TWENDS - Two Ends

In the two-player game “Two Ends”, an even number of cards is laid out in a row. On each card, face up, is written a positive integer. Players take turns removing a card from either end of the row and placing the card in their pile. The player whose cards add up to the highest number wins the game. Now one strategy is to simply pick the card at the end that is the largest — we’ll call this the greedy strategy. However, this is not always optimal, as the following example shows: (The first player would win if she would first pick the 3 instead of the 4.)

3 2 10 4

You are to determine exactly how bad the greedy strategy is for different games when the second player uses it but the first player is free to use any strategy she wishes.


There will be multiple test cases. Each test case will be contained on one line. Each line will start with an even integer n followed by n positive integers. A value of n = 0 indicates end of input. You may assume that n is no more than 1000. Furthermore, you may assume that the sum of the numbers in the list does not exceed 1,000,000.


For each test case you should print one line of output of the form:

In game m, the greedy strategy might lose by as many as p points.

where m is the number of the game (starting at game 1) and p is the maximum possible difference between the first player’s score and second player’s score when the second player uses the greedy strategy. When employing the greedy strategy, always take the larger end. If there is a tie, remove the left end.


4 3 2 10 4
8 1 2 3 4 5 6 7 8
8 2 2 1 5 3 8 7 3

In game 1, the greedy strategy might lose by as many as 7 points.
In game 2, the greedy strategy might lose by as many as 4 points.
In game 3, the greedy strategy might lose by as many as 5 points.

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da_201501181: 2017-05-21 14:02:18

NOTE: "If there is a tie, remove the left end." caused me 1 WA..
Try Treats for cows (TRT) Similar kind of...!!

vengatesh15: 2017-04-05 18:00:03

don't forget to consider i.e in greedy approach if both end have same value then take the left end that cost me 2 WA

nilabja16180: 2017-03-21 19:37:31

DP with RECURSION, AC IN ONE GO! 0.00 sec

Last edit: 2017-03-21 19:38:14
ruben_ash: 2017-02-28 17:49:22

Solve for the good strategy with dp + memoization, and try to find the greedy answer in O(1).

gautam: 2017-02-23 10:40:18

bcz of == got too many wa..but finally Ac ..;)

starbot: 2017-02-11 13:45:06

used map for memoization..TLE, then used array..AC

mkfeuhrer: 2017-01-30 18:12:01

Try spoj TRT before this :-) AC :-)

epsilonalpha: 2016-12-22 01:16:24

0.02s in C++! AC in one go!
Recursion + Memoization is the way to go. Do take care of the equality case where the left end is to be removed. :)

siddharth_0196: 2016-10-05 13:22:41

Recursion + Memoization = AC!
A must do problem for beginners like me!

alaa_alrayes96: 2016-09-27 22:00:14

recursion + momization make sure that if equal remove from the left side
got AC

Added by:Camilo Andrés Varela León
Time limit:0.100s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ERL JS-RHINO NODEJS PERL6 VB.NET
Resource:East Central North America 2005