How to solve this question using dp .
Problem- link
Please anybody explain the soln.
see Ashishgup solution . dp cant be more neat than his solution
I saw but unable to understand this single line means why everyone is using suffix array even if we are processing from left to right :(
ans = max(dp(i + 1), a[i] + suf[i + 1] — dp(i + 1));
My logic was:
DP(i) stores the maximum value a particular player can get if he starts at the ith index and goes till the end of the array.
One possibility is, I retain my turn and skip the element, thus going to dp(i+1).
Other possibility is, I take the ith element (and get a score of a[i]) and lose my turn. If I lose my turn, then the score I get is:
sum of remaining elements — the maximum score that the other player can get if he starts at index i + 1.
That is, suf[i+1] — dp(i+1).
Hence the line: ans = max(dp(i + 1), a[i] + suf[i + 1] — dp(i + 1)).
see Ashishgup solution . dp cant be more neat than his solution
I saw but unable to understand this single line means why everyone is using suffix array even if we are processing from left to right :(
ans = max(dp(i + 1), a[i] + suf[i + 1] — dp(i + 1));
My logic was:
DP(i) stores the maximum value a particular player can get if he starts at the ith index and goes till the end of the array.
One possibility is, I retain my turn and skip the element, thus going to dp(i+1).
Other possibility is, I take the ith element (and get a score of a[i]) and lose my turn. If I lose my turn, then the score I get is:
sum of remaining elements — the maximum score that the other player can get if he starts at index i + 1.
That is, suf[i+1] — dp(i+1).
Hence the line: ans = max(dp(i + 1), a[i] + suf[i + 1] — dp(i + 1)).