I am working on the following Python list exercise from codingbat.com:
Given an array of ints, return the sum of the first 2 elements in the array. If the array length is less than 2, just sum up the elements that exist, returning 0 if the array is length 0. Examples:
sum2([1, 2, 3]) → 3 sum2([1, 1]) → 2 sum2([1, 1, 1, 1]) → 2
My solution below works:
def sum2(nums):
if len(nums)>=2:
return nums[0] + nums[1]
elif len(nums)==1:
return nums[0]
return 0
But I wonder if there's any way to solve the problem with fewer conditional statements.
If you can't use sum, one possible solution uses exceptions:
totalsum = 0
try:
totalsum += nums[0]
totalsum += nums[1]
except IndexError:
pass
return totalsum
Catch the error and short-circuit the summation if an element doesn't exist. Easier to ask forgiveness than permission, as they say.
try this:
def sum2(nums):
if len(nums) == 1:
return nums[0]
else:
return sum(nums [:2])
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