'N' positive integers are placed in a circle in such a way such that for each valid i, A(i) and A(i+1) are adjacent, and A(1) and A(N) are also adjacent.

We want to repeat the following operation exactly N−1 times (until only one number remains): :-)

1)Select two adjacent numbers. Let's denote them by a and b.

2)Score a+b points.

3)Erase both a and b from the circle and insert a+b in the space between them.

What is the minimum number of points we can score ?

Example:-

[10 10 1]

[10,10,1]→[10,11] , Score: 11

[10,11]→[21], Score: 21

Total Score: 11+21 = 32

How can I solve this problem using dynamic programming ?

My approach is to brute-force and try all possible ways.

try greedy method and figure it out yourself :)

U have to select the pair with minimum sum with updations . So its better to compute the minimum adjacent sum and update it after doing step no 2.

example :
5
7 3 4 1 5

Solution :

7 3 (4+1) 5 // Sum = 5
7 3 5 5
7 (3+5) 5 // Sum = 5+ 8=13
7 8 5
8 (7+5) // Sum = 13+12=25
(Addition of 7,5 is possible because it is circular array)
8 12
(8+12) // Sum = 25 +20=45
Final answer = 45.
Hope this helps.

greedy approach is not working

Deja vu: http://www.cplusplus.com/forum/beginner/256890/

MikeyBoy wrote:
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