Max Average Subarray I Solution

💡 Day 48 of LeetCode Problem Solved! 🔧 🌟128. Longest Consecutive Sequence🌟 🔗 Solution Code: https://lnkd.in/gyN5ZJBr Task: Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time.   Example 1: Input: nums = [100,4,200,1,3,2] Output: 4 Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4. Example 2: Input: nums = [0,3,7,2,5,8,4,6,0,1] Output: 9 Example 3: Input: nums = [1,0,1,2] Output: 3 #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

💡 Day 42 of LeetCode Problem Solved! 🔧 🌟1876. Substrings of Size Three with Distinct Characters🌟 Task : • A string is good if there are no repeated characters. • Given a string s, return the number of good substrings of length three in s. • Note that if there are multiple occurrences of the same substring, every occurrence should be counted. • A substring is a contiguous sequence of characters in a string.   Example 1: Input: s = "xyzzaz" Output: 1 Explanation: There are 4 substrings of size 3: "xyz", "yzz", "zza", and "zaz". The only good substring of length 3 is "xyz". Example 2: Input: s = "aababcabc" Output: 4 Explanation: There are 7 substrings of size 3: "aab", "aba", "bab", "abc", "bca", "cab", and "abc". The good substrings are "abc", "bca", "cab", and "abc". #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

🚀 Day 58 of #100DaysOfCode Solved 165. Compare Version Numbers on LeetCode 🔗 🧠 Key Insight: Version strings are split by "." into multiple revisions. We compare each revision numerically (not lexicographically). Example: "1.01" = "1.1" (leading zeros don’t matter) ⚙️ Approach (Split + Compare): 1️⃣ Split both versions using "." 🔹 version1 → s1[] 🔹 version2 → s2[] 2️⃣ Traverse till max length of both arrays 3️⃣ For each index i: 🔹 num1 = i < s1.length ? parseInt(s1[i]) : 0 🔹 num2 = i < s2.length ? parseInt(s2[i]) : 0 4️⃣ Compare: 🔹 if num1 < num2 → return -1 🔹 if num1 > num2 → return 1 5️⃣ If all equal → return 0 ⏱️ Time Complexity: O(n + m) 📦 Space Complexity: O(n + m) (for split arrays) #100DaysOfCode #LeetCode #DSA #Strings #TwoPointers #Java #InterviewPrep #CodingJourney

Day 68 — LeetCode Progress Problem: Check if an array is “special” — meaning every pair of adjacent elements has different parity (one even, one odd). Required: Given an integer array, return true if no two adjacent elements have the same parity, otherwise return false. Idea: If two neighboring numbers are both even or both odd, the condition fails immediately. So just compare parity of adjacent elements throughout the array. Approach: Traverse the array starting from index 1 For each element: Compare it with the previous element If both have same parity → return false If the loop completes → return true Time Complexity: O(n) Space Complexity: O(1) Clean, simple, and a good reminder that not every problem needs something fancy — sometimes it’s just about spotting the pattern. #LeetCode #DSA #Java #Arrays #ProblemSolving #CodingJourney

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

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 67 of #LeetCode Challenge ✅ Problem Solved: Check If Two String Arrays are Equivalent 💡 What I learned today: • Learned how to compare two string arrays without joining them • Understood how to traverse multiple strings using pointers • Improved handling of indices across arrays and strings • Realized the importance of edge cases to avoid runtime errors 🧠 Approach: • Used four pointers to track positions in both arrays and strings • Compared characters one by one • Moved to the next string when current string ends • Ensured both arrays are fully traversed at the end 📊 Key Takeaway: Efficient solutions avoid extra space — comparing character by character is better than building new strings 🔥 Consistency + small improvements every day = big progress #Day67 #LeetCode #CodingJourney #DSA #Java #ProblemSolving #Consistency

💡 Day 41 of LeetCode Problem Solved! 🔧 🌟28. Find the Index of the First Occurrence in a String🌟 Task : • Given two strings needle and haystack, return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Example 1: Input: haystack = "sadbutsad", needle = "sad" Output: 0 Explanation: "sad" occurs at index 0 and 6. The first occurrence is at index 0, so we return 0. Example 2: Input: haystack = "leetcode", needle = "leeto" Output: -1 Explanation: "leeto" did not occur in "leetcode", so we return -1. #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

🚀 𝐃𝐚𝐲 89/100 – 𝐈𝐬 𝐒𝐮𝐛𝐬𝐞𝐪𝐮𝐞𝐧𝐜𝐞 Today’s problem was 𝐈𝐬 𝐒𝐮𝐛𝐬𝐞𝐪𝐮𝐞𝐧𝐜𝐞 — a simple yet fundamental problem to understand two-pointer technique. 🔍 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠: We can efficiently check if one string is a subsequence of another using two pointers. 💡 𝐂𝐨𝐫𝐞 𝐈𝐝𝐞𝐚: Traverse both strings Move pointer of s only when characters match Always move pointer of t  𝐖𝐡𝐲 𝐢𝐭 𝐰𝐨𝐫𝐤𝐬? We try to match characters of s in order within t without disturbing sequence. ⚡ 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡: Initialize two pointers i (for s) and j (for t) If characters match → move both Else → move only j If i reaches end of s → subsequence exists ⏱️ 𝐓𝐢𝐦𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(𝐧) 📦 𝐒𝐩𝐚𝐜𝐞 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲: 𝐎(1) #Day89 #100DaysOfCode #Java #DSA #LeetCode #TwoPointers #CodingJourney

Taking the Two-Pointer technique to LeetCode! Today's challenge: Remove Duplicates from a Sorted Array. After practicing the Two-Pointer pattern for reversing arrays and swapping 0s and 1s, I applied it to a classic LeetCode problem. The challenge? Remove duplicates from an array in-place with O(1) extra memory. Because the array is already sorted, all duplicates are grouped together. • The Slow Pointer (The Writer): Keeps track of where the next unique element should be placed. • The Fast Pointer (The Reader): Scans ahead through the array to find new, unique numbers. • The Logic: If the Reader finds a duplicate, it just skips it. But the moment the Reader finds a new number, the Writer records it at the front of the array and steps forward. #DSA #Java #LeetCode #RemoveDuplicate