### GlydonNeedsDedication's blog

By GlydonNeedsDedication, history, 3 months ago,

for the Problem 1849C

The SOLUTION I wrote, it is having some out of bounds error in the below code portion.

    s += (char)('0' ^ '1' ^ s.back());
s = ("" + (char)('0' ^ '1' ^ s[0])) + s; // this line giving error 'out of bounds' though working fine in LOCAL
n = s.size();


I am not able detect the exact reason.

Same purpose If I try to do like below, its working as expected

// working as expected
s += (char)('0' ^ '1' ^ s.back());
reverse(all(s));
s += (char)('0' ^ '1' ^ s.back());
reverse(all(s));
n = s.size();


By GlydonNeedsDedication, history, 2 years ago,

As I didn't get the solution provided in the editorial of 1585D - Yet Another Sorting Problem and how it can be done with O(N)

I had come up with this logic O(NlogN):

1st-> If there is a duplicate element, we can sort the whole array with the help of those two using the 3-cycle technique so always "YES"

2nd-> if there is no duplicate present, simply take one element at a time and place that element into its perfect position (the position where it should be if it was sorted) using the 3 cycles technique.
When 2 elements are remaining and they are not sorted it means they can't be sorted as you require 3 elements to sort!

There might be better solutions available but I guess it's easy to understand and observe also.

By GlydonNeedsDedication, history, 2 years ago,

I saw this Solution of Problem B by maroonrk using Topological sort in contest Round #758 (Div.1 + Div. 2)
Can anybody explains the intuition please?

By GlydonNeedsDedication, history, 2 years ago,

For a practice problem of bitmask in Topic Stream Mashup: Bitwise Operations