this is my approach , it passes the given testcase , but fails all others.
I am trying to solve questions but i am finding the recurion difficult.
Not able to do the question
Hey @chetan_aggarwalbX1
your approach isnt correct
say
4
100 1000 10 90
Here I will pick 90 first so that opponent don’t get access to 1000
So this means its not always the greatest we pick .’
logic:
Remember both A and B are equally smart and we want A to get maximum.
if coins are indexed from i to j and both have to choose one out of two ends.
case 1: if A chooses a[i], now opponent can choose either a[i+1] or a[j] whatever is maximum out of two. which means A will get Minimum in his next chance based on opponent choice.
case 1.a: if opponent choose a[i+1], player will chose either a[i+2] or or a[j]
case 1.b: if opponent choose a[j], player will chose either a[i+1] or or a[j-1]
So A will get minimum out of case 1.a and case 1.b…as opponent is equally smarter and choose such a coin which give him maximum but minimum to player A.
case 2: if A chooses a[j], now opponent can choose either a[i] or a[j-1] whatever is maximum out of two. which means A will get Minimum in his next chance based on opponent choice.
case 2.a: if opponent choose a[i], player will chose either a[i+1] or or a[j-1]
case 2.b: if opponent choose a[j-1], player will chose either a[i] or or a[j-2]
So A will get minimum out of case 2.a and case 2.a…as opponent is equally smarter and choose such a coin which give him maximum but minimum to player A.
This will give you a recursive case in which final answer is max(case1,case2)
Here check this for reference
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