Generating Valid IP Addresses from Digits

🚀 Day 21 of my #100DaysOfCode Journey Today, I solved the LeetCode question Contains Duplicate II. The task is to check if there are two equal elements in the array such that their index difference is at most k. ✅ Steps: First, iterate through each element in the array For every element, check only the next k elements Compare values within this limited range If a match is found → return true 💻 My Approach: I used a sliding window + brute force approach (without using HashMap/HashSet). Instead of checking the whole array, I optimized it by limiting the inner loop to k distance only. 🌟 Learning Takeaways: Optimizing brute force can avoid TLE Understanding constraints (like k distance) helps reduce complexity Multiple approaches exist — choosing based on space/time is key #DSA #Java #LeetCode #CodingJourney #100DaysOfCode

🚀 Day 11 of #100DaysOfCode Solved LeetCode Problem #27 – Remove Element today! 💻 🔹 Approach: Used the Two Pointer Technique One pointer (i) iterates through the array Another pointer (ptr) keeps track of valid elements If nums[i] != val, assign it to nums[ptr] and increment ptr 🔹 Time Complexity: O(n) 🔹 Space Complexity: O(1) 💡 Key Learning: In-place array modification using two pointers helps optimize space and improves efficiency. 📌 Consistency is the key — one step closer every day! #LeetCode #DSA #Java #CodingJourney #Day11 #100DaysOfCode

🚀 Day 69 — LeetCode Practice 🚀 Today’s focus was on string manipulation and careful indexing. ✅ Problem solved today: 🔹 Find Common Characters 💡 Key learnings from today: • Understood how to find common characters across multiple strings • Learned the importance of character frequency counting • Practiced handling nested loops efficiently • Realized how small mistakes in indexing can affect the entire logic • Improved attention to detail while working with strings Initially, I made a mistake by using the wrong variable inside charAt(). This caused incorrect indexing, which can either lead to wrong output or even StringIndexOutOfBoundsException. After fixing it, the logic worked perfectly by correctly comparing characters and tracking their minimum frequency across all words. This problem reinforced an important lesson: Sometimes errors are not in logic — but in small details like indexing and variable usage. Accuracy matters as much as logic. Step by step, getting better 💪 On to Day 70 🚀 #Day69 #DSA #LeetCode #ProblemSolving #Java #CodingJourney #Consistency #FutureEngineer

Day 101 - LeetCode Journey Solved LeetCode 856: Score of Parentheses ✅ Looks tricky at first, but turns out to be a neat pattern problem. Instead of using a stack, used depth tracking + bit manipulation to calculate score efficiently. Core idea: Every "()" contributes 2^depth So just track depth and add score at the right moment. Key learnings: • Understanding pattern inside parentheses • Using depth instead of extra space • Bit manipulation for optimization ⚡ • Clean O(n) solution without stack ✅ All test cases passed ⚡ O(n) time | O(1) space Simple logic, powerful result 💡 #LeetCode #DSA #Stack #BitManipulation #Java #CodingJourney #ProblemSolving #InterviewPrep

100 Days of Code Day - 24 🔁 Swap Nodes in Pairs – LeetCode Problem Solved a classic Linked List problem today! 💡 👉 Given a linked list, swap every two adjacent nodes without modifying values — only changing the links. ✅ Example: Input: [1,2,3,4] Output: [2,1,4,3] 💭 Approach: Used a dummy node and pointer manipulation to efficiently swap nodes in O(n) time and O(1) space. 📌 Key Learning: Understanding pointer handling is crucial for mastering Linked Lists. #LeetCode #Java #DSA #LinkedList #CodingJourney #ProblemSolving

🚀 Day 63 of #LeetCode Practice Solved: Count Equal and Divisible Pairs in an Array (Easy) 🔹 Problem Summary: Given an array and an integer k, find the number of pairs (i, j) such that: • nums[i] == nums[j] • (i * j) is divisible by k 🔹 Approach: • Used two nested loops to check all pairs • Compared elements to find equal values • Checked if (i * j) % k == 0 • Counted valid pairs 🔹 Key Learning: Brute-force helps build a strong foundation and improves problem understanding before optimization. 💡 Consistency, patience, and small daily improvements lead to big results over time. #LeetCode #DSA #Java #CodingJourney #Consistency #KeepLearning

🚀 Day 18 of #100DaysOfCode Solved LeetCode 334 – Increasing Triplet Subsequence 📈 Today’s problem was all about optimizing from brute force to a smart greedy approach. Instead of checking all triplets, I tracked the smallest and second smallest values while iterating once through the array. 💡 Key Insight: If we find a number greater than both first and second, an increasing triplet exists! 🔍 What I learned: Greedy approach can reduce time complexity drastically Maintaining two variables is enough to detect a triplet Writing clean and efficient O(n) solutions ⚡ Performance: Runtime: 2 ms (Beats 99.28%) Memory: Solid optimization #DSAwithEdSlash #LeetCode #Java #CodingJourney #ProblemSolving

🚀 Day 17 of My LeetCode Journey 📌 Solved: LeetCode #11 – Container With Most Water Today’s problem was a great example of applying the Two Pointer Approach efficiently. 🔍 Problem Summary: Given an array of heights, find two lines that together with the x-axis form a container that holds the maximum amount of water. 💡 Approach Used: Started with two pointers at both ends Calculated area using: min(height[left], height[right]) × width Moved the pointer with smaller height to maximize area ⚡ Key Learning: Instead of checking all pairs (O(n²)), we can optimize to O(n) using the two-pointer technique. 🧠 Time Complexity: O(n) 📦 Space Complexity: O(1) 🔥 Consistency is the key — learning something new every day! #LeetCode #DSA #Java #CodingJourney #ProblemSolving #100DaysOfCode

Day 20/100 🚀 | LeetCode Grind Cracked “Guess Number Higher or Lower” using Binary Search 🎯 Optimized the approach by narrowing down the search space efficiently with each guess — classic divide & conquer in action. 💡 Key Takeaway: When the search space is sorted and bounded, Binary Search is your best friend. ⚡ Runtime: 0 ms (Beats 100%) 📊 Efficient and clean implementation Consistency > Intensity. Showing up every day 💪 #Day20 #100DaysOfCode #LeetCode #DSA #BinarySearch #CodingJourney #Java #ProblemSolving

🚀 Day 73 — LeetCode Practice 🚀 Today’s problem: Find the Distance Value Between Two Arrays 💡 What I learned today: • Practiced nested loops and comparison logic • Strengthened understanding of absolute difference using Math.abs() • Learned how to use a boolean flag to optimize checking conditions • Improved handling of edge cases and conditions 🧠 Key idea: For each element in arr1, check all elements in arr2 → if no element satisfies the condition |arr1[i] - arr2[j]| ≤ d → then count it. ✨ Step by step, logic is becoming clearer and faster. #Day73 #LeetCode #Java #ProblemSolving #DSA #CodingJourney #Consistency