Binary Search Solves Single Element in Sorted Array

🚀 Day 22 of #100DaysOfCode Solved 34. Find First and Last Position of Element in Sorted Array on LeetCode 🔍 🧠 Key insight: To find both boundaries of a target in a sorted array, a single binary search isn’t enough. Running two binary searches—one for the leftmost and one for the rightmost occurrence—gives an optimal solution. ⚙️ Approach: 🔹Perform a binary search to find the first occurrence 🔹Perform another binary search to find the last occurrence 🔹Adjust search boundaries after finding the target to expand left/right ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #CodingJourney

🚀 Day 11 of #100DaysOfCode Solved Find Minimum in Rotated Sorted Array on LeetCode using Binary Search 🔄 🧠 Key insight: In a rotated sorted array, at least one half is always sorted. By comparing the middle element with the right boundary, we can determine which half contains the minimum and safely discard the other. ⚙️ Approach: 🔹Use binary search on the array 🔹Compare nums[mid] with nums[right] 🔹If nums[mid] > nums[right] → minimum lies in the right half 🔹Else → minimum lies in the left half (including mid) ⏱️ Time Complexity: O(log n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 27/100 – LeetCode Challenge 🚀 Problem: Sort List Approach: Applied Merge Sort on the linked list Used slow and fast pointers to find the middle Recursively sorted both halves Merged the sorted halves Time Complexity: O(n log n) Space Complexity: O(log n) (recursion stack) Key takeaway: Merge sort is the most efficient sorting technique for linked lists, as it avoids random-access operations required by other algorithms. #LeetCode #100DaysOfCode #DSA #Java #LinkedList #MergeSort #ProblemSolving

🚀 Day 9 of #100DaysOfCode Solved Search in Rotated Sorted Array II on LeetCode using Modified Binary Search (Handling Duplicates) 🔎 🧠 Key insight: With duplicates present, it’s sometimes impossible to identify the sorted half directly. If nums[left] == nums[mid], we safely move left++ to shrink the search space. ⚙️ Approach: 🔹Apply binary search 🔹Handle duplicates by skipping equal boundary values 🔹Identify sorted half and check if target lies within it 🔹Continue search accordingly ⏱️ Time Complexity: O(log n) average, O(n) worst case (due to duplicates) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 3 of Daily DSA 🚀 Solved LeetCode 1: Two Sum Approach: Used a HashMap to store numbers with their indices. For each element, checked if the complement (target - current) already exists. Complexity: • Time: O(n) • Space: O(n) Performance: Runtime: 2 ms (Beats 99.15%) Memory: 47.34 MB Focusing on writing clean and efficient solutions before over-optimizing. Consistency > Intensity 💯 #DSA #LeetCode #Java #ProblemSolving #Consistency

🚀 #100DaysOfCode – Day 12 Solved Search in Rotated Sorted Array II on LeetCode 🔄🔍 🧠 Key insight: When duplicates are present, it’s not always possible to directly identify the sorted half. If nums[left] == nums[mid], we can safely move left++ to reduce the search space and continue. ⚙️ Approach: 🔹Use modified binary search 🔹Handle duplicates by skipping equal boundary elements 🔹Identify the sorted half 🔹Check if the target lies in that range and adjust pointers accordingly ⏱️ Time Complexity: Average: O(log n) Worst case (many duplicates): O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #CodingJourney

🚀 Day 24 of #100DaysOfCode Solved 167. Two Sum II – Input Array Is Sorted on LeetCode 🔢👉👈 🧠 Key insight: Because the array is already sorted, we can avoid hash maps and use a two-pointer approach to find the target sum efficiently. ⚙️ Approach: 🔹Initialize two pointers at the start and end of the array 🔹Calculate the sum of values at both pointers 🔹If the sum is too large → move the right pointer left 🔹If the sum is too small → move the left pointer right 🔹Stop when the target is found ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #TwoPointers #Java #ProblemSolving #LearningInPublic #CodingJourney

🚀 Day 82 of #100DaysOfCode 🔍 Problem Solved: Find Minimum in Rotated Sorted Array Today I worked on a binary search based problem where we need to find the minimum element in a rotated sorted array in O(log n) time. 💡 Key Insight: If the array is already sorted (nums[low] <= nums[high]), then the first element is the minimum. Otherwise: If left half is sorted → minimum must be in right half If right half is unsorted → minimum lies there By carefully adjusting low and high, we narrow down the search space efficiently. 🧠 What I practiced: Modified Binary Search Identifying sorted halves Avoiding unnecessary comparisons #Java #DSA #BinarySearch #LeetCode #ProblemSolving #CodingJourney

🚀 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 74 of #100DaysOfCode Today I solved Search in Rotated Sorted Array from LeetCode using a modified Binary Search approach. 🧠 Problem Insight: Even though the array is rotated, one half is always sorted. By identifying the sorted half, we can decide where to move — left or right. ⚡ Key Learning: Instead of thinking “array is rotated ”, Think → “One side is still sorted ” 💻 Approach: Use Binary Search Check which half is sorted Verify if target lies in that half Reduce search space accordingly ⏱ Complexity: Time: O(log n) Space: O(1) #Day74 #100DaysOfCode #DSA #BinarySearch #Java #CodingJourney