Two Sum LeetCode Solution in Java

🚀 Day 4 of #100DaysOfCode Solved Single Element in a Sorted Array on LeetCode using Binary Search ⚡ 🧠 Key insight: In a sorted array where every element appears twice except one, index parity (even/odd) helps decide which half to search. ⚙️ Approach: 🔹Apply binary search 🔹Compare mid with its adjacent element 🔹Use index parity to move left or right 🔹Narrow down until the single element is found ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 13 of Daily DSA 🚀 Solved LeetCode 33: Search in Rotated Sorted Array ✅ Approach: Used modified Binary Search to handle rotation efficiently. At each step, identified the sorted half of the array and checked whether the target lies within that range. Based on this, adjusted the search boundaries to maintain logarithmic performance. This problem is a great example of how classic binary search logic can be extended with small observations. ⏱ Complexity: • Time: O(log n) • Space: O(1) 📊 LeetCode Stats: • Runtime: 0 ms (Beats 100%) ⚡ • Memory: 43.97 MB (Beats 45.43%) #DSA #LeetCode #Java #BinarySearch #ProblemSolving #DailyCoding #Consistency

Day 7 of Daily DSA 🚀 Solved LeetCode 167: Two Sum II – Input Array Is Sorted Approach: Used the two-pointer technique leveraging the sorted nature of the array. Moved pointers inward based on the comparison with the target to find the answer in one pass. Complexity: • Time: O(n) • Space: O(1) (constant extra space) LeetCode Stats: • Runtime: 2 ms (Beats 96.10%) • Memory: 48.54 MB (Beats 38.87%) This problem highlights how understanding input constraints can turn a brute-force solution into an optimal one. #DSA #LeetCode #Java #TwoPointers #ProblemSolving #Consistency

🚀 Day 99 of #100DaysOfLeetCode 📌 Problem: 67. Add Binary 📊 Difficulty: Easy 💡 Key Insight: Binary addition follows the same rules as decimal addition — add digits from right to left while maintaining a carry. The only difference is everything is in base 2. 🛠️ Approach: Use two pointers starting from the end of both strings Add corresponding bits along with a carry Append the result bit (sum % 2) Update carry (sum / 2) Reverse the final string to get the answer ⏱️ Complexity: Time: O(n) Space: O(n) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

Day 60/100 – LeetCode Challenge ✅ Problem: #67 Add Binary Difficulty: Easy Language: Java Approach: Reverse Iteration with Carry Time Complexity: O(max(n, m)) Space Complexity: O(max(n, m)) Key Insight: Binary addition follows same rules as decimal: Sum bits + carry Result bit = sum % 2 New carry = sum / 2 Solution Brief: Iterated from rightmost bits of both strings. Tracked carry and built result from right to left using StringBuilder. Reversed final string for correct order. #LeetCode #Day60 #100DaysOfCode #Binary #Java #Algorithm #CodingChallenge #ProblemSolving #AddBinary #EasyProblem #StringManipulation #BitManipulation #DSA

🚀 Day 13 of #100DaysOfCode Solved Remove Linked List Elements on LeetCode 🔗 🧠 Key insight: While traversing a linked list, careful pointer updates are needed—especially when the head node itself matches the value to be removed. ⚙️ Approach: 🔹Handle cases where the head contains the target value 🔹Traverse the list using a pointer 🔹Skip nodes whose value matches the given target by adjusting next pointers ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 7/100 – LeetCode Challenge Problem: Palindrome Number Today’s problem focused on number manipulation without converting the integer into a string. Approach: If number is negative → return false Reverse the digits using modulo (% 10) and division (/ 10) Compare reversed number with original number Logic Used: Extract last digit: rem = x % 10 Build reversed number: rev = rev * 10 + rem Remove last digit: x = x / 10 Complexity: Time: O(log₁₀ n) Space: O(1) Key takeaway: Problems involving digits can often be solved mathematically without string conversion. #100DaysOfCode #LeetCode #DSA #Java #ProblemSolving #CodingJourney

🚀 Day 18 of #100DaysOfCode Solved Remove Duplicates from Sorted List II on LeetCode 🔗 🧠 Key insight: When duplicates appear in a sorted linked list, we need to remove all occurrences, not just one. Using a dummy (sentinel) node makes it easier to handle cases where duplicates start at the head. ⚙️ Approach: 🔹Create a dummy node pointing to the head 🔹Traverse the list with two pointers 🔹If a duplicate sequence is found, skip the entire block 🔹Otherwise, move forward normally ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 15 of Daily DSA 🚀 Solved LeetCode 162: Find Peak Element ✅ Approach: Used Binary Search to efficiently find a peak element. At each step, compared the middle element with its neighbors to decide which side must contain a peak. This works because at least one peak always exists in the array. ⏱ Complexity: • Time: O(log n) — binary search on array • Space: O(1) — no extra space used 📊 LeetCode Stats: • Runtime: 0 ms (Beats 100%) ⚡ • Memory: 44.20 MB (Beats 75.11%) A great example of how binary search can be applied beyond just sorted arrays. #DSA #LeetCode #Java #BinarySearch #ProblemSolving #DailyCoding #Consistency

Solved one of the classic Binary Tree problems on LeetCode: Validate Binary Search Tree. At first glance, the problem seems straightforward just check if left <root< right. But the real challenge is ensuring that every node respects the global BST constraints, not just its immediate parent. Problems like this remind me how important it is to think beyond local conditions and consider global invariants in tree structures. #LeetCode #DataStructures #Java #BinaryTree #DSA #ProblemSolving #CodingJourney