Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
Nested cycle solution in Clear category for Count Inversions by mike269
def count_inversion(sequence):
"""
Count inversions in a sequence of numbers
"""
count = 0
for i, x in enumerate(sequence[:-1]):
for y in sequence[i+1:]:
if x > y: count += 1
return count
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert count_inversion((1, 2, 5, 3, 4, 7, 6)) == 3, "Example"
assert count_inversion((0, 1, 2, 3)) == 0, "Sorted"
assert count_inversion((99, -99)) == 1, "Two numbers"
assert count_inversion((5, 3, 2, 1, 0)) == 10, "Reversed"
Sept. 26, 2014
Comments:
