Java Substring Challenge Solved with Sliding Window Approach

Day 35 of Consistency 💻🔥 Solved Longest Substring Without Repeating Characters — a classic sliding window problem that really tests optimization skills. ✨ Key Takeaways: Mastered the sliding window technique Used HashMap for efficient lookup Improved understanding of two-pointer approach ⚡ Performance: Runtime: 5 ms Beat: 87%+ submissions Every day is about getting sharper, not just solving problems. #LeetCode #DataStructures #Algorithms #CodingJourney #Consistency #Java #ProblemSolving

Problem of the Day 33 : Longest Cycle in a Directed Graph Solved a challenging graph problem where each node has at most one outgoing edge. Used a smart approach with a HashMap to track visit order and detect cycles efficiently. Learned how simple traversal + proper tracking can solve complex problems in O(V) time. #DSA #Graphs #Coding #ProblemSolving #Java

Day 60 of #100DaysOfLeetCode 💻✅ Solved #389. Find the Difference problem in Java. Approach: • Initialized a variable to store the sum • Subtracted ASCII values of all characters in string s • Added ASCII values of all characters in string t • The remaining value represents the extra character • Converted the result back to character and returned it Performance: ✓ Runtime: 3 ms (Beats 38.82% submissions) ✓ Memory: 43.34 MB (Beats 30.69% submissions) Key Learning: ✓ Learned a clever use of ASCII values for problem solving ✓ Practiced using math-based approach instead of extra data structures ✓ Improved understanding of string manipulation techniques Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #Math #ProblemSolving #CodingJourney #100DaysOfCode

Day 82 of DSA Journey Solved Isomorphic Strings today! This problem looks simple, but it teaches an important concept — one-to-one mapping between characters. 🔍 What I learned: Each character in one string must map to exactly one character in the other No two characters should map to the same character Consistency across the entire string is key 💡 Approach: Used two HashMaps to track mappings in both directions: s → t t → s This avoids conflicts and ensures correctness. 🧠 Example: "paper" → "title" ✅ "foo" → "bar" ❌ ✨ Key takeaway: Always think about bidirectional mapping when dealing with transformation problems. Small problem, but powerful concept! #DSA #Java #CodingJourney #100DaysOfCode #ProblemSolving

💻 Interesting Java problem: Find the closest target in a circular array Given a words[] array, a target, and a startIndex, the task is to find the shortest circular distance to the target. Leetcode Problem :https://lnkd.in/g3RbdChE Core idea: For every matching target: find direct distance → Math.abs(i - startIndex) find circular distance → n - dist take the minimum. This is a neat example of how circular arrays require us to think beyond normal linear traversal. Small twist, smart logic. 🚀 #Java #Algorithms #DSA #CodingInterview #ProblemSolving #LeetCode

Day 73 of #100DaysOfLeetCode 💻✅ Solved #389. Find the Difference problem in Java. Approach: • Initialized a variable to store the sum • Added ASCII values of all characters in string t • Subtracted ASCII values of all characters in string s • The remaining value represents the extra character • Converted the result back to character and returned it Performance: ✓ Runtime: 1 ms (Beats 99.89% submissions) 🚀 ✓ Memory: 42.77 MB (Beats 97.03% submissions) Key Learning: ✓ Practiced optimized string traversal using enhanced for-loop ✓ Reinforced understanding of ASCII-based solutions ✓ Learned how mathematical approaches can simplify problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #Math #ProblemSolving #CodingJourney #100DaysOfCode

🚀 LeetCode Problem of the Day 📌 Problem: Minimum Distance to Target Element Given an integer array nums (0-indexed) and two integers target and start, we need to find an index i such that: nums[i] == target |i - start| is minimized 👉 Return the minimum absolute distance. 💡 Key Idea: We simply scan the array and track the minimum distance whenever we find the target. 🧠 Approach Traverse the array Whenever nums[i] == target, compute abs(i - start) Keep updating the minimum result 💻 Java Solution class Solution { public int getMinDistance(int[] nums, int target, int start) { int res = Integer.MAX_VALUE; for (int i = 0; i < nums.length; i++) { if (nums[i] == target) { res = Math.min(res, Math.abs(i - start)); } } return res; } } 📊 Complexity Analysis ⏱ Time Complexity: O(n) → single pass through the array 🧠 Space Complexity: O(1) → no extra space used ⚡ Simple linear scan, optimal solution — sometimes brute force is already the best solution! #LeetCode #Coding #Java #ProblemSolving #DataStructures #Algorithms #InterviewPrep

#Day77 of my second #100DaysOfCode Finally wrapped up binary search on 1D arrays, now moving to binary search on answers. DSA • Solved Sqrt(x) (LeetCode 69) — finding the integer square root of a number without using built-in functions – Brute: iterate until square exceeds target → O(n) – Optimal: binary search on answer space → O(log n) • Key idea: instead of searching an index, we search for the answer itself • Learned how to narrow down the range based on mid² comparison Nice shift in thinking — binary search feels much more powerful now. #DSA #BinarySearch #LeetCode #Algorithms #Java #100DaysOfCode #WomenWhoCode #BuildInPublic #LearningInPublic

#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

Day 69 of #100DaysOfLeetCode 💻✅ Solved #349. Intersection of Two Arrays problem in Java. Approach: • Iterated through each element of the first array • Checked if the element exists in the second array • Used a temporary array to store intersection elements • Ensured no duplicates by checking already added elements • Copied the result into a final array of correct size Performance: ✓ Runtime: 4 ms (Beats 35.60%) ✓ Memory: 44.41 MB (Beats 96.71%) Key Learning: ✓ Practiced array traversal using nested loops ✓ Learned how to handle duplicates manually ✓ Improved understanding of set-like operations without using extra data structures Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #ProblemSolving #CodingJourney #100DaysOfCode