Check if Two String Arrays are Equivalent in Java

📝 Day 12/30 – LeetCode #344 (Reverse String) | Java Although reversing a string seems simple, the constraint here is to do it in-place without using extra memory. Using built-in functions would defeat the purpose of the problem. By applying a two-pointer approach, I swapped characters from both ends of the array until they met in the middle. This resulted in a clean O(n) solution with O(1) space. This problem reinforced how even simple tasks can test understanding of constraints and memory efficiency. #LeetCode #Java #DSA #TwoPointers #Strings #Consistency #LearningInPublic

🚀Java practice - Day 72 Completed! 👍 Problem: Check if All Characters Have Equal Number of Occurrences Language: Java Today’s problem focused on validating whether all characters in a string appear the same number of times. A simple concept, but it required careful frequency tracking and comparison.✨ #Day72 #Java #LeetCode #Strings #HashMap #ProblemSolving #DailyCoding #Consistency #100DaysOfCode

LeetCode Practice - 1047. Remove All Adjacent Duplicates In String ✅ Key idea Instead of using a real stack: 🔷We use a character array 🔷And an integer pointer (top) 🔷This behaves exactly like a stack. 🧩 What we maintain 🔷char[] stack → stores final characters 🔷top → points to the last inserted character 🔷top = -1 → stack is empty #LeetCode #Java #StringHandling #CodingPractice #ProblemSolving #DSA #DeveloperJourney #TechLearning

📝 Day 11/30 – LeetCode #26 (Remove Duplicates from Sorted Array) | Java A brute-force approach using extra space would work, but the constraint here is to modify the array in-place. Since the array is already sorted, duplicates always appear next to each other. Using a two-pointer approach, one pointer tracks the position of the last unique element while the other scans through the array. Whenever a new value is found, it is placed at the next valid position. This results in an O(n) solution with O(1) extra space. This problem reinforced how understanding input constraints can significantly simplify the solution. #LeetCode #Java #DSA #TwoPointers #Arrays #ProblemSolving #Consistency #LearningInPublic

DSA journey 🚀 📌 LeetCode #1464 – Maximum Product of Two Elements in an Array 💻 Language: Java 🔹 Approach: Traverse the array once to find the largest and second largest elements Update both values in a single loop Apply the formula: (max − 1) × (secondMax − 1) ⏱ Time Complexity: O(n) 🧩 Space Complexity: O(1) Small optimizations, big wins 💡 Consistency over perfection 💯 #DSA #Java #LeetCode #ProblemSolving #LearningInPublic #DSAWithedSlash 🚀

🔥Day 91 of #100DaysOfLeetCode Problem: 1653. Minimum Deletions to Make String Balanced Difficulty: Medium Key Insight: A balanced string must have all 'a's before all 'b's. Any 'b' before an 'a' creates a violation. Approach: Preprocess the string using: - prefixB[i]: number of 'b's strictly before index i - suffixA[i]: number of 'a's strictly after index i For every index i, treat it as a split point and compute: deletions = prefixB[i] + suffixA[i] The minimum over all splits is the answer. Time Complexity: O(n) Space Complexity: O(n) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

🚀 LeetCode Problem Solved: Two Sum II (Input Array is Sorted) Implemented an optimized solution in Java using the Two Pointer technique. Since the array is already sorted, we can efficiently adjust pointers based on the current sum and find the required pair without using extra space. ✅ Time Complexity: O(n) ✅ Space Complexity: O(1) Source Code : https://lnkd.in/ew2_GMw9 This problem is a great example of how understanding constraints (sorted input) helps in designing efficient solutions. #LeetCode #Java #DSA #TwoPointers #ProblemSolving #CodingJourney

#day284 of #1001daysofcode problem statement (2943): Maximize Area of Square Hole in Grid. The key was realizing that the largest square depends on the longest sequence of consecutive removed bars — both horizontally and vertically. Key takeaway: -> Sorting helps detect consecutive patterns easily   -> Consecutive gaps matter more than total removals   -> The smallest side decides the maximum square   #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode #1001daysofcode

LeetCode Practice - 896. Monotonic Array 🔷Logic Behind “Monotonic Array” (Java) ✔An array is called monotonic if it moves in only one direction: 📌Either it never decreases (increasing or equal) 📌Or it never increases (decreasing or equal) ✔So the idea is very simple: 📌Check if the array follows at least one of these directions. 🔍 Step 1 — Use two flags We take two boolean variables: boolean increasing = true; boolean decreasing = true; 🔄 Step 2 — Compare neighboring elements We loop through the array and compare every pair: What does this mean? If nums[i] > nums[i+1] → array is going down, so it cannot be increasing If nums[i] < nums[i+1] → array is going up, so it cannot be decreasing We keep eliminating possibilities. 🧪 Example 1 [1, 2, 2, 3] Comparisons: 1 ≤ 2 → ok for increasing 2 ≤ 2 → ok 2 ≤ 3 → ok No violation for increasing, so: increasing = true decreasing = false ➡ Result = true 🧪 Example 2 [1, 3, 2] Comparisons: 1 < 3 → not decreasing 3 > 2 → not increasing So: increasing = false decreasing = false ➡ Result = false ✅ Final decision ✔return increasing || decreasing; ✔If either increasing or decreasing is still true, the array is monotonic. #LeetCode #Java #CodingPractice #ProblemSolving #DSA #Array #DeveloperJourney #TechLearning

Day 13/100 – LeetCode Challenge 🚀 Problem: Binary Search Approach: Maintained left and right pointers Compared middle element with target Reduced the search space by half each step Time Complexity: O(log n) Space Complexity: O(1) Key takeaway: Binary search is the foundation for solving problems with logarithmic efficiency. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #InterviewPrep #100DaysOfLeetCode