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O(N^2) solution in Clear category for Count Inversions by redteer
def count_inversion(sequence):
"""
Count inversions in a sequence of numbers
"""
countNum=0
for i in range(0,len(sequence)):
for j in range(i,len(sequence)):
if sequence[i]>sequence[j]:
countNum+=1
return countNum
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. 5, 2014
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