Solved Palindrome Partitioning with Backtracking in Java

Day 29/100 – #100DaysOfCode 🚀 | #Java #LeetCode #BinaryTree ✅ Problem Solved: Verify Preorder Serialization of a Binary Tree 🌲 🧩 Problem Summary: Given a string representing a preorder serialization of a binary tree, determine if it’s valid. Example: "9,3,4,#,#,1,#,#,2,#,6,#,#" → ✅ valid "1,#" → ❌ invalid 💡 Approach Used: Used the slot-counting method 🧠 Each node uses one slot and non-null nodes create two new slots. Process the preorder string, ensuring slots never go negative and exactly zero remain at the end. ⚙️ Time Complexity: O(n) 📦 Space Complexity: O(1) ✨ Takeaway: This problem teaches how binary tree structure can be validated with simple counting logic — no need to rebuild the tree! 🌱 #Java #LeetCode #BinaryTree #100DaysOfCode #ProblemSolving #CodingChallenge

Day 54/100 – #100DaysOfCode 🚀 | #Java #SlidingWindow #TwoPointers ✅ Problem Solved: Count the Number of Substrings With Dominant Ones (LeetCode) 🧩 Problem Summary: You’re given a binary string. A substring is dominant if: number of 1s > number of 0s × k (where k is given). Return the total count of such substrings. 💡 Approach Used: Used an optimized Sliding Window + Two Pointers approach: Traverse with a right pointer. Adjust the left pointer whenever the substring stops being dominant. All valid windows contribute (left + 1) substrings. This avoids brute-force O(N²) and makes the solution efficient. ⚙️ Time Complexity: O(N) 📦 Space Complexity: O(1) ✨ Takeaway: Sliding window transforms heavy substring problems into clean linear-time solutions. #Java #LeetCode #SlidingWindow #TwoPointers #ProblemSolving #CodingChallenge #100DaysOfCode

Day 39/100 – #100DaysOfCode 🚀 | #Java #LeetCode #HashMap ✅ Problem Solved: Minimum Index Sum of Two Lists 🧩 Problem Summary: Given two string arrays, find the common strings with the smallest index sum (index in list1 + index in list2). If multiple answers exist, return all of them. 💡 Approach Used: ➤ Stored the elements of the first list in a HashMap with their indices. ➤ Iterated the second list and checked for matches. ➤ Tracked the minimum index sum and collected all strings that match that sum. This avoids nested loops and improves efficiency. ⚙️ Time Complexity: O(n + m) 📦 Space Complexity: O(n) ✨ Takeaway: HashMaps continue to be one of the most effective tools for reducing time complexity through direct lookups 🔍⚡ #Java #LeetCode #HashMap #DataStructures #ProblemSolving #100DaysOfCode #CodingJourney

Day 40/100 – #100DaysOfCode 🚀 | #Java #LeetCode #TwoPointers #DynamicProgramming ✅ Problem Solved: Longest Palindromic Substring 🟣 🧩 Problem Summary: Given a string, return the longest palindromic substring within it. The substring must read the same forward and backward. 💡 Approach Used: Expand Around Center Technique Every palindrome is centered around: A single character (odd length) Or between two characters (even length) For each index: Expand left and right while characters match. Track the longest substring during expansions. This avoids unnecessary recomputation and works efficiently. ⚙️ Time Complexity: O(n²) 📦 Space Complexity: O(1) ✨ Takeaway: Sometimes, the cleanest approach comes from observing patterns — in this case, symmetry around a center 🎯 #Java #LeetCode #TwoPointers #ExpandAroundCenter #ProblemSolving #100DaysOfCode #CodingJourney

Day 45/100 – #100DaysOfCode 🚀 | #Java #Hashing #SlidingWindow ✅ Problem Solved: Contains Duplicate III (LeetCode 220) 🧩 Problem Summary: Given an integer array and integers k and t, determine if there exist two distinct indices i and j such that: |i - j| ≤ k |nums[i] - nums[j]| ≤ t 💡 Approach Used: Used a Sliding Window + TreeSet to maintain numbers in a sorted structure. Steps: Traverse the array and maintain a window of at most size k. For each element x, find if there exists another element in the set such that |x - y| ≤ t. This is done using ceiling() to find the closest value ≥ x. Insert the element in TreeSet and remove the element that slides out. ⚙️ Time Complexity: O(n log k) 📦 Space Complexity: O(k) ✨ Takeaway: This problem teaches how ordered data structures like TreeSet help efficiently handle range queries in sliding window scenarios. #Java #TreeSet #SlidingWindow #LeetCode #ProblemSolving #100DaysOfCode #CodingChallenge

#100DaysOfCode – Day 85 Delete the Middle Node of a Linked List Problem Statement: Given the head of a singly linked list, delete the middle node and return the modified list. My Approach: Used two-pointer technique (slow & fast) to identify the middle node in a single traversal. Maintained a prev pointer to skip the middle node without extra space. Carefully handled the edge case when the list has only one node. Complexity: Time: O(N) Space: O(1) Linked lists are all about pointers and precision a small mistake (like an extra semicolon ";") can make a huge difference. Understanding how slow and fast pointers move gives deep insight into efficient traversal patterns. #LeetCode #100DaysOfCode #Java #LinkedList #ProblemSolving #takeUforward #CodingJourney #DataStructures #CodeNewbie

Day 49/100 – #100DaysOfCode 🚀 | #Java #DynamicProgramming #Optimization ✅ Problem Solved: Palindrome Partitioning II (LeetCode 132) 🧩 Problem Summary: Given a string s, partition it such that every substring is a palindrome, and return the minimum number of cuts needed. 💡 Approach Used: This is an optimization of yesterday’s backtracking idea, but brute-force is too slow. So, used Dynamic Programming. Key Steps: Precompute Palindromes Used DP to mark isPalindrome[i][j] → whether substring s[i..j] is a palindrome. Minimum Cuts DP dp[i] = minimum cuts needed for substring s[0..i]. If s[0..i] is palindrome → dp[i] = 0 Else → minimize over all dp[j-1] + 1 where s[j..i] is palindrome. ⚙️ Time Complexity: O(N²) 📦 Space Complexity: O(N²) ✨ Takeaway: This problem highlights how precomputation + DP drastically reduces complexity compared to brute-force recursion. #Java #LeetCode #DynamicProgramming #DP #Palindrome #ProblemSolving #100DaysOfCode #CodingChallenge

🚀 LeetCode #58 – Length of Last Word (Java Solution) Today I solved a classic string problem that tests how well you handle edge cases and string traversal logic. 🧩 Problem Statement: Given a string s consisting of words and spaces, return the length of the last word in the string. A word is defined as a maximal substring consisting of non-space characters only. 🧠 Dry Run Example Input: " fly me to the moon " ➡ After skipping spaces → "fly me to the moon" ➡ Last word = "moon" ➡ Output: 4 ⚙️ Time & Space Complexity Time Complexity: O(n) — we traverse the string once (from the end to the start). Space Complexity: O(1) — no extra space used apart from a few variables. 💡 Key Takeaways: ✅ Learned how to handle trailing spaces properly before counting. ✅ Strengthened understanding of string traversal in reverse. ✅ Explored an alternative approach using trim() + split("\\s+") for readability. ✅ Remembered to test with edge cases like "a" or "Hello " to ensure correctness. Every small problem like this helps in building stronger fundamentals 💪 #Java #LeetCode #ProblemSolving #CodingJourney #LearningInPublic #75DaysOfCode #DataStructures #Algorithms

🎯 LeetCode 394 – Decode String Today I solved a really interesting Stack + String based problem! 💡 Problem Summary: We are given an encoded string containing patterns like 3[a2[c]] where: Numbers represent how many times the substring should repeat. Nested patterns are allowed. The goal is to decode the string correctly. 🧠 Approach: I used two stacks: One to store counts (how many times to repeat) One to store previously formed strings We iterate through the string: Build numbers when digits appear When [ appears, push state to stacks When ] appears, pop and reconstruct substring Else, just append characters normally ⏱️ Time Complexity: O(n) 💾 Space Complexity: O(n) #LeetCode #Java #DSA #Stack #StringManipulation #CodingJourney #LearnEveryday #100DaysOfCode 🚀

🔹 Day 46: Is Subsequence (LeetCode #392) 📌 Problem Statement: Given two strings s and t, return true if s is a subsequence of t, or false otherwise. A subsequence is formed by deleting some (possibly none) characters from the original string without disturbing the relative order of the remaining characters. ✅ My Approach: I used a two-pointer technique — one pointer iterates through string t, and the other tracks progress through string s. Each time a matching character is found, the pointer for s moves forward. If we reach the end of s, it means all its characters appeared in sequence within t. 📊 Complexity: Time Complexity: O(n) Space Complexity: O(1) ⚡ Submission Stats: Runtime: 2 ms (Beats 68.53%) Memory: 41.32 MB (Beats 87.28%) 💡 Reflection: A simple yet elegant problem that highlights how pointer movement can efficiently handle string comparisons without extra memory usage. ✨ #LeetCode #Java #Strings #TwoPointers #100DaysOfCode #Day46