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dp[curr][(abs(l-d))]why do you think it is enough to know only abs(d-L) ? If d = 5 you have to consider both cases with L = 4, and with L = 5 not only one of them. Usedp[curr][360 + l - d]UPD: there is mistake in the text. "both cases with L = 4, and with L = 6" abs(4-5) == abs(6-5)
i guess dp[curr][abs(l-d)] takes care of this. if d=5,then l=4,5,6 and the dp states will be dp[9][1],dp[10][0],dp[11][1]. if i am incorrect please correct me!
in the example below func reaches island 1138 with L = 92 first, and for abs(100-92) == 8 remembers result 1. Then func reaches island 1138 with L = 108 and returns answer 1 because abs(100-108) == 8. It's wrong.
n = 2, d = 100
1138
1138+109
your answer is 1, but it's 2.
Thanks so much ..got it:)