Optimizing Sticker Problem with DP and Memoization

𝐃𝐚𝐲 87/100 – 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🚀 Problem: 228. 𝐒𝐮𝐦𝐦𝐚𝐫𝐲 𝐑𝐚𝐧𝐠𝐞𝐬  Today I solved a problem where we need to summarize consecutive numbers in a sorted unique array into ranges. 🔑 𝐈𝐝𝐞𝐚: Traverse the array and keep extending the range while consecutive numbers continue. Once the sequence breaks, close the range and store it. 💡 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Start with the first element as start Move forward while nums[i] + 1 == nums[i+1] If range exists → "start->end" Else → single number "start" 𝐊𝐞𝐲 𝐈𝐧𝐬𝐢𝐠𝐡𝐭: Efficient single pass solution (O(n)) by grouping consecutive elements on the fly. #LeetCode #Java #ProblemSolving #DSA #100DaysOfCode #CodingJourney

Day 92 of #100DaysOfLeetCode 💻✅ Solved #739. Daily Temperatures problem in Java. Approach: • Used Monotonic Stack to track indices • Compared current temperature with stack top • Updated result when a warmer day was found • Stored indices for future comparisons Performance: ✓ Runtime: 60 ms (Beats 79.01% submissions) 🚀 ✓ Memory: 107.89 MB (Beats 17.53% submissions) Key Learning: ✓ Learned Monotonic Stack technique ✓ Improved handling of next greater element problems ✓ Strengthened stack-based problem solving Learning one problem every single day 🚀 #Java #LeetCode #DSA #Stack #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Day 61/100 Today’s problem: Check Balanced String 📌 Problem Summary: A string is called balanced if the sum of digits at even indices is equal to the sum at odd indices. 💡 What I learned: - How to handle index-based logic - Difference between even & odd indexing - Clean iteration and condition checking 🧠 Approach: - Traverse the string - Maintain two sums → even index & odd index - Compare both sums ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(1) Consistency is the key — small steps every day lead to big results 💪 #Day61 #CodingJourney #Java #DSA #ProblemSolving #Consistency #KeepLearning #100DaysOfCode

LeetCode — Problem 189 | Day 3 💡 Problem: Rotate Array Given an array, rotate it to the right by k steps. 🧠 My Approach: - Used reverse technique for in-place rotation - First reversed the entire array - Then reversed first k elements - Finally reversed remaining elements - Handled k using k = k % n This problem gave a good understanding of: ✔️ Array manipulation ✔️ In-place optimization (O(1) space) ✔️ Reverse logic #LeetCode #DSA #Java #CodingJourney

✅ LeetCode Top Interview 150 – Day 91 Today I solved Clone Graph Key Idea : Use DFS/BFS traversal to clone each node Maintain a HashMap to store original → cloned node mapping For each node, create a copy and recursively clone its neighbors Reuse already cloned nodes from map to avoid duplication and cycles This approach ensures a correct deep copy of the graph #DSA #Java #Leetcode #Day91

🚀 Day 64/100 Today’s problem: Find all strings that are substrings of another word 🧠 What I learned: - How to compare strings using nested loops - Using ".contains()" to check substrings efficiently - Importance of breaking early to optimize performance - Strengthening problem-solving with brute-force approach 💡 Key Insight: Sometimes simple solutions (O(n²)) are enough when constraints are small. No need to overcomplicate! 🔁 Consistency > Perfection #Day64 #DSA #Java #CodingJourney #Consistency #KeepLearning #100DaysOfCode

Day 81 - Balanced Binary Tree Checked whether a binary tree is height-balanced using a bottom-up approach. Approach: • Recursively compute height of left and right subtrees • If height difference > 1 → not balanced • Use -1 as a signal to stop early Time Complexity: O(n) Space Complexity: O(h) #Day81 #LeetCode #BinaryTree #DSA #Java #CodingJourney #ProblemSolving

🔥 Day 58 of #100DaysOfCode Solved – Check Balanced String 🔍 What I did: Traversed the string and calculated the sum of digits at even and odd indices separately, then compared both sums to determine if the string is balanced. 💡 Key Learning: Practiced string traversal and indexing Improved understanding of even vs odd index logic Learned how to convert char to integer (c - '0') 🎯 Takeaway: Even simple index-based problems can strengthen your fundamentals and attention to detail. #Day58 #LeetCode #Java #ProblemSolving #CodingJourney #100DaysOfCode #DSA

📘 DSA Journey — Day 27 Today’s focus: Binary Search on modified arrays. Problem solved: • Search in Rotated Sorted Array (LeetCode 33) Concepts used: • Binary Search • Identifying sorted halves • Conditional search space reduction Key takeaway: This problem extends binary search to a rotated sorted array, where the array is not fully sorted but divided into two sorted parts. At each step, we: • Find the mid element • Check which half (left or right) is sorted • Decide whether the target lies in the sorted half • Eliminate the other half This allows us to still achieve O(log n) time complexity. Continuing to strengthen fundamentals and problem-solving consistency. #DSA #Java #LeetCode #CodingJourney

🚀 Day 48 of My #LeetCode Journey Today’s problem: 2615. Sum of Distances 💡 Key Idea: Instead of calculating distances between equal elements using brute force (O(n²)), I used: HashMap to group indices of same values Prefix Sum to efficiently compute distances This reduced the complexity to O(n) 🔥 🧠 What I Learned: How prefix sums can optimize distance calculations Efficient handling of repeated elements Writing clean and optimized code using Java ⚡ Approach: Store indices of each number Use prefix sums to calculate left & right distances Combine both to get final answer 📈 Time Complexity: O(n) 📦 Space Complexity: O(n) Consistency is key. Small progress every day leads to big results 💪 #Day48 #Java #FullStackDeveloper #CodingJourney #100DaysOfCode #DSA #LeetCode