Get a number in an unordered straight that follow a start value from an interval

I search a fast method to perform my problem.

imagine ordered seats numeroted from 1 to 8, imagine they are people on seats [ 2, 6, 5, 3 ]. i want to get back the second (interval +2) people after the seat number 4 (start value)

for examples :

with this array : [2, 5, 8, 7, 1] , i started with value 3 and i move +2 times, the third next number in the list is 5, the second is 7, the method must return this value

with the sam开发者_Python百科e [2, 5, 8, 7, 1] , i started from 7 and i move +3 times here the method must return to the minimal value. trought 8.. 1.. 2.., result : 2

with [1, 3], start 4, count +2, result 3

with [5, 3, 9], start 3, count +1, result 5

with [5, 3, 9], start 3, count +2, result 9

I hope someone will understand my problem. thanks you

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Sort your list, use bisect to find the starting index, then mod the result of the addition by the length of the list.

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So, this is basically just an example implementation of Ignacio's algorithm in Python:

from bisect import bisect

def circular_highest(lst, start, move):
    slst = sorted(lst)
    return slst[(bisect(slst, start) - 1 + move) % len(lst)]

print circular_highest([2, 5, 8, 7, 1], 3, 2)
print circular_highest([2, 5, 8, 7, 1], 7, 3)
print circular_highest([1, 3], 4, 2)
print circular_highest([5, 3, 9], 3, 1)
print circular_highest([5, 3, 9], 3, 2)

Output:

7
2
3
5
9