LeetCode 719: Kth Smallest Pair Distance Solution

Worked on a challenging problem: “Subarrays with K Different Integers” Key takeaway: Instead of directly solving for exactly K distinct elements, I learned a smarter approach: 👉 count(at most K) − count(at most K−1) 🔹 Concepts I practiced: Sliding Window technique HashMap for frequency tracking Two-pointer approach 🔹 What stood out: The idea of counting all valid subarrays ending at each index using (r - l + 1) was really powerful. It completely changed how I think about subarray problems. Always learning, one problem at a time. #DataStructures #Algorithms #Java #LeetCode #SlidingWindow #LearningJourney #ProblemSolving

🚀 Day 45 of #100DaysOfCode 🧩 Problem Solved: Longest Common Prefix Today’s challenge was to find the common prefix among a list of strings. I used an efficient approach by leveraging sorting. 💡 Approach: Sort the array of strings Compare the first and last strings Build the prefix character by character 🧠 Why this works: After sorting, the first and last strings are the most different. Their common prefix will be the answer for all strings. ⚡ Key Learning: Smart thinking > brute force. Using sorting can reduce unnecessary comparisons and simplify logic. 🏷️ Tags: #100DaysOfCode #Day45 #LeetCode #Java #DSA #CodingJourney #ProblemSolving #Algorithms #String #Array 📈 Staying consistent and improving step by step! 🚀

🚀 Day 9 of #100DaysOfDSA Solved LeetCode 239 – Sliding Window Maximum using Brute Force 💡 Problem: Given an array and a window of size k, find the maximum element in each window. Example: nums = [1,3,-1,-3,5,3,6,7], k = 3 Output: [3,3,5,5,6,7] 🧠 Approach (Brute Force): Traverse each window of size k Find maximum element for every window manually 💻 Code (Java): class Solution { public int[] maxSlidingWindow(int[] nums, int k) { int n = nums.length; int[] result = new int[n - k + 1]; for (int i = 0; i <= n - k; i++) { int max = nums[i]; for (int j = i; j < i + k; j++) { if (nums[j] > max) { max = nums[j]; } } result[i] = max; } return result; } } 📊 Complexity: Time: O(n * k) Space: O(1) 🔥 Learning: This approach is simple but not efficient for large inputs. Next step: optimize using Deque (O(n)) #LeetCode #DSA #Java #Coding #InterviewPrep #100DaysOfCode trying to understand the dequeue approach. 🍁 Saidhanya Sree

Day 33/75 — Sort an Array (Merge Sort) Today’s problem required sorting an array in O(n log n) time without using built-in sort. Approach: • Divide array into two halves • Recursively sort both halves • Merge sorted halves Key idea: return merge(left, right); Merge step ensures elements are placed in sorted order. Time Complexity: O(n log n) Space Complexity: O(n) This problem reinforced the concept of Divide & Conquer. 33/75 🚀 #Day33 #DSA #MergeSort #Java #Algorithms #LeetCode

🚀 Day 3/100 – LeetCode Challenge 🔍 Problem: Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be inserted to maintain the sorted order. 💡 Key Concepts Used: • Binary Search • Time Complexity Optimization – O(log n) • Efficient searching in sorted arrays 📚 What I Learned: • How binary search reduces search time significantly compared to linear search • Handling edge cases when the target element is not present • Determining the correct insertion index while maintaining sorted order hashtag #100DaysOfCode #LeetCode #Java #DataStructures #Algorithms #BinarySearch #CodingJourney #ProblemSolving

#Day75 of my second #100DaysOfCode Today’s problem was more about spotting patterns than writing code. DSA • Solved Single Element in a Sorted Array (LeetCode 540, Medium) – Brute: linear scan checking pairs → O(n) – Optimal: binary search using index pattern → O(log n) • Key idea: elements appear in pairs, and the single element breaks this pattern • Used position-based logic to decide which half to search This one really showed how observing patterns can simplify the approach. #DSA #BinarySearch #LeetCode #Algorithms #Java #100DaysOfCode #WomenWhoCode #BuildInPublic #LearningInPublic

🚀 #100DaysOfCode | Day 48 🔍 Solved: Find Peak Element Today I worked on an interesting Binary Search problem where the goal was to find a peak element in an array. 💡Key Insight: By comparing the middle element with its next element, we can determine the direction of the peak. 📌 Approach: ✔ Applied Binary Search for O(log n) efficiency ✔ Compared mid with next element ✔ Moved towards the increasing side to find peak ✔ Reduced search space step by step Why this works: Since adjacent elements are not equal, there will always be at least one peak. By comparing neighboring elements, we can decide the direction and reduce the search space efficiently. 🎯 What I Learned: This problem showed how Binary Search can be used beyond searching—especially for identifying patterns like peaks in arrays. #Java #DSA #LeetCode #BinarySearch #CodingJourney #ProblemSolving #TechSkills 🚀

Day 58/100 | #100DaysOfDSA 🔄🔍 Today’s problem: Search in Rotated Sorted Array A twist on Binary Search with a rotated array. Key idea: Even though the array is rotated, one half is always sorted. Approach: • Use binary search • Find mid element • Check which half is sorted (left or right) • Decide if target lies in the sorted half • Narrow the search accordingly Why it works: At every step, we eliminate half of the search space just like standard binary search. Time Complexity: O(log n) Space Complexity: O(1) Big takeaway: Understanding the structure of the problem helps adapt classic algorithms like binary search. Rotation doesn’t break order — it just shifts it. 🔥 Day 58 done. #100DaysOfCode #LeetCode #DSA #Algorithms #BinarySearch #Arrays #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity