# level 4:Sum of Intervals(2019-12-07)
###### tags: `Codewars` `python`

Write a function called sumIntervals/sum_intervals() that accepts an array of intervals, and returns the sum of all the interval lengths. Overlapping intervals should only be counted once.

Intervals
Intervals are represented by a pair of integers in the form of an array. The first value of the interval will always be less than the second value. Interval example: [1, 5] is an interval from 1 to 5. The length of this interval is 4.

Overlapping Intervals
List containing overlapping intervals:
<pre>
[
[1,4],
[7, 10],
[3, 5]
]
</pre>
The sum of the lengths of these intervals is 7. Since [1, 4] and [3, 5] overlap, we can treat the interval as [1, 5], which has a length of 4.
<pre>
sumIntervals( [
[1,2],
[6, 10],
[11, 15]
] ); // => 9

sumIntervals( [
[1,4],
[7, 10],
[3, 5]
] ); // => 7

sumIntervals( [
[1,5],
[10, 20],
[1, 6],
[16, 19],
[5, 11]
] ); // => 19
</pre>

my code:
<pre>
def sum_of_intervals(intervals):
number = []
maxn = 0
minn = 0
for i in intervals:
if i[1] > maxn:
maxn = i[1]
if i[0] < minn:
minn = i[0]
for i in range(0, maxn - minn):
number.append(0)
for i in intervals:
for j in range(i[0]-minn, i[1]-minn):
number[j] = 1
return sum(number)
</pre>