Window algorithm: • Count vowels in the first window of size k • Slide the window forward one character at a time • Add the new character and remove the old one from the count • Track the maximum vowels seen so far This approach optimizes the solution to O(n) time complexity. 📌 Example: Input: s = "abciiidef", k = 3 Output: 3 Explanation: The substring "iii" contains 3 vowels, which is the maximum. 💡 Key Concepts Practiced: ✔ Sliding Window Technique ✔ String Traversal ✔ Efficient Problem Solving Consistent DSA practice on LeetCode is helping me improve my algorithmic thinking and coding efficiency. #LeetCode #DSA #SlidingWindow #ProblemSolving #CodingPractice #Java #LearningInPublic #Programming

🚀 Day 7 of #100DaysOfDSA 🔍 Problem: Valid Anagram (LeetCode 242) Today I solved a classic string problem — checking whether two strings are anagrams. 💡 Key Idea: Instead of sorting (O(n log n)), I used a frequency count approach (O(n)). 👉 Logic: Create an array of size 26 (for a–z) Traverse both strings together Increase count for s, decrease for t If all values become 0 → strings are anagrams ✅ 🧠 Why it works: Same characters with same frequency cancel each other out. ⏱️ Complexity: Time: O(n) Space: O(1) : Always think about optimizing from brute force (O(n²)) to optimal (O(n)). #DSA #Java #Coding #LeetCode #Programming #SoftwareEngineering 🍁 Saidhanya Sree

LeetCode Problem || Count Submatrices with Top-Left Element and Sum Less Than k (3070) 🚀 🔍 Approach : I implemented a 2D Prefix Sum technique to efficiently calculate the sum of submatrices. Instead of recalculating sums again and again, I built a presum matrix where each cell stores the cumulative sum from the top-left corner. Key Idea: Use inclusion-exclusion principle: Current = Grid Value + Left + Top − Top-Left If the sum at any point exceeds k, break early to optimize performance. ⚡ Time Complexity: O(m × n) ⚡ Space Complexity: O(m × n) #LeetCode #DSA #Java #Coding #ProblemSolving #LearningJourney

Day 37 Implemented a Queue using two Stacks to maintain FIFO order by transferring elements when needed. This approach ensures amortized O(1) time complexity for push, pop, and peek operations. #LeetCode #Java #DataStructures #Stack #Queue #CodingPractice #ProblemSolving #Algorithms #DSA #Programming

🚀 Day 525 of #750DaysOfCode 🚀 💡 LeetCode 1980: Find Unique Binary String Today’s problem was an interesting one involving binary strings and a clever observation. We are given n unique binary strings of length n, and the task is to return any binary string of length n that does not exist in the given array. 🔎 Approach I Used I applied a concept similar to Diagonalization. Traverse the array from 0 → n-1 Look at the i-th character of the i-th string Flip it (0 → 1 or 1 → 0) Append the flipped character to build a new string This guarantees the newly created string differs from every string in the list at least at one position, ensuring it is unique and not present in the array. ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(n) This approach works efficiently because the constraints are small and we only need to ensure one position difference with each string. Problems like this highlight how a simple observation can eliminate brute force completely. #leetcode #coding #programming #java #datastructures #algorithms #problemSolving #developer #codingchallenge #750DaysOfCode

🚀 Day 16/100 Days of Code Challenge Solved the classic “Sort Colors” problem (LeetCode 75) today! 🎯 🔹 Problem Summary: Given an array containing 0s, 1s, and 2s (representing colors), sort them in-place without using built-in sort. 🔹 Approach Used: Maintained three pointers: low, mid, high Efficiently partitioned the array in a single pass 🔹 Key Learning: ✔️ In-place sorting can be done in O(n) time ✔️ Using pointers smartly avoids extra space 💡 Time Complexity: O(n) 💡 Space Complexity: O(1) 🔹 Code Highlight : Used swapping and pointer movement to organize elements in one traversal. Consistency is the key 🔑 — one step closer to mastering DSA! #Day16 #100DaysOfCode #DSA #Java #CodingChallenge #LeetCode #ProblemSolving #TechJourney

🚀 Day 22/60 — LeetCode Discipline Problem Solved: Reverse Integer (Revision) Difficulty: Medium Today’s practice focused on revisiting the classic integer manipulation problem — Reverse Integer. At first glance the problem appears straightforward: reverse the digits of a number. However, the real challenge lies in carefully handling edge cases such as integer overflow and maintaining correctness within the 32-bit signed integer range. Instead of using built-in conversions, the solution iteratively extracts digits and rebuilds the reversed number while ensuring that the result stays within valid bounds. 💡 Focus Areas: • Strengthened digit extraction using modulo operations • Practiced iterative number reconstruction • Handled integer overflow edge cases • Reinforced careful boundary condition checks • Focused on writing efficient and clean logic ⚡ Performance Highlight: Achieved ~99.9% runtime efficiency on submission. Revisiting foundational problems like this continues to sharpen attention to detail and improve algorithmic discipline. #LeetCode #60DaysOfCode #100DaysOfCode #DSA #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering #Programming #Developers #TechCareers #Java

Solved Today’s LeetCode Daily Challenge! Problem: Check if Strings Can Be Made Equal With Operations II At first glance, it looks like a swapping problem… but the real trick is understanding the constraint You can only swap characters if the distance between indices is even. Key Insight: Characters at even indices can only swap among even positions Characters at odd indices can only swap among odd positions So instead of simulating swaps (which is messy ), I used a smarter approach: Count frequency of characters at even indices Count frequency at odd indices Compare both strings If both match → Possible Else → Not possible Time Complexity: O(n) Space Complexity: O(1) Takeaway: Sometimes problems look complex, but a small observation can simplify everything. Consistency + pattern recognition = #LeetCode #DSA #Java #Coding #ProblemSolving #TechJourney #100DaysOfCode

🚀 Day 85 of #100DaysOfCode Today I practiced a simple yet useful array prefix minimum problem. 🔹 Problem Given an array cost, return a new array where each element represents the minimum cost encountered from the start up to that index. 🔹 Approach I used a running minimum technique: Maintain a variable min to track the smallest value seen so far. Traverse the array. Update min using Math.min(min, cost[i]). Store the current minimum in the result array. 🔹 Time Complexity ⏱ O(n) – Single pass through the array. 🔹 Space Complexity 📦 O(n) – For storing the result array. 🔹 Key Learning This is a classic prefix computation pattern where we keep track of information while traversing the array. #DSA #Java #Programming #CodingJourney #100DaysOfCode #SoftwareEngineering