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CUTANDPASTE - Cut and Paste

  • Title: Cut and Paste
  • Judge / source: CSES
  • Original URL: https://cses.fi/problemset/task/2072
  • Secondary topics: Implicit treap, Split/merge sequence surgery, Subtree size as position
  • Difficulty: hard
  • Subtype: Cut a substring and paste it to the end many times
  • Status: solved
  • Solution file: cutandpaste.cpp

Why Practice This

This is the cleanest in-repo anchor for the first implicit-treap lane.

The statement gives exactly the right signal:

  • the sequence is mutable
  • positions keep changing after every operation
  • each operation is structural, not just arithmetic on a fixed index line

So the real lesson is:

  • represent the sequence as one treap
  • split by position, not by explicit key
  • rearrange pieces with merges

Recognition Cue

Reach for an implicit treap when:

  • sequence edits shift many later positions
  • you need insert / erase / cut / paste by current position
  • storing explicit indices would force too many updates
  • the natural operation is "first k elements vs the rest"

The strongest smell here is:

  • "cut positions a..b, then paste that block somewhere else"

Problem-Specific Transformation

Treat the current string as one implicit treap whose in-order traversal is the current character order.

For one operation (a, b) from the statement:

  • convert to zero-based half-open interval [a - 1, b)
  • split into three pieces:
  • prefix [0, a - 1)
  • middle [a - 1, b)
  • suffix [b, n)

Then rebuild in the order:

prefix + suffix + middle

So the statement is no longer:

  • shift characters inside one array or string

It becomes:

  • two splits
  • two merges

on one implicit treap.

Core Idea

Each node stores:

  • one character
  • one random priority
  • subtree size

The subtree size is what gives implicit position.

If split(root, k) returns:

  • left = first k characters
  • right = the rest

then Cut and Paste is just:

(a, bc) = split(root, l)
(b, c) = split(bc, r - l)
root = merge(merge(a, c), b)

The important invariant is:

the in-order traversal of the treap is always the current string

So after each split/merge sequence, the structure changes shape but the represented string remains exactly what the operation intends.

Complexity

Let n be the initial string length and m the number of operations.

  • build: O(n log n) in the straightforward starter-style construction
  • each cut/paste operation: expected O(log n)
  • final output traversal: O(n)
  • memory: O(n)

This comfortably fits the CSES limits.

Pitfalls / Judge Notes

  • The statement uses one-based inclusive intervals.
  • Keep internal split positions zero-based.
  • split(root, k) means first k elements vs the rest, not "up to index k inclusive".
  • Do not try to update explicit character indices after every operation.
  • The answer is the final in-order traversal after all operations, not intermediate outputs.

Reusable Pattern

Solutions