Verify two numbers from a list will add up to a number in the same list using Python
0
$begingroup$
I am trying to solve this question:
- Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.
I came up with:
def func(l):
sum_set = set() # using a set to store sums of two numbers.
for index in range(len(l)-1):
counter = 1
while counter + index < len(l):
sum = l[index] + l[index+counter]
sum_set.add(sum)
counter += 1
for number in l:
if number in sum_set:
return True
return False
print(func([1,2,3,4]))
print(func([1,2,1,4]))
print(func([1,1,1]))
print(func([1]))
print(func([1,2,5,9]))
print(func([2,1,1]))
print(func([1,-1,0]))
From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:
- I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g.
l = [1,1,1,1,2]should be optimised tol = [1,1,2]
Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?
python
asked 5 mins ago
Yu ZhangYu Zhang
302312
$endgroup$
add a comment |
0
$begingroup$
I am trying to solve this question:
- Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.
I came up with:
def func(l):
sum_set = set() # using a set to store sums of two numbers.
for index in range(len(l)-1):
counter = 1
while counter + index < len(l):
sum = l[index] + l[index+counter]
sum_set.add(sum)
counter += 1
for number in l:
if number in sum_set:
return True
return False
print(func([1,2,3,4]))
print(func([1,2,1,4]))
print(func([1,1,1]))
print(func([1]))
print(func([1,2,5,9]))
print(func([2,1,1]))
print(func([1,-1,0]))
From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:
- I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g.
l = [1,1,1,1,2]should be optimised tol = [1,1,2]
Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?
python
asked 5 mins ago
Yu ZhangYu Zhang
302312
$endgroup$
add a comment |
0
0
0
$begingroup$
I am trying to solve this question:
- Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.
I came up with:
def func(l):
sum_set = set() # using a set to store sums of two numbers.
for index in range(len(l)-1):
counter = 1
while counter + index < len(l):
sum = l[index] + l[index+counter]
sum_set.add(sum)
counter += 1
for number in l:
if number in sum_set:
return True
return False
print(func([1,2,3,4]))
print(func([1,2,1,4]))
print(func([1,1,1]))
print(func([1]))
print(func([1,2,5,9]))
print(func([2,1,1]))
print(func([1,-1,0]))
From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:
- I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g.
l = [1,1,1,1,2]should be optimised tol = [1,1,2]
Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?
python
asked 5 mins ago
Yu ZhangYu Zhang
302312
$endgroup$
I am trying to solve this question:
- Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.
I came up with:
def func(l):
sum_set = set() # using a set to store sums of two numbers.
for index in range(len(l)-1):
counter = 1
while counter + index < len(l):
sum = l[index] + l[index+counter]
sum_set.add(sum)
counter += 1
for number in l:
if number in sum_set:
return True
return False
print(func([1,2,3,4]))
print(func([1,2,1,4]))
print(func([1,1,1]))
print(func([1]))
print(func([1,2,5,9]))
print(func([2,1,1]))
print(func([1,-1,0]))
From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:
- I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g.
l = [1,1,1,1,2]should be optimised tol = [1,1,2]
Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?
python
python
asked 5 mins ago
Yu ZhangYu Zhang
302312
asked 5 mins ago
Yu ZhangYu Zhang
302312
asked 5 mins ago
Yu ZhangYu Zhang
302312
asked 5 mins ago
Yu ZhangYu Zhang
302312
asked 5 mins ago
Yu ZhangYu Zhang
302312
302312
add a comment |
add a comment |
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