can one miss one round and participate in other??

The page says "The countdown to Round 1A 2019 is on", but I can't find a countdown anywhere. Am I missing something? I hope I'm waiting in the right place, not that it would make any difference even if I was late, I just want to see the problems.

I know this is a new interface by Google, but I hope they make it a bit more exciting next year. One wouldn't know that an international event is shortly going to take place with thousands of programmers waiting anxiously to compete. Ten years ago, the Topcoder chatroom before a big event had a dynamic atmosphere and was a great place to meet programming friends and hang out. I miss that... :'-(

where can we find link to the dashboard?

is it only me unable to see the problems??

Are the questions loading for anyone?

somehow i am tired of their Shakespeare language

Why my rank is decreased from 200+ to 180+？

Is it possible to see country-wise standings?

My solutions to first two:

A lazy way:

• I'm too sleepy to try to construct an approach.
• Would random work well here?
• Write a fully recursive brute-force, where on every step we try every possible move in a random order.
• Try to find solutions for all cases.
• Switch seeds until all cases pass (too lazy to do it separately) -- 4th attempt works.
• Now we precomputed all the answers.
• Encode it so that it fits in 100KB (using 2 letters A-T for coordinates just about works).
• PROFIT

I tried using the method described here with a Knight's Tour, why does my large for Pylons fail???

C++ Code

Editorial: https://codejam.withgoogle.com/2018/challenges/0000000000051635/analysis

For C I first constructed a trie of all of the strings reversed and then calculated the answer based on this dfs:

int getans(tnode* t) {
    int rem = t->cnt; // cnt is number of strings that pass through this node
    int res = 0;
    M00(i, 26) if(t->next[i] != nullptr) {
        int k = getans(t->next[i]);
        rem -= k;
        res += k;
    }
    if(rem >= 2 && t->c != ' ') res+=2; // second condition makes sure this is not the root node (which means common suffix has length 0)
    return res;
}

I'm not sure why this solution works. Can anyone tell me?

EDIT: NVM I UNDERSTAND

My solution for PYLONS was to divide the matrix into steps of 2s and 3(if odd). We can start at $$$mat[0][0]$$$ and follow a pattern.

The pattern for $$$2*m$$$ is $$$mat[0][i]$$$ $$$- \gt $$$ $$$mat[1]$$$ $$$[$$$$$$(i+3)mod\ m$$$$$$]$$$ $$$- \gt $$$ $$$mat[0][i+1]$$$.

For $$$3*m$$$, you can do $$$mat[0][i] - \gt mat[1][(i+3)mod\ m] - \gt mat[2][i] - \gt mat[0][i+1]$$$.

You'll have to take care of some corner cases, like when $$$max(n,m) \lt =4$$$

I can't find any counter example for my solution for Pylons (simple test case). https://gist.github.com/kronos/d8ccb2941cd057c28d9bc0192b9a6652 if you see any issues in my solution — please, let me know.

Good hacks for Alien Rhyme? I solved the visible but got WA for hidden and don't know what to do

https://ideone.com/e.js/mKXWib i cant find a mistake-pylons

can someone give a counter-case for my solution to third problem?

Solution for alien rhyme:

• Make an unordered_map, where key is all possible suffix for each of the word and value is a vector that stores the words for which we got that suffix

• iterate through the map, and for all keys(suffixes) that have more than 1 element in their vector (i.e. there are more than 1 word that have that suffix), push those keys in a new vector.

• This new vector would be sorted by length of string in decreasing order

• for each of suffix in vector, choose 2 words from map that are not yet used.

• Finally print the count of all chosen words.

On C, I didn't do trie/hashmap, just reversed and sorted the words and put in a linked list, and then from len 50 to 1 deleted adjacent words if they shared the same prefix length len, and that prefix wasn't the previously deleted words one.

Can someone post solution for C where u dont use a trie ? (Actual Code)

pylon can be solved in two ways. O(r*c) make 2*n,3*n split few case will fail have to hard code them in my case 4*4,4*6,6*4,6*6 https://shashankmishracoder.wordpress.com/2019/04/13/google-code-jam-2019-round1a-pylon/

another way is to use dfs and backtracks for depth r*c https://shashankmishracoder.wordpress.com/2019/04/13/google-code-jam-2019-round-1a-pylon-dfs-and-backtracking-approach/

can someone find a error in my solutions for pylons

[Link]-(https://pastebin.com/1UWXk9GA)