solution.py

# 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:

# [
#    [1, 4],
#    [7, 10],
#    [3, 5]
# ]
# 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.

# Examples:
# 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

# sumIntervals( [
#    [0, 20],
#    [-100000000, 10],
#    [30, 40]
# ] ) => 100000030
# Tests with large intervals
# Your algorithm should be able to handle large intervals.
# All tested intervals are subsets of the range [-1000000000, 1000000000].


def sum_of_intervals(intervals):
    intervals = sorted(intervals, key=lambda x: x[0])
    interval_start, result = intervals[0][0], 0
    for i in intervals:
        if i[1] >= interval_start:
            result += i[1] - [interval_start, i[0]][i[0] > interval_start]
            interval_start = i[1]

    return result