Joined May 2024
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Solved 13 Oct. #LeetCode problem of the day: 632. Smallest Range Covering Elements from K Lists in #Java using Priority Queue Let's dive into the approach to find the smallest range that covers elements from each sorted list! (1/9)
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I applied to 50 Jobs, What’s Wrong with my Resume? (Freelancer looking for Backend Intern Role)
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Replying to @snitchindia
@snitchindia This is ridiculous! Ordered 5 items worth ₹5804, returned all & paid ₹300 return charges - refund should be ₹5504. But you gave only ₹4911! Why? Because you refunded a ₹1007 shirt twice (₹932 after charges) instead of refunding my ₹1600 shirt.
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Your support is a joke - refused to check, called it “cancellation,” & ended the chat. This isn’t just about money - it’s about the wasted time, frustration, and sheer inconvenience you caused. Is this how you treat customers?
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Solved 14 Oct. #LeetCode problem of the day: 2530. Maximal Score After Applying K Operations in #Java using Priority Queue Let's dive into the approach to break down overlapping intervals into non-overlapping groups. (1/9)
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Time Complexity: The overall complexity is O(k log n), where n is the size of the array. This comes from the cost of heap operations for each of the k operations. (8/9)
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Thinking Process: By leveraging a max-heap to always grab the largest number and minimizing it after each operation, we ensure that the solution remains optimal within the given constraints. (9/9)
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Priority Queue Approach: We can maintain a priority queue to keep track of the smallest elements across all lists. Initially, we insert the first element of each list into the queue. (3/9)
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Time Complexity: The overall complexity is O(n log k), where n is the total number of elements across all lists, and k is the number of lists due to maintaining the priority queue. (8/9)
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Thinking Process: By leveraging the sorted order of each list and using a priority queue to track the smallest and largest elements, we efficiently narrow down the smallest possible range! (9/9)
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