🚀 Day 87 – #100DaysOfCode Today I solved a problem on Maximum Number of String Pairs using HashSet. 🔹 Problem: Given an array of strings, find the number of pairs where one string is the reverse of another. 🔹 Approach I Used: Traverse through each word in the array. Reverse the current string using StringBuilder. Check if the reversed string already exists in a HashSet. If it exists → we found a pair ✅ Otherwise → store the current word in the set. 💡 Key Idea: Using a HashSet allows constant time lookup O(1) to quickly check if the reverse string already appeared. ⚡ Time Complexity: O(n) ⚡ Space Complexity: O(n) #DSA #Java #CodingJourney #LeetCode #ProblemSolving #100DaysOfCode
Java Problem Solving: Maximum Number of String Pairs with HashSet
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🚀 Day 41/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 3427. Sum of Variable Length Subarrays Used a nested loop approach to calculate subarray sums for each index by dynamically determining the starting point using max(0, i - nums[i]). ⏱️ Time Complexity: O(n²) 📦 Space Complexity: O(1) Strengthening understanding of subarray problems and index-based range calculations. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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🚀 Day 46/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2315. Count Asterisks Used a boolean flag approach to track whether the current position is inside a pair of '|' and count only the valid '*' outside those sections. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) Strengthening understanding of state-based string traversal and conditional counting. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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🚀 Day 47/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 448. Find All Numbers Disappeared in an Array Used an in-place marking technique by treating indices as a hash map and marking visited numbers as negative to identify missing elements. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) (excluding output list) Strengthening understanding of array manipulation and in-place hashing tricks. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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Day 2 of my #365DaysCodingChallenge | CodeOjas Journey Today, I solved 3 interesting array-based problems using Java: 🔹 Unique Subarray Sum Pairs – Found pairs of subarrays with equal sums using hashing techniques. 🔹 Array Rotation – Rotated array efficiently using optimized reversal approach. 🔹 Shifted Array – Implemented element shifting with proper handling of wrap-around using modular arithmetic. 💡 Approach: Focused on breaking problems into smaller parts, using concepts like subarrays, indexing, and optimization techniques. ⚡ Complexity: Most solutions were optimized to O(n) time with minimal space. 📌 Takeaway: Strong understanding of array manipulation and problem breakdown is key to solving DSA problems efficiently. Building consistency and improving every day 💪🔥 📂 Code: [Your GitHub Link] #DSA #Java #CodeOjas #365DaysOfCode #CodingJourney #ProblemSolving
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𝗬𝗼𝘂 𝗵𝗮𝘃𝗲 𝗯𝗲𝗲𝗻 𝘂𝘀𝗶𝗻𝗴 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 𝘄𝗿𝗼𝗻𝗴. It is not just for sorted arrays. The real definition — "Eliminate HALF the search space with each decision." That one shift in thinking unlocks 20+ LeetCode problems. Swipe to see the template, live code, and 7 problems you can now solve with one pattern. 👇 Save this. 🔖 #DSA #BinarySearch #Java #LeetCode #CodingInterview #DebugWithPurpose
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🚀 Day 31/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 69. Sqrt(x) Used the Binary Search approach to find the integer square root without using built-in power functions. The search space is reduced by checking mid * mid against x. ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) Strengthening problem-solving skills with binary search patterns on answer space. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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💡 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
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🚀 Day 52/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 3174. Clear Digits Used a stack-like approach with StringBuilder to remove the closest non-digit character whenever a digit is encountered. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) Strengthening understanding of string processing and stack-based logic simulation. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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🚀 Day 48/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 1534. Count Good Triplets Used a brute-force triple nested loop to check all possible triplets and validate the given conditions using absolute differences. ⏱️ Time Complexity: O(n³) 📦 Space Complexity: O(1) Strengthening understanding of nested loop patterns and condition-based filtering. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency
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🚀 Solved an interesting problem today: “Minimum Stickers to Spell Word” At first, I tried a greedy approach… failed ❌ Then I realized this is actually a DP + memoization problem. 💡 Key Insight: Instead of trying all combinations blindly, we reduce the target string step by step. 🧠 Approach: - Convert each sticker into frequency map - Recursively try reducing target - Use memoization to avoid recomputation ⏱️ Time Complexity: Optimized significantly with caching 🔥 This problem improved my thinking about state reduction. #DSA #CodingInterview #Java #LeetCode
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