Kth Largest Element in Array with PriorityQueue

🔹 Day 85 – #100DaysOfLeetCode Problem: 3010. Divide an Array Into Subarrays With Minimum Cost I Difficulty: Easy Key Insight: The cost of a subarray depends only on its first element. Since the first subarray always starts at index 0, the problem reduces to selecting the two smallest possible starting elements from the remaining array. Approach: Fix the first subarray cost as nums[0] Find the smallest and second smallest values in nums[1…n-1] Add them to get the minimum total cost Time Complexity: O(n) Space Complexity: O(1) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

🧠 Day 76 of #100DaysOfLeetCode 📌 Problem: Combination Sum II 📊 Difficulty: Medium 💡 Key Insight: Sorting the array helps handle duplicates. While backtracking, duplicates are skipped only in the “not pick” path to ensure unique combinations. 🛠️ Approach: Sort the candidates array Use backtracking to explore combinations Each number can be used once → move to index + 1 Skip duplicate elements to avoid repeated combinations ⏱️ Complexity: Time: O(2^n) Space: O(n) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

🔥 Day 82 of #100DaysOfLeetCode ✅ Problem: Number of Longest Increasing Subsequence ✅ Difficulty: Medium 💡 Key Insight: Instead of tracking only the LIS length, we also maintain the number of ways each LIS can be formed. For every element, we check all previous smaller elements and update both length and count dynamically. 🛠 Approach: Use two arrays: dp[i] → Length of LIS ending at index i cnt[i] → Number of LIS ending at index i If a longer subsequence is found → update length and copy count. If another subsequence of the same length is found → add the counts. Finally, sum counts for indices having the maximum LIS length. ⏱ Complexity: Time: O(n²) Space: O(n) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

#day280 of #1001daysofcode problem statement (0085): Maximal Rectangle. For each row, I converted the matrix into a histogram of consecutive 1s and didn't change the Largest Rectangle in Histogram code which i uploaded earlier. Key takeaway: -> Break problems into familiar subproblems   -> Reusability of concepts matters more than memorizing solutions   -> In histogram-based solutions, resetting height on '0' is crucial   #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode #1001daysofcode

🚀 Day 147 | Frequency Counting Across Multiple Strings Today’s problem was a clean exercise in comparing character frequencies across multiple inputs. 🧩 Problem Solved: 1002. Find Common Characters 🔍 Approach: • Initialized a frequency array using the first word. • For each subsequent word, counted character frequencies and kept the minimum count for each character. • Constructed the result by adding each character as many times as its final minimum frequency. ✨ Key Insight: Using minimum frequency across all strings ensures only truly common characters are included. 📚 Topics: Hashing · Strings · Frequency Counting 💻 Platform: LeetCode #LeetCode #DSA #ProblemSolving #DailyCoding #Strings #Hashing #Java #Consistency

🌡️ Daily Temperatures – Stack Based Problem (LeetCode 739) Solved Daily Temperatures using Monotonic Stack approach. Learned how to efficiently find the next greater element in future days and calculate waiting days in O(n) time. 📌 Key Concepts: Monotonic Stack Next Greater Element Pattern Time Optimization (O(n) vs O(n²)) #DSA #LeetCode #MonotonicStack #Java #ProblemSolving #CodingJourney

day 92/100 #leetcodegrinding The number of rotations equals the index of the minimum element in the array. Using binary search, we can find this efficiently in O(log n) time. 🧠 What I focused on: Identifying the pivot (minimum element) Handling rotated vs non-rotated cases Applying binary search with correct boundaries Writing a clean and efficient Java solution This problem really strengthens understanding of rotated arrays and search logic. Keeping the streak alive 🚀 #DSA #ProblemSolving #Java #BinarySearch #Arrays #LeetCode #CodingPractice #DailyLearning

🔹 Day 97 – LeetCode Practice 📌 Problem: Divide Array Into Equal Pairs (LeetCode #2206) 📊 Difficulty: Easy 🧠 Problem Overview: You’re given an integer array containing 2n elements. The goal is to check whether the array can be divided into n pairs such that: Every element is used exactly once Both elements in each pair are equal ✅ Approach Used: Sorted the array to bring identical elements together. Traversed the array while counting occurrences of each number. Verified that every number appears an even number of times, ensuring valid pairs. 📈 Submission Results: Status: Accepted ✅ Runtime: 8 ms Memory Usage: 46.94 MB 💡 Reflection: This problem is a great reminder that sorting can simplify pairing logic significantly. Once the array is ordered, validating pairs becomes straightforward and efficient. #LeetCode #ProblemSolving #Arrays #Java #DSA #CodingPractice #Consistency

📅 Day 74 of #100DaysOfLeetCode 🧩 Problem: 3315. Construct the Minimum Bitwise Array II 📊 Difficulty: Medium 🧠 Key Insight For each prime number nums[i], we need to find the minimum integer x such that: x | (x + 1) == nums[i] x | (x + 1) is always odd Therefore, if nums[i] is even, no valid x exists → answer is -1 For odd nums[i], the minimum x can be formed by flipping the lowest 0-bit in nums[i] and setting all lower bits to 1 ⚙️ Approach Initialize the answer array For each number nums[i]: If nums[i] is even → assign -1 Otherwise: Find the lowest bit position where nums[i] has 0 Clear that bit Set all lower bits to 1 Store the resulting value as the minimum valid x ⏱ Complexity Time: O(n) Space: O(1) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

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