This week I learned.. Hey, <spoiler summary="Hamming Code"> Hamming code is a 7 digit long code in which digit d0-d4 represent the outermost or in dependent cells and d5-d7 represent the dependent cells. It has a unique property that if you change any digit in a valid hamming code you could find out which digit you changed. b5=(b2+b3+b4)%2 b6=(b1+b3+b4)%2 and b7=b7=(b1+b2+b4)%2 Now suppose if b2 was changed you know b2 was changed because it is independent, but suppose if b5 is changed then you can know which unique(in dependent cell) was changed. </spoiler> <spoiler summary="Topological sorting"> Question alien sort on topological sorting. The question is follows, you are given a set of strings sorted by some manner, you have to find by which manner they are sorted. As we know that if things are sorted so obviously arr[0]<arr[1], so we can check where these two strings differ, and create a dependency based on it, because arr[0] was before arr[1] for a reason, we check what made ar...

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Is my logic correct for the problem related to Kadanes's algorithm? The original problem is : [problem:1373D] <spoiler summary="My Approach"> First we only need to take subarrays with even size as it will change index parity. Now, we take sum of numbers in eve indexes and odd indexes and if ever sum in eve indexes go above odd indexes we can safely discard the subarray as it will only cause a decrease in the total sum. Now as every element has two neighbors except the boundary elements we need to check for both neighbors. </spoiler> Now the approach was correct if we have to reverse only one subarray. But what if, we can reverse a number of subarray, suppose K. How would that be solved? I came up with the following approach, is it correct? <spoiler summary="Approach"> As we can exclude some subarrays from making our answer smaller, we would exclude the subarray with: sum of eve elements > sum of odd elements We can make an array which will contain: arr[2*i+1]-arr[2*i] that is, the difference of consecutive odd and even index...

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Problem regarding Codeforces time limit Hey, I tried solving problem [problem:96B] and it gave me very strange TLE. My submission [submission:207541848] got TLE at n=10^9, but the same code passes the same test cases on the codeforces custom test compiler with a time of 1300ms (approximated). As the time limit is 2second, Why am i getting TLE on main test?   Now the rest of the solution is right and fits under the time limit. I just made a special case with n=10^9 and passed the main test. [submission:207553081] Please help me figuring out what's happening wrong. Thank you white_square

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Why I am still newbie with answers. <spoiler summary="Background"> I have solved around 500 problems on various platform and given around 70 contests(all websites) from the day I started my competitive programming journey which was around 8 months ago </spoiler> After doing so much practice I am still not able to turn green, so I asked myself what I am doing wrong and so here is the list of answers I found. I don't sit seriously for the contest for whole 2 hours, after just solving and apparently debugging the first question my mind gives up and I cannot think straight for the 2nd question. I try to concentrate and somehow after several wrong answers on pretest 2 and 3, I sometimes solve question 2. Now to become pupil either I should solve 2 questions fast or solve question 3, depending on the type of contest. I neither solve question 2 fast nor solve question 3. As a matter of fact I have never read question 3 in any contest(div2 +) , even after solving question 2. My rate of up solving questions is around...

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Need help with B-181 Question:- [problem:181B] <spoiler summary="Question"> You are given n points on a plane. All points are different. Find the number of different groups of three points (A, B, C) such that point B is the middle of segment AC.The groups of three points are considered unordered, that is, if point B is the middle of segment AC, then groups (A, B, C) and (C, B, A) are considered the same. 3 ≤ n ≤ 3000 - 1000 ≤ xi, yi ≤ 1000 </spoiler> Solution that got accepted-: [submission:178183074] Solution that got rejected-: [submission:178183367] <spoiler summary="Other solution using same concept"> [submission:178183562] [submission:178182133] [submission:178145749] </spoiler> Summary: The logic of both code is same, I've tried using HashMap and HashSet but the solution does not work. *I do not understand why don't the code using HasHmap/HashSet work.* Any help is appreciated. Thank you.

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