LeetCode 58: Length of Last Word in String

Day 39 of Daily DSA 🚀 Solved LeetCode 1446: Consecutive Characters ✅ Problem: The power of a string is the maximum length of a non-empty substring that contains only one unique character. Given a string s, return its power. Rules: * Substring must be non-empty * Substring must contain only one unique character * Return the maximum such length Approach: Used a simple linear scan to track the current streak of consecutive identical characters and update the maximum. Steps: 1. Initialize max and count both to 1 2. Iterate from index 1 onwards 3. If current character equals previous → increment count 4. Else → reset count to 1 5. Update max at every step 6. Return max ⏱ Complexity: • Time: O(n) • Space: O(1) 📊 LeetCode Stats: • Runtime: 29 ms (Beats 2.02%) • Memory: 45.33 MB Sometimes the simplest sliding window — just two variables — is all you need to solve a problem cleanly. #DSA #LeetCode #Java #SlidingWindow #Strings #CodingJourney #ProblemSolving

Classical Backtracking Problem with a Classical Approach Solved Word Search (LeetCode 79) using a clean and intuitive DFS + Backtracking strategy Approach (Simple & Clear): •Start from every cell that matches the first character of the word. •Use DFS (Depth-First Search) to explore all 4 directions (up, down, left, right). •Mark the current cell as visited (by replacing it temporarily) to avoid revisiting. •If the full word is matched → return true immediately. •Backtrack by restoring the cell when the path doesn’t work. Key Insight: We explore all possible paths, but prune early when characters don’t match — this keeps the solution efficient. ⏱️ Time Complexity: 👉 O(m × n × 4^L) •m × n → starting points •4^L → exploring 4 directions for each character of length L 📌 Space Complexity: 👉 O(L) (recursion stack) 🔥 Clean recursion + smart backtracking = problem cracked! #DSA #Backtracking #LeetCode #CodingInterview #Java #ProblemSolving

Day 38 of Daily DSA 🚀 Solved LeetCode 1897: Redistribute Characters to Make All Strings Equal ✅ Problem: Given an array of strings, determine if you can make every string equal by moving any character from one string to any position in another string. Rules: * You can pick two distinct indices i and j and move any character from words[i] to any position in words[j] * Any number of operations are allowed * Return true if all strings can be made equal, false otherwise Approach: Used a HashMap to count the total frequency of every character across all strings. If every character's count is divisible by the number of strings, equal redistribution is possible. Steps: 1. Concatenate all strings into one using StringBuilder 2. Count the frequency of each character using a HashMap 3. For every character count, check if it is divisible by the number of words 4. If any count is not divisible → return false 5. Otherwise → return true ⏱ Complexity: • Time: O(n * m) — n = number of words, m = average word length • Space: O(1) — at most 26 characters in the map 📊 LeetCode Stats: • Runtime: 12 ms (Beats 17.93%) • Memory: 46.63 MB A brilliant insight — redistribution is just a divisibility check! No complex simulation needed. #DSA #LeetCode #Java #HashMap #Strings #CodingJourney #ProblemSolving

✨ Day 42 of 90 – Pattern Mastery Journey 🧠 Pattern : Alphabet Hash Pattern 💡 Approach: ✔ Created an n × n matrix using nested loops ✔ Printed alphabets only when row index equals column index (i == j) ✔ Filled all other positions with `#` ✔ Used ASCII logic `(char)('A' + i - 1)` to generate characters 🚀 This problem helped me understand how **diagonal conditions work in matrices** and how simple conditions can create clean structured patterns. #PatternMasteryJourney #Java #CodingJourney #ProblemSolving

Day 97/200 – LeetCode Challenge Solved “Largest Rectangle in Histogram” (Hard) today. This problem is a great example of how powerful the Monotonic Stack technique can be. Instead of brute force, we efficiently determine how far each bar can extend to compute the maximum rectangle area. Using a monotonic increasing stack to track indices. Identifying left and right boundaries for each bar. Every day is making data structures feel more intuitive! #Day96 #LeetCode #Java #CodingJourney #ProblemSolving

🚀 Day 63 of #100DaysOfLeetCode ✅ Problem Solved: Unique Binary Search Trees (LeetCode 96) Today’s problem was a great example of how Dynamic Programming and mathematical patterns (Catalan Numbers) come together. 🔍 Key Insight: For every node chosen as root, the number of unique BSTs is: 👉 Left Subtrees × Right Subtrees This leads to the recurrence: dp[n] = Σ (dp[left] × dp[right]) 💡 What I learned: Breaking problems into smaller subproblems makes complex structures easier Recognizing patterns like Catalan Numbers is a game changer DP is not just about arrays, it's about thinking smart ⚡ Result: ✔️ Runtime: 0 ms (Beats 100%) ✔️ Clean and optimized solution Consistency is slowly turning into confidence 💪 #LeetCode #DataStructures #DynamicProgramming #CodingJourney #ProblemSolving #Java #100DaysOfCode

🚀 Day 36 of my DSA Journey 📌 Problem: 560. Subarray Sum Equals K Given an array and a value k, we need to find how many subarrays (continuous parts of the array) add up to k. 💡 My Approach (Simple Explanation): Instead of checking all subarrays (which is slow), I used the prefix sum + HashMap trick. Keep adding elements to get a running sum Check if (current sum - k) was seen before If yes, it means a valid subarray exists Store prefix sums in a map to reuse quickly This makes the solution efficient ⚡ ✅ Result: Accepted ⏱ Runtime: 0 ms ✨ Small optimizations like this really show how powerful concepts can simplify problems! #Day36 #DSA #CodingJourney #Java #LeetCode #ProblemSolving

🚀 LeetCode Problem || Count Submatrices With Equal Frequency of X and Y(3212) Today I worked on a problem where we need to count submatrices based on a condition involving characters in a grid. 🧩 Problem Idea We are given a grid containing: 'X' → treated as +1 'Y' → treated as -1 '.' → treated as 0 We need to count submatrices where: ✔ The total sum is 0 ✔ And at least one 'X' is present #Java #DSA #LeetCode #Matrix #Algorithms #ProblemSolving #CodingJourney

#100DaysOfCode – Day 2 Today I worked on a DSA problem based on arrays: Check if an array is sorted and rotated 🔍 Approach: Instead of finding the exact rotation point, I focused on identifying a pattern: In a sorted and rotated array, the order should break at most once. So, I checked how many times an element is greater than the next element while traversing the array in a circular manner. ✔️ If the count of such breaks is 0 or 1 → valid ❌ If it’s more than 1 → not a sorted rotated array 🧠 Key Takeaway: This problem taught me how pattern observation can simplify logic and avoid unnecessary complexity. Sometimes the best solution is not the most obvious one! 📈 Staying consistent and improving step by step 💪 #100DaysOfCode #DSA #DataStructures #Algorithms #Java #CodingJourney #ProblemSolving #LeetCode #Consistency