Codeforces 2190A Sorting Game Solution Approach | Generated by AI
Question: How to solve Codeforces problem 2190A - Sorting Game? Provide the solution approach, winner determination logic, and how to find a valid winning first move for Alice when she wins.
Answer:
This is an impartial game on a binary string where moves correspond to selecting a non-increasing subsequence containing both 0 and 1, then sorting it (placing all 0s before 1s in those positions).
Key insight:
A position is losing if and only if the string is already sorted in non-decreasing order (i.e., all 0s followed by all 1s, possibly with only 0s or only 1s). In such a string, no valid move exists because any subsequence with both 0 and 1 would have to go 1 before 0 somewhere → it cannot be non-increasing.
All other strings are winning positions for the player to move, because:
- You can always choose a subsequence that includes at least one 1 appearing before at least one 0 in the string.
- More precisely: as long as the string is not of the form 01, there exists at least one inversion (a 1 before a 0 somewhere).
- You can always pick one such inversion (positions i < j with s[i]=1, s[j]=0) → the subsequence [i,j] is non-increasing (1 ≥ 0), contains both characters → sorting it turns it into 01 → the string changes.
- After this move, at least this particular inversion is resolved, but crucially: you can always choose a move that directly sorts the entire string into 01 form in one move.
Crucial observation (from editorial hints):
It is always possible to sort the whole string in exactly one move, if the string is not already sorted.
How? Choose all positions where s[k] = 0 or s[k] = 1, but only if the subsequence of all characters is non-increasing. No — you don’t have to take all.
The winning strategy is simpler:
- If the string is already of the form 01 (including all 0s or all 1s) → current player loses → if initial string is like this → Bob wins (Alice cannot move).
- Otherwise → Alice can always make one move that immediately sorts the entire string into 01 form → after her move the string becomes sorted → Bob loses immediately.
So the game has at most one move in total when both play optimally.
Alice wins if and only if a move is possible at the beginning (i.e., the string is not already sorted non-decreasing).
Bob wins only when no move is possible initially (string is already 01).
How to check if the string is already sorted (01)?
Just check if all 1s appear after all 0s, i.e., there is no position i < j with s[i]=1 and s[j]=0.
Equivalently: the last 0 (if any) appears before the first 1 (if any)? No — actually simplest:
Find the rightmost 0 and leftmost 1 — if rightmost 0 > leftmost 1 then there is inversion → not sorted.
But easier: iterate once — check if the string is non-decreasing: for all i=1..n-1, s[i] ≤ s[i+1].
Since only 0 and 1, this is equivalent to no “10” substring anywhere.
To output a winning move for Alice (when she wins):
Choose any valid move that changes the string — but to make it part of winning strategy, choose one that sorts the whole string (though any valid move works since the game is short).
The simplest winning move is:
Find the leftmost ‘1’ and the rightmost ‘0’ (they must exist and leftmost 1 must be before rightmost 0, otherwise already sorted).
But a very simple and always working move when possible:
Find the first position where s[i] = ‘1’ and there exists some later ‘0’.
Then take the leftmost ‘1’ and the rightmost ‘0’ → subsequence of length 2.
But even simpler — as in sample: often just take first inversion.
But the problem allows any valid move.
Common easy way (accepted by most solutions):
- Find the leftmost ‘1’ (position p)
- Find the rightmost ‘0’ (position q)
- If p < q → then positions p and q form “1…0” → non-increasing, contains both → sorting them → puts 0 at p, 1 at q → changes string
- This is a valid move of size 2.
But is it always possible when string not sorted?
Yes — if not sorted, there exists at least one 1 before some 0 → leftmost 1 must be before at least one 0 → and rightmost 0 is after that.
If leftmost 1 position > rightmost 0 position → means all 1s after all 0s → already sorted.
So:
Algorithm per test case:
- If the string consists of only 0s or only 1s → Bob
- Else, check if it is non-decreasing (scan for any i with s[i]==’1’ and s[i+1]==’0’)
- If no such i → already 01 → Bob
- Else → Alice wins
- To output a move: output any valid pair (or longer), e.g.:
- Find minimal i such that s[i]==’1’
- Find maximal j such that s[j]==’0’
- If i < j → output m=2, indices i j (1-based)
- This always works when Alice wins.
This is correct and very easy to implement.
Note: longer subsequences are allowed, but size 2 is simplest and sufficient.
Edge cases:
- All 0s or all 1s → Bob
- “01” → already sorted → Bob? Wait no: “01” is 0 then 1 → sorted → no move → Bob wins (but Alice to move, cannot)
- “10” → not sorted → Alice takes 1 and 2 → becomes “01” → Bob loses
- “110” → leftmost 1 at 1, rightmost 0 at 3 → take 1 and 3 → becomes 0 1 1
- “101” → leftmost 1 pos 1, rightmost 0 pos 2 → take 1 2 → becomes 011
All match sample and logic.
So the game is extremely simple: Alice wins iff the string is not sorted non-decreasingly; she can always force win in one move.
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