indy256's blog

Zalando CodeSprint on HackerRank
By indy256, 8 years ago, In English

Hi everyone! I am glad to announce that Zalando CodeSprint on HackerRank is scheduled on June 4, 2016, 12:00 PM CEST. Contest duration is 24 hours, but don't be scared by that. There will be 8 problems to solve.

Sign up here: https://www.hackerrank.com/zalando-codesprint

The prizes are:

  • Rank 1: Pebble Watch, 200 Euro gift voucher of your choice, Hackathon Hoodie
  • Rank 2-5: 200 Euro gift voucher of your choice, Hackathon Hoodie
  • Rank 6-20: Hackathon Hoodie
  • Vote: I like it
  • +32
  • Vote: I do not like it

»
8 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Could you explain what kind of problems it'll feature?

→ Reply
  • »
    »
    8 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    No exact information about problem types and their difficulty is available. Most probable format is 7 usual algorithmic challenges and 1 marathon-style challenge.

    → Reply
    • »
      »
      »
      8 years ago, # ^ |
      ← Rev. 2   Vote: I like it +5 Vote: I do not like it

      Sorry, was wrong. Actually 8 algorithmic challenges.

      → Reply
»
8 years ago, # |
  Vote: I like it +14 Vote: I do not like it

The hardest problem was literally this one: http://codeforces.me/blog/entry/10355

→ Reply
  • »
    »
    8 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    Sorry, I misread T.T

    → Reply
    • »
      »
      »
      8 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      Did you look at the link? Are you sure sqrt-decomposition is shorter than it? BTW, copy-past treap implementation is easier than sqrt-decomposition

      → Reply
    • »
      »
      »
      8 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Can you share your idea?

      → Reply
      • »
        »
        »
        »
        8 years ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        Let the original array be v[].

        Let p[] be a permutation such that p[i] is the original position of element i.

        You can see that at any moment, p[] will consist of one or more interval of consecutive numbers. Then let a[] be the set of these intervals.

        For each query, we have to cut at most 2 intervals in a[], then put them to the front.

        Then after queries, rebuilt the original array using a[].

        If we do it like that, at any moment, a[] will have at most intervals, and we can cut any intervals, put them to the front in

        → Reply
»
8 years ago, # |
  Vote: I like it 0 Vote: I do not like it

is this contest going to be rated?

→ Reply