Competitive-Programming | My solutions from different contests and online judges
kandi X-RAY | Competitive-Programming Summary
kandi X-RAY | Competitive-Programming Summary
There are solutions from 2009 until now; The old ones doesn't have the solution explanation. Almost all of them in c++.
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QUESTION
I have using standard input() to read the string in the competitive programming but its throws NZEC error. The reason is the huge data set so we have to use I/O optimization. I referred to below link: https://www.geeksforgeeks.org/fast-i-o-for-competitive-programming-in-python/
...ANSWER
Answered 2021-Mar-07 at 09:34stdin helps to read input faster
QUESTION
let's say I have a BIT(Fenwick Tree) with non-negative values and I want to find the smallest index in it for given cumulative frequency in O(logN).
Now, I can do it O(log^2(N)) like this.
...ANSWER
Answered 2020-Jul-15 at 09:31Here is my implementation of a lower_bound
-like function for a Fenwick tree with 0-based indexing:
QUESTION
I have read this: https://www.topcoder.com/community/competitive-programming/tutorials/binary-search. I can't understand some parts==>
What we can call the main theorem states that binary search can be used if and only if for all x in S, p(x) implies p(y) for all y > x. This property is what we use when we discard the second half of the search space. It is equivalent to saying that ¬p(x) implies ¬p(y) for all y < x (the symbol ¬ denotes the logical not operator), which is what we use when we discard the first half of the search space.
But I think this condition does not hold when we want to find an element(checking for equality only) in an array and this condition only holds when we're trying to find Inequality for example when we're searching for an element greater or equal to our target value.
Example: We are finding 5 in this array.
...ANSWER
Answered 2019-Sep-08 at 15:02We CAN use that theorem when looking for an exact value, because we
only use it when discarding one half. If we are looking for say 5,
and we find say 6 in the middle, the we can discard the upper half,
because we now know (due to the theorem) that all items in there are > 5
Also notice, that if we have a sorted sequence, and want to find any element
that satisfies an inequality, looking at the end elements is enough.
QUESTION
I am trying to develop a good intuition of Dynamic Programming questions but I am not able to understand a particular aspect of the questions.
I will take an example of Coin Change problem as provided on leetcode https://leetcode.com/problems/coin-change/
In many tutorials, a bottoms-up approach is mentioned such as in this one - https://www.topcoder.com/community/competitive-programming/tutorials/dynamic-programming-from-novice-to-advanced/
In this approach, we start from the optimal solution and then build up the array towards our solution. Example - We find the optimal solution of finding sum 2 then 3 and so on. In the end, we will have our solution. This is the approach which I understand.
I am having trouble wrapping my head around another approach of recursion with memoization. I have written a backtracking approach to the problem but am not sure how to apply memoization to it.
...ANSWER
Answered 2019-Jul-27 at 19:49Before finding recursive solution for DP, try identifying sub-problems of the problem in question. Since, each sub problem is same as parent problem and same algorithm will be applied.
Let's take the example for coin change where you are given list of denominations, d[] and sum, S and we need to find minimum number of denominations, count (of denominations) for summing up to S. If we want to define a solution (method) for finding count that would be int findMinDenom(int[] d, int S). At this point we don't know what implementation it would be but we do know what parameters are required for the problem which is d and S.
Keeping in mind that sub-problems would also have same solution but with lower sum. Hence, we try to implement in such way that findMinDenom solves for each of sub-problems. This will lead us to a recursive solution where we call same method with lower sum, s.
QUESTION
Fast I/O recommends the use of following two line during programming competitions to speed up IO:
...ANSWER
Answered 2019-Jul-05 at 21:51how do I ensure that the buffer doesn't get overflowed,
The output buffer doesn't "overflow". When it gets full, it is automatically flushed, i.e. its contents are written out and its length is reset to 0. This is the case whether cin
/ cout
are tied or not.
cin and cout work properly without blocking
You normally want operations on cin
/ cout
to block. But again, blocking vs. non-blocking I/O has nothing to do with whether cin
/ cout
are tied.
and buffer gets flushed properly when I am not using std::endl. Does the use of "\n" automatically handles it?
Outputting '\n'
only flushes the buffer if the stream is in line-buffered mode. cout
is automatically put in line-buffered mode if output goes to a terminal; otherwise it is block buffered (i.e. it only gets flushed when it runs full).
In a programming competition cout
usually goes to a pipe or log file, so it will be block buffered and '\n'
doesn't cause a flush. However, in that situation it also doesn't matter whether prompts are displayed before input is read (which is the normal use case for tied cin
/ cout
). Just make sure you produce the right output and let the I/O library worry about buffering. The buffer is automatically flushed when it runs full, when the stream is closed, and when your program exits. No output is lost (unless your program crashes, but then you have other things to worry about).
QUESTION
I read in geeksforgeeks and added the template to it so that it can work for every integer data types such as long
, long long
and int
.
ANSWER
Answered 2019-Jun-30 at 14:53the output is shown as 0 and not the input number.
the test in
QUESTION
I was going though this article and it explains
It is also possible to iterate over all the subsets of a particular subset (represented by a bit pattern), provided that you don’t mind visiting them in reverse order (if this is problematic, put them in a list as they’re generated, then walk the list backwards). The trick is similar to that for finding the lowest bit in a number. If we subtract 1 from a subset, then the lowest set element is cleared, and every lower element is set. However, we only want to set those lower elements that are in the superset. So the iteration step is just i = (i - 1) & superset.
I'm not able to understand it despite re-reading several times. Could someone explain with some example?
...ANSWER
Answered 2018-Oct-16 at 09:38If we have some set represented as a bitmask, for example if we have the universe:
QUESTION
http://codeforces.com/contest/462/problem/A that is a problem and I want to solve it by c++ can some one explain to me what he want here specially at the line of the question
and that is the answer but I can`t understand it https://github.com/ahmedalbarbary/Competitive-programming/blob/master/Codeforces%20-%20not%20mine%20codes/problem-%20462A%20-%20Appleman%20and%20Easy%20Task
thanks for helping me
...ANSWER
Answered 2017-Jul-25 at 11:53You have to count for every cell [i,j]
how many of cell [i,j-1], cell[i,j+1], cell[i-1,j], cell[i+1,j] (i.e. the adjacent cells, i.e. neighbour cells) contain an 'o'
.
If count is an even number (for every cell [i,j]) the result is "yes", else "no". (Thus, the test may be finished when the first odd count is detected.) Thereby, 0 is counted as even number as well (of course).
The possible issue:
For border cells, some of the tests have to be skipped to prevent out-of-bound access.
The solution uses a trick for this: It stores input beginning from indices [1,1] instead of indices [0,0]. This leaves "unused" border cells around the actual input matrix. The advance: no tests for cells to skip are necessary.
As the memory is filled with '*'
before the unused border cells will not have any negative effect for counting.
Why the trick is used:
The platform will hopefully perform read access to border cells and counting faster than the check (for every cell) whether cells have to be skipped (to prevent out-of-bound access).
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