Vedant Rawale’s Post

Leetcode weekly contest 409: https://lnkd.in/gM-PF4yv Code: https://lnkd.in/gjah9GRy Question 4: We can separate the circle into alternating groups like: [blue, red, blue, red] red [red, blue, red, blue] ... And then, we treat them like ranges in the circle. query(1, size = n) = multiple(from n to max_size) - count( from n to max_size) * (n - 1); #wecommit100xshare

Hi connections, Got 3 correct submissions in today's LeetCode biweekly contest which could have been 4 but due to time constraints and pressure I missed it by a margin. Q1. Brute force: Checked if already present and if was ans^=nums[ i ]. Q2. 1. Created a vector occ and stored all indices where x ocurred in nums 2. Iterated through queries using for each loop and val = query<=occ.size()?occ[query-1]:-1; Q3. Data structure used map. Simply check if ball is already coloured if yes decrease count of oldCol and increase that of newCol. If not present add newCol to map and check if it already occured. Q4. I tried using a map so as to store the obstacles in increasing order and iterate through until *it-tmp>=len or it==st.end(). But some modification is needed in case where there are no first type of queries. Overall I felt the contest was somewhat on the easier side. Hopefully next time I can get all 4. Thanks🙃

🔄 Day 58 of #100DaysOfCodeChallenge 🔄 Today, I tackled the "Swap Nodes in Pairs" problem. The challenge was to swap every two adjacent nodes in a linked list and return the modified list. If the list has an odd number of nodes, the last node remains in its original position. To solve this, I used an iterative approach that allows for swapping nodes in pairs efficiently. Here’s a step-by-step breakdown of the approach: Initialize a Dummy Node: Create a dummy node that points to the head of the list. This helps handle edge cases, such as when the list starts with only one node. Use a prev pointer initialized to dummy to help manage the connections during the swap. Traverse the List: Traverse the list two nodes at a time. For each pair, adjust the pointers to swap the nodes: Point prev.next to the second node of the pair. Point the first node's next to the node after the pair. Point the second node's next to the first node. Update Pointers: Move the prev pointer to the end of the swapped pair to continue swapping the next pair. Return the Result: Return dummy.next as the new head of the modified list.

Nice feature to EventCatalog coming... the ability to do code diffs in code snippets. In this example the OrderPlaced event has changed from 0.1.0 to v0.1.1. This can show what changes were made, and let you document why 🤩

Clean Code tip: Avoid complex one-liners. Ternary operators might work for simple conditions: ✅ var examResult = score >= 50 ? "Passed" : "Failed"; But once you have a complex condition or nested logic, it gets harder to read, extend and debug: ❌ var discountCode = order.IsFirstOrder() || IsBlackFriday(date) ? (order.Amount > 50 || product.IsOnSale ? "BIGSAVE20" : "EASYOFFER10") : (customer.LoyaltyPoints > 1000 && customer.LastActive <= 365 ? "LOYALTY15" : "STANDARD5"); Don't try to minimize the number of lines. Instead, minimize the time needed to maintain your code.

Did you know with Sentry's event fingerprinting you can group errors with greater granularity? If your quering an API service the stack trace is generally the same even if the outgoing request is very different Add it to your Sentry.init beforeSend function.

Hi everyone, I am happy to share that I participated in LeetCode Weekly Contest 402 and solved all 4 questions in 54 minutes and got LCUS Rank 411. Intuition: 1 & 2) First determine how pairing can be done. In array for x if x%24 is 12 or 0 it can be paired with itself only, so nC2 for these case else for all cases x%24 can be paired with 24-(x%24). So freq(x%24) * freq(24-(x%24)) pairs, but in second case we are counting (a to b) and (b to a) as pairs so divide it by 2. Answer is: first case + second case. 3) Make a sorted pair vector of {x,freq(x)} of nums. DP problem where at each step you have to check pick score of x that is x*freq(x)+ maximum dp score of elements less than x-2. Take maximum of all dp results that is the answer. 4) Suppose you are given an array nums first calculate a dx array where dx[i] = nums[i]-nums[i-1] (i!=0), for peak dx[i] > 0 && dx[i+1] < 0. Make a segment tree of n length and put 1 if nums[i] is peak else put 0. Now every time you update just update the dx[i] & dx[i+1] & nums[i] as rest will be unaffected by update of nums[i] and check for dx[i-1 ... i+1] for new peak values and update them in segment tree. For query [a,b] query segment tree for [a+1, b-1] as first and last elements should not be counted for peak.

🐛🐛 Fun bug of the day 🐛🐛 Given a spreadsheet and multiple columns to check, how would you find the non-unique rows on the values in the given columns? And how would you scale this for sheets with dozens of columns and 10M+ rows? This seems relatively simple, but we started running into issues where the validation was consistently freezing on row 14,680,064 regardless of how large the sheet is. What could be going on here? (Hint: 14680064 = 0xE00000) Here’s our code (in Rust) and I’ll leave the solution in the comments:

LeetCode Weekly contest 400 Solved all 4 problems 😊 Problem A: Just count the numbers of entries and subtract the number of exits. Submission Link: https://lnkd.in/gBjbcvsU Problem B: Used a custom comparator for this one and sorted the given meetings using it (Custom comparator : Ascending order of start time and descending order of end time) Submission Link: https://lnkd.in/gCKPryvm Problem C: Keep track of the smallest and closest element to any star by using priority_queue and used a custom comparator to do this. Submission Link: https://lnkd.in/g6WG_fqM Problem D: Solved it using 2 approaches i). Keep track of the 'on' bits and 'off' bits and apply the sliding window logic using the fact that 'AND' of 2 numbers can not be greater than those 2 numbers. Submission Link: https://lnkd.in/gEFTAsvb ii) Store the 'AND' operations of elements in segment tree then find the optimal answer by fixing start point at ith position and getting end point by binary search. Submission Link: https://lnkd.in/gzPZgn9h

"Flip perspectives, find unity." cout<<”HELLO WORLD<<endl; You can get the explanation here -  https://lnkd.in/gpHRdtpJ A medium problem today Leetcode POTD – 1072 – Flip columns for maximum number of equal rows -  https://lnkd.in/ga6gMDYc We are given with mxn matrix with os or 1s We need to flip any number of columns to make maximum number of equal rows Let us now go with a small example [1 0], so to make the row equal we need to flip the second or first column, then we can get a row equal But remember we need to find the maximum number of rows that can be made equal NOT flips… So let us go with another example [1 0] [0 1] If we flip the second or first column you will get two equal rows… Can we find a way to store these rows in a data structure Let us find that way and approach Consider the previous example, if we store the first row as 0 and the second row also as 0 As both of them are storing the same that means 2.. so by flipping some columns we can get these rows as equal… Use a string for storing the rows equality And a hashmap to count the frequency of these strings.. You can find my explanation here -  https://lnkd.in/gpHRdtpJ