I have a unsorted list. I'm asked to print k-times number of positive values in a segment.The boarders for segment start with 1 not 0. So, if the boarder [1,3] it means that we should find all positives among elements with indices [0,1,2]
For example,
2 -1 2 -2 3
4
1 1
1 3
2 4
1 5
The answer needs to be:
1
2
1
3
Currently, I'm creating a list with length as original
where i
equals 1 if original
is positive and 0 if original
is negative or zero. After that I sum for this segment:
lst = list(map(int, input().split()))
k = int(input())
neg = [1 if x > 0 else 0 for x in lst]
for i in range(k):
l,r = map(int, input().split())
l = l - 1
print(sum(neg[l:r]))
Despite the fact that it's the fastest code that I created so far, it is still too slow for this task. How would I optimize it (or make it faster)?
If I understand you correctly, there doesn't seem to be a lot of room for optimization. The only thing that comes to mind really is that the lst
and neg
steps could be combined, which would save one loop:
positive = [int(x) > 0 for x in input().split()]
k = int(input())
for i in range(k):
l, r = map(int, input().split())
print(sum(positive[l-1:r]))
We can just have bool
s in the positive
list, because bool
is just a subclass of int
, meaning True
is treated like 1
and False
like 0
. (Also I would call the list positive
instead of neg
.)
The complexity is still O(n) though.