Abhishek Saini

stop watching reels. read books. walk 10k steps. explore curiosity. learn new things. be grateful. do something. practice doing nothing. turn notifications off. lift weights. learn logic. learn probability. don't badmouth people. ignore the trolls. don't be jealous. be competitive. go in depth. think at a higher level. track what you want to improve. walk while talking. walk when feeling down. make a smiling face when sad. iterate. read and write. take risks. don't take the moment for granted. don't take yourself too seriously. talk to people. be open to being wrong. make money. money isn't everything. care about something deeply. don't say yes to everything. listen more. do hard things. question your beliefs. observe. sleep well. eat well. laugh. experience life.

- note to self

Codeforces Round 944
Got India Rank 2 and World Rank 43.

Sharing video solutions for all problems.
Find the link in the first comment.

Leetcode Contest 408 - Better problems than recent Leetcode contests

Problem A - Sum one-digit and two-digit numbers separately and see if they are the same or not.

Problem B - Only squares of primes numbers are such numbers. So we just need to count how many prime squares are between these two numbers.

Problem C - We calculate all such substrings starting from each left index one by one. So the issue is this can take (N * N) which isn't good enough. What we can do is we don't iterate all substrings starting from this index, we just move from '0' to '0'. Why this is enough because whenever zeroes exceed sqrt(N), we can stop moving because now we know more dominant substrings are not possible. Hence we are done in N * sqrt(N).

Problem D - Consider each circle as a node of a graph and all 4 sides as a node of a graph. Now to add edges to this graph, we can check which circles intersect (for example - if the distance between centers should be less than the radius sum). Similarly, we add edges between sides and circles as well. Now in this graph, if the left edge node and the right edge node are connected (and a few more similar cases), we can see no path will exist between those two corners.

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Interviewer - Best sorting algorithm : Quick Sort or Merge Sort

Gambhir - Bubble Sort