Given a row of n black houses, you seek to paint some of them orange so as to maximize the total profit, subject to the constraint that painting orange three (or more) houses in-a-row is prohibited. The profit for painting house i orange is profits(i], a positive integer. Illustrative example. For the following input, the optimal solution is to paint houses 1, 2, 5, 6, and 8 orange, which earns a total profit of 45 = 5 + 13 + 11 + 14 + 2. 2 3 4 5 6 7 8 profits[i] 5 13 6 3 11 14 4 Algorithm. Consider the following partial implementation of a dynamic programming solution to the problem, where opt[i] represents the maximum profit obtainable from painting only houses 1 through i: 2.

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Given a row of n black houses, you seek to paint some of them orange so as to maximize the
total profit, subject to the constraint that painting orange three (or more) houses in-a-row is
prohibited. The profit for painting house i orange is profits[i],a positive integer.
Illustrative example. For the following input, the optimal solution is to paint houses 1, 2, 5, 6,
and 8 orange, which earns a total profit of 45 = 5 + 13 + 11 + 14 + 2.
2
3
4
5
6
7
8
profits[i]
5
13
11
14
4
2
Algorithm. Consider the following partial implementation of a dynamic programming solution
to the problem, where opt[ij represents the maximum profit obtainable from painting only
houses 1 through i:
Transcribed Image Text:Given a row of n black houses, you seek to paint some of them orange so as to maximize the total profit, subject to the constraint that painting orange three (or more) houses in-a-row is prohibited. The profit for painting house i orange is profits[i],a positive integer. Illustrative example. For the following input, the optimal solution is to paint houses 1, 2, 5, 6, and 8 orange, which earns a total profit of 45 = 5 + 13 + 11 + 14 + 2. 2 3 4 5 6 7 8 profits[i] 5 13 11 14 4 2 Algorithm. Consider the following partial implementation of a dynamic programming solution to the problem, where opt[ij represents the maximum profit obtainable from painting only houses 1 through i:
Its worst-case running time is
O (n)
O(n log n)
O (n²)
O O(6"), where o = ty5
1+v5
%3D
Transcribed Image Text:Its worst-case running time is O (n) O(n log n) O (n²) O O(6"), where o = ty5 1+v5 %3D
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