LeetCode Day 28: k-th Lexicographical Happy String

🚀 Day 31 of #LeetCode Challenge 🔍 Problem: Decode the Slanted Ciphertext Today’s problem was all about understanding patterns and matrix traversal in a clever way! 💡 Key Idea: * The encoded string is formed by writing characters in a matrix row-wise * The trick is to read diagonally (↘️ direction) to decode the original message * No need to build a 2D matrix — we can directly calculate indices! 🧠 What I Learned: * How to convert 1D string into virtual 2D matrix * Diagonal traversal techniques * Importance of trimming trailing spaces in string problems ⚡️ Approach: * Calculate number of columns → cols = n / rows * Traverse diagonally from each column * Append characters using index formula i * cols + j * Remove extra spaces at the end 💻 Language: Java 🎯 Time Complexity: O(n) 📦 Space Complexity: O(n) Consistency is the key 🔑 — one problem at a time, getting better every day! #Day31 #LeetCode #Java #CodingJourney #ProblemSolving #100DaysOfCode

📌 LeetCode Daily Challenge — Day 14 Problem: 1415. The k-th Lexicographical String of All Happy Strings of Length n Topic: Backtracking, String   📌 Quick Problem Sense: A happy string uses only ['a', 'b', 'c'] and no two adjacent characters are the same. Given n (length) and k (position), return the k-th happy string in lexicographical order — or "" if fewer than k exist. You need to generate candidates in order and stop at the k-th one. Backtracking is your entire strategy!   🧠 Approach (Simple Thinking): 🔹 At each position you have at most 2 choices — any character from ['a','b','c'] except the previous one 🔹 Characters are tried in alphabetical order — this naturally gives lexicographic ordering 🔹 Use backtracking to build strings character by character, skipping invalid choices 🔹 Every time a string of length n is completed, increment a counter 🔹 When counter hits k, that's your answer — return immediately, no need to explore further 🔹 If backtracking exhausts all possibilities and counter never reached k, return "" 🔹 Total happy strings = 3 * 2^(n-1) → if k exceeds this, return "" instantly   ⏱️ Time Complexity: Each position has at most 2 choices, depth is n → O(2^n) We stop early at the k-th string, so often much faster in practice 📦 Space Complexity: Recursion stack depth = n, current string buffer = n → O(n) No extra arrays storing all strings!   I wrote a full breakdown with dry run, analogy, and step-by-step code walkthrough here: https://lnkd.in/gFrtgPFf If you solved it iteratively or used a math trick to directly compute the k-th string without generating all of them, drop it in the comments, always curious to see how others think about it 💬 See you in the next problem 👋 #LeetCode #DSA #Backtracking #CodingInterview #Java

Day 33 of Daily DSA 🚀 Solved LeetCode 58: Length of Last Word ✅ Problem: Given a string s, return the length of the last word (a sequence of non-space characters). Approach: First, trim() the string to remove trailing spaces Traverse the string from the end Count characters until a space is encountered 💡 Key Insight: Instead of splitting the string (which uses extra space), we can simply iterate backwards for an efficient solution. ⏱ Complexity: • Time: O(n) • Space: O(1) 📊 LeetCode Stats: • Runtime: 0 ms (Beats 100%) ⚡ • Memory: 43.17 MB A simple yet important problem to strengthen string traversal techniques. #DSA #LeetCode #Java #Strings #ProblemSolving #CodingJourney #Consistency

🚀 09/04/26 — Alternating Precision: Mastering String Merging Today I tackled Merge Strings Alternately (LeetCode 1768), focusing on efficient string construction and pointer management. This problem is an excellent exercise in handling sequences of different lengths using a unified traversal strategy. 🧵 Merge Strings Alternately Logic The goal is to merge two strings by adding letters in alternating order, starting with the first string. Any remaining letters from the longer string are appended to the end. StringBuilder Strategy: I used StringBuilder for the result to ensure efficient appends, avoiding the overhead of string concatenation in Java. The Interleaved Loop: I initialized two pointers, i and j, at 0. A single while loop runs as long as either string has remaining characters (i < word1.length() || j < word2.length()). Conditional Appending: Inside the loop, I check each pointer individually. If a pointer is still within its string's bounds, I append the character and increment that pointer. Automatic Handling: This structure naturally handles cases where one string is longer, as the conditions safely skip the shorter string once it is exhausted. Complexity Metrics: Time Complexity: O(a + b), where a and b are the lengths of word1 and word2, as every character is visited once. Space Complexity: O(a + b) to store the merged result in the StringBuilder. 📈 Consistency Report Today’s session further solidifies the pointer-based intuition I've been developing. The logic used here—managing multiple indices within a single loop—mirrors the "Two Sum" and "String Search" patterns I mastered earlier in the roadmap. Achieving a 95.66% beat rate proves that choosing the right tools, like StringBuilder, is just as important as the algorithm itself. My 1ms alternating merge implementation is attached below! 📄👇 #DSA #Java #LeetCode #StringManipulation #TwoPointers #Complexity #Consistency #LearningInPublic

𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐇𝐚𝐫𝐝 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝: 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐒𝐭𝐫𝐢𝐧𝐠 𝐰𝐢𝐭𝐡 𝐋𝐂𝐏 𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧 𝐋𝐢𝐧𝐤 :https://lnkd.in/gPQ5WExk 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐋𝐢𝐧𝐤 : https://lnkd.in/gYpvYnaW Today I worked on an interesting 𝐡𝐚𝐫𝐝 𝐩𝐫𝐨𝐛𝐥𝐞𝐦: 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐒𝐭𝐫𝐢𝐧𝐠 𝐰𝐢𝐭𝐡 𝐋𝐂𝐏. The problem provides an 𝐋𝐂𝐏 (𝐋𝐨𝐧𝐠𝐞𝐬𝐭 𝐂𝐨𝐦𝐦𝐨𝐧 𝐏𝐫𝐞𝐟𝐢𝐱) 𝐦𝐚𝐭𝐫𝐢𝐱 and asks us to construct the lexicographically smallest string that satisfies this matrix.  𝐊𝐞𝐲 𝐈𝐝𝐞𝐚 Instead of constructing substrings directly, the approach is: 1️⃣ Start assigning characters from 'a' onward. 2️⃣ If lcp[i][j] > 0, it means the substrings starting at i and j share the same first character. 3️⃣ Propagate the same character across positions where the LCP value indicates a match. 4️⃣ Finally, validate the entire LCP matrix using the rule: lcp[i][j] = (word[i] == word[j]) ? 1 + lcp[i+1][j+1] : 0 If any condition fails, no valid string exists.  𝐖𝐡𝐚𝐭 𝐈 𝐥𝐞𝐚𝐫𝐧𝐞𝐝 • How LCP matrices represent relationships between substrings • Constructing strings using greedy character assignment • Validating constraints using dynamic programming relations • Importance of verifying generated structures against given matrices Problems like this really sharpen string manipulation + matrix reasoning skills. Always fun pushing through Hard-level challenges! #LeetCode #DataStructures #Algorithms #CodingPractice #Java #ProblemSolving

Day 4 of my DSA Journey 🚀 Today’s challenge: The classic Palindrome Number problem. For today's solution, I focused on an intuitive approach using String manipulation in Java. Sometimes the best way to solve a problem is to break it down into steps that make logical sense right out of the gate! 🧠 My Approach: Handle the edge case: Negative numbers can never be palindromes, so return false immediately. Convert the integer to a String. Use a for loop to iterate backwards through the string, building a new reversed string character by character. Compare the original string with the reversed string to determine if it's a palindrome. ⚡ Key Learnings & Java Gotchas: String Iteration: In Java, strings aren't exactly like arrays, so I had to use .charAt(i) instead of standard bracket notation [i] to grab individual characters. The Comparison Trap: I learned the crucial difference between == and .equals(). In Java, == compares memory locations, but .equals() checks if the actual text values are the same. A huge "aha!" moment for handling Strings! Edge Case Handling: Always check for negative numbers first to save processing time. 📌 Complexity: Time: O(n) — where n is the number of digits, since we loop through the string once. Space: O(n) — for storing the string representations of the number. Every bug is a lesson, and every solved problem is a step forward. On to the next one! 💪 #DSA #DataStructures #Algorithms #ProblemSolving #CodingJourney #Java #TechJourney#DSAJourney #LeetCode #Coding #LearnInPublic #GrowthMindset

🚀 Day 25 of my #100DaysOfCode Journey Today, I solved the LeetCode problem Valid Perfect Square. Problem Insight: Given a positive integer, the task is to determine whether it is a perfect square without using built-in functions like sqrt(). Approach: Used Binary Search to efficiently find whether a number has an integer square root. Initialized search range from 0 to num Calculated mid and checked mid * mid If equal → return true If square is smaller → move right (low = mid + 1) If square is larger → move left (high = mid - 1) Used long for multiplication to avoid overflow issues. Time Complexity: O(log n) — efficient binary search approach Takeaway: Binary Search is not just for arrays — it can be applied to mathematical problems to optimize performance and avoid brute force. #DSA #Java #LeetCode #CodingJourney #100DaysOfCode #BinarySearch

The Two-Pointer streak continues! Today’s LeetCode Problem: Is Subsequence. After using multiple pointers to sort arrays and move zeroes, I applied the exact same pattern to string manipulation today. The challenge was figuring out if a shorter string (s) is a valid subsequence of a longer string (t) without disturbing the relative order of the characters. Instead of generating all possible subsequences (which would be a massive O(2ⁿ) performance drain), the Two-Pointer approach solves this beautifully in a single pass. Knocking this out in O(n) time and O(1) space is another great reminder of how incredibly versatile this algorithmic pattern is for both arrays and strings. #DSA #Java #LeetCode #IsSebsequenceProblem

🚀 Tackling algorithmic challenges one substring at a time! Just solved the “Number of Substrings Containing All Three Characters” problem in Java using a sliding window approach. ✨ Key takeaway: Efficient solutions often come from thinking in terms of windows and counts rather than brute force. 💻 Output validated with test cases — clean, optimized, and accepted! #Java #CodingChallenge #ProblemSolving #DataStructures #Algorithms #LeetCode #ProgrammingJourney

🚀 LeetCode Practice Update Solved “1415. The k-th Lexicographical String of All Happy Strings of Length n” today. It’s an easy–medium problem for anyone comfortable with Backtracking, but a great exercise to strengthen recursive thinking and constraint-based generation. 🧠 Approach The idea is to generate all valid “happy strings” using backtracking and then simply pick the k-th string in lexicographical order. A happy string means: It only contains characters 'a', 'b', 'c' No two adjacent characters are the same ⚙️ Strategy 1. Start with an empty string and recursively build all possible strings of length n. 2. At each step, try adding characters from 'a' → 'c' to maintain lexicographical order. 3. Before adding a character, check the constraint: If the last character of the current string is the same as the character we want to add → skip it. 4. If the string length becomes n, add it to the list of valid happy strings. 5. After generating all valid strings: If k > total strings, return an empty string. Otherwise return the (k-1)th index from the list. 🔁 Key Backtracking Step After exploring a choice, we remove the last character to restore the previous state and try the next option. This pattern of choose → explore → undo (backtrack) is the heart of backtracking problems. 💡 Problems like this really show how powerful backtracking can be when generating all valid combinations under constraints. #LeetCode #Backtracking #Java #DSA #ProblemSolving #CodingJourney