What wrong with this logic

#1

It is passing only one test case…
What is wrong here

#2

@ashishnnnnn your logic is incorrect, you are using greedy approach which will give WA.

#3

Consider both possibilities, and take max when it’s your turn and min when its opponents turn!

#4

But here players are playing optimally … then why should i consider min…when its opponents turn

#5

this is the reason itself, when it’s your turn, you try to maximize your coins, but when its opponents turn, he tries to maximize his coins (or minimize your coins!)

#6

Do we need The help of D.P to solve this question…
Because as you said that when it’s my turn i should select maximum…which i am doing…

#7

DP optimization is possible,
you are doing it greedily!
by max I mean this max(arr[i] + solve(i+1,j), arr[j] + solve(i,j-1) )
here I don’t compare arr[i] and arr[j]
I try both options ans then take maximum out of them!
try to think this way

#8

Hii here by doing this max(arr[i] + solve(i+1,j), arr[j] + solve(i,j-1) ) we are only considering about the first palyer that he should be come up with maximum and other one minimum… sorry if i am missing something… Not able to get it

#9

See we first have to know whose turn it is to pick coin, after that we do this:
if((i+j)%2) // piyush’s turn
dp[i][j] = max(a[i]+dp[i+1][j],a[j]+dp[i][j-1])
else //Nimit’s turn
dp[i][j] = min(dp[i+1][j],dp[i][j-1])
here dp[i][j] is maximum coins collected by piyush when coins remaining from index i to index j.
so answer is dp[0][n-1]
If you still can’t get it, I would recommend you to unlock editorial.

#10

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