Java Binary Search Peak Element Solution

Day 77 of #100DaysOfLeetCode 💻✅ Solved #153. Find Minimum in Rotated Sorted Array problem in Java. Approach: • Used Binary Search to find minimum element • Compared mid with right to decide direction • Reduced search space until pointers converged Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 43.62 MB (Beats 78.18% submissions) Key Learning: ✓ Strengthened Binary Search on rotated arrays ✓ Learned how to identify unsorted portion efficiently ✓ Improved optimization over linear search Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Day 92 of #100DaysOfLeetCode 💻✅ Solved #739. Daily Temperatures problem in Java. Approach: • Used Monotonic Stack to track indices • Compared current temperature with stack top • Updated result when a warmer day was found • Stored indices for future comparisons Performance: ✓ Runtime: 60 ms (Beats 79.01% submissions) 🚀 ✓ Memory: 107.89 MB (Beats 17.53% submissions) Key Learning: ✓ Learned Monotonic Stack technique ✓ Improved handling of next greater element problems ✓ Strengthened stack-based problem solving Learning one problem every single day 🚀 #Java #LeetCode #DSA #Stack #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Day 69 of #100DaysOfLeetCode 💻✅ Solved #349. Intersection of Two Arrays problem in Java. Approach: • Iterated through each element of the first array • Checked if the element exists in the second array • Used a temporary array to store intersection elements • Ensured no duplicates by checking already added elements • Copied the result into a final array of correct size Performance: ✓ Runtime: 4 ms (Beats 35.60%) ✓ Memory: 44.41 MB (Beats 96.71%) Key Learning: ✓ Practiced array traversal using nested loops ✓ Learned how to handle duplicates manually ✓ Improved understanding of set-like operations without using extra data structures Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Day 48 of My #LeetCode Journey Today’s problem: 2615. Sum of Distances 💡 Key Idea: Instead of calculating distances between equal elements using brute force (O(n²)), I used: HashMap to group indices of same values Prefix Sum to efficiently compute distances This reduced the complexity to O(n) 🔥 🧠 What I Learned: How prefix sums can optimize distance calculations Efficient handling of repeated elements Writing clean and optimized code using Java ⚡ Approach: Store indices of each number Use prefix sums to calculate left & right distances Combine both to get final answer 📈 Time Complexity: O(n) 📦 Space Complexity: O(n) Consistency is key. Small progress every day leads to big results 💪 #Day48 #Java #FullStackDeveloper #CodingJourney #100DaysOfCode #DSA #LeetCode

🔥 𝗗𝗮𝘆 𝟵𝟰/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟭𝟱𝟯𝟵. 𝗞𝘁𝗵 𝗠𝗶𝘀𝘀𝗶𝗻𝗴 𝗣𝗼𝘀𝗶𝘁𝗶𝘃𝗲 𝗡𝘂𝗺𝗯𝗲𝗿 | 🟢 Easy | Java Marked as Easy — but the optimal solution is pure binary search brilliance. 🎯 🔍 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 Given a sorted array, find the kth missing positive integer. Linear scan works — but can we do O(log n)? 💡 𝗧𝗵𝗲 𝗞𝗲𝘆 𝗜𝗻𝘀𝗶𝗴𝗵𝘁 At index i, the value arr[i] should be i+1 in a complete sequence. So missing numbers before arr[i] = arr[i] - 1 - i This lets us binary search on the count of missing numbers! ⚡ 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 ✅ If arr[mid] - 1 - mid < k → not enough missing numbers yet, go right ✅ Else → too many missing, go left ✅ After the loop, left + k gives the exact answer 𝗪𝗵𝘆 𝗹𝗲𝗳𝘁 + 𝗸? After binary search, left is the index where the kth missing number falls beyond. left numbers exist in the array before that point, so the answer is left + k. No extra passes needed. ✨ 📊 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 ⏱ Time: O(log n) — vs O(n) linear scan 📦 Space: O(1) This is a perfect example of binary searching on a derived condition, not just a value. A real upgrade from the naive approach. 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gVYcjNS6 𝟲 𝗺𝗼𝗿𝗲 𝗱𝗮𝘆𝘀. 𝗧𝗵𝗲 𝗳𝗶𝗻𝗶𝘀𝗵 𝗹𝗶𝗻𝗲 𝗶𝘀 𝗿𝗶𝗴𝗵𝘁 𝘁𝗵𝗲𝗿𝗲! 🏁 #LeetCode #Day94of100 #100DaysOfCode #Java #DSA #BinarySearch #Arrays #CodingChallenge #Programming

Day 90 of #100DaysOfLeetCode 💻✅ Solved #42. Trapping Rain Water problem in Java. Approach: • Used Two Pointer technique • Maintained leftMax and rightMax boundaries • Calculated trapped water based on smaller side • Accumulated water while traversing Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 47.70 MB (Beats 72.36% submissions) Key Learning: ✓ Mastered Two Pointer approach for complex problems ✓ Learned optimal water trapping strategy ✓ Improved understanding of boundary-based calculations Learning one problem every single day 🚀 #Java #LeetCode #DSA #TwoPointers #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

🔥 𝗗𝗮𝘆 𝟵𝟯/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟭𝟲𝟬𝟴. 𝗦𝗽𝗲𝗰𝗶𝗮𝗹 𝗔𝗿𝗿𝗮𝘆 𝗪𝗶𝘁𝗵 𝗫 𝗘𝗹𝗲𝗺𝗲𝗻𝘁𝘀 𝗚𝗿𝗲𝗮𝘁𝗲𝗿 𝗧𝗵𝗮𝗻 𝗼𝗿 𝗘𝗾𝘂𝗮𝗹 𝗫 | 🟢 Easy | Java A self-referential condition — x elements must be ≥ x. Elegant problem, elegant solution. 🎯 🔍 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 Find x such that exactly x elements in the array are ≥ x. Return -1 if no such x exists. ⚡ 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗙𝗿𝗲𝗾𝘂𝗲𝗻𝗰𝘆 𝗖𝗼𝘂𝗻𝘁 + 𝗦𝘂𝗳𝗳𝗶𝘅 𝗦𝘂𝗺 ✅ Cap all values at n (array length) — anything larger contributes the same way ✅ Build a frequency count array of size n+1 ✅ Traverse from right to left, accumulating a running suffix sum ✅ When suffix sum == current index i → x = i is the answer! 💡 𝗪𝗵𝘆 𝗰𝗮𝗽 𝗮𝘁 𝗻? x can never exceed n (can't have more elements than the array size). So values above n are equivalent — capping them avoids index overflow and keeps the logic clean. 📊 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 ⏱ Time: O(n) — two passes 📦 Space: O(n) — frequency array No sorting. No binary search. Just a clever frequency count + suffix accumulation. Sometimes the cleanest approach is right under your nose. 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gm2c4-6x 𝟳 𝗺𝗼𝗿𝗲 𝗱𝗮𝘆𝘀. 𝗦𝗼 𝗰𝗹𝗼𝘀𝗲 𝘁𝗼 𝟭𝟬𝟬! 💪 #LeetCode #Day93of100 #100DaysOfCode #Java #DSA #Arrays #FrequencyCount #CodingChallenge #Programming

Day 82 of #100DaysOfLeetCode 💻✅ Solved #11. Container With Most Water problem in Java. Approach: • Used Two Pointer technique • Calculated area using min height and width • Moved pointer with smaller height inward • Tracked maximum area throughout Performance: ✓ Runtime: 5 ms (Beats 83.17% submissions) ✓ Memory: 77.35 MB (Beats 53.65% submissions) Key Learning: ✓ Mastered Two Pointer optimization technique ✓ Learned efficient area calculation strategy ✓ Improved decision-making for pointer movement Learning one problem every single day 🚀 #Java #LeetCode #DSA #TwoPointers #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

📘 DSA Journey — Day 33 Today’s focus: Binary Search for next greater element. Problem solved: • Find Smallest Letter Greater Than Target (LeetCode 744) Concepts used: • Binary Search • Upper bound concept • Circular handling Key takeaway: The goal is to find the smallest character strictly greater than a given target in a sorted array. This is a classic upper bound problem: We use binary search to find the first element greater than the target. During search: • If letters[mid] > target, store it as a possible answer and move left • Else move right An important edge case: If no character is greater than the target, we return the first element (circular behavior). Continuing to strengthen binary search patterns and problem-solving consistency. #DSA #Java #LeetCode #CodingJourney

📘 DSA Journey — Day 28 Today’s focus: Binary Search for minimum in rotated arrays. Problem solved: • Find Minimum in Rotated Sorted Array (LeetCode 153) Concepts used: • Binary Search • Identifying unsorted half • Search space reduction Key takeaway: The goal is to find the minimum element in a rotated sorted array. Using binary search, we compare the mid element with the rightmost element: • If nums[mid] > nums[right] → minimum lies in the right half • Else → minimum lies in the left half (including mid) This works because the rotation creates one unsorted region, and the minimum always lies in that region. By narrowing the search space each time, we achieve O(log n) time complexity. This problem highlights how slight modifications in array structure still allow binary search to work efficiently with the right observations. Continuing to strengthen binary search patterns and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney