```
def spaced_out(list_of_num):
if len(list_of_num) > 3:
list_of_num = list_of_num[:3]
print("keeping only first three elements")
```**(1)**
if abs(list_of_num[0]-list_of_num[1]) == abs(list_of_num[1]-list_of_num[2]): **(2)**
return "Spaced Out"
else:
return "NOT Spaced Out"

# Spaced Out?

Upasana | October 23, 2019 | 1 min read | 15 views | Python Coding Problems

Given 3 different integer numbers, determine if the difference between the smallest number and the middle number is the same as the difference between the middle number and the largest number.

The numbers can be entered in any order. If the differences are exactly the same the program should indicate the numbers are "Spaced Out" otherwise the numbers are "NOT Spaced Out".

1 |
In case, the input is more than three numbers then we will be considering only first three numbers as given in problem statement. |

2 |
We are checking if the array is spaced out or not by using `abs` function in python |

Input

`print(spaced_out([1,3,5,7]))`

Output

keeping only the first three elements Spaced Out

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