Short Problem Definition:
Insertion Sort is a simple sorting technique which was covered in previous challenges. Sometimes, arrays may be too large for us to wait around for insertion sort to finish. Is there some other way we can calculate the number of times Insertion Sort shifts each elements when sorting an array?
If ki is the number of elements over which the ith element of the array has to shift, then the total number of shifts will be k1 +k2 +…+kN.
Link
Insertion Sort Advanced Analysis
Complexity:
time complexity is O(N\*log(N))
space complexity is O(1)
Execution:
There are more solutions with nlogn time for this challenge. I decided to use Binary Indexed Trees as they are a data structure I am not that familiar with. The other solution includes a modified merge-sort that is posted as the problem editorial. BITs are effective for computing cumulative frequencies in log(N) time and are therefore perfectly suited for this problem. They assume a full tree and therefore are bound to the maximal range defined in the problem specification.
EDIT 2020: The solution no longer works. I added a new solution below that is based on Codility “ArrayInversionCount” Solution
Solution:
#!/usr/bin/py
class BinaryIndexedTree():
def __init__(self, sz):
self.sz = sz
self.tree = [0] * sz
def update(self, idx, val):
while (idx <= self.sz):
self.tree[idx] += val
idx += (idx & -idx)
def read(self, idx):
sum = 0
while (idx > 0):
sum += self.tree[idx]
idx -= (idx & -idx)
return sum
if __name__ == '__main__':
t = input()
for _ in range(t):
n = input()
arr = map(int, raw_input().split())
cnt = 0
bit = BinaryIndexedTree(10**7+1)
for val in reversed(arr):
bit.update(val, 1)
cnt += bit.read(val-1)
print cnt
#!/bin/python
import math
import os
import random
import re
import sys
def mergeSort(alist):
inversion_count = 0
n = len(alist)
if n == 1:
return 0
mid = len(alist) // 2
lefthalf = alist[:mid]
righthalf = alist[mid:]
n1 = len(lefthalf)
n2 = len(righthalf)
inversion_count += mergeSort(lefthalf)
inversion_count += mergeSort(righthalf)
i = 0
j = 0
k = 0
while i < n1 and j < n2:
if lefthalf[i] <= righthalf[j]:
alist[k] = lefthalf[i]
i = i + 1
else:
alist[k] = righthalf[j]
j = j + 1
inversion_count += n1 - i
k = k + 1
while i < n1:
alist[k] = lefthalf[i]
i = i + 1
k = k + 1
while j < n2:
alist[k] = righthalf[j]
j = j + 1
k = k + 1
return inversion_count
if __name__ == '__main__':
t = int(raw_input())
for t_itr in xrange(t):
n = int(raw_input())
arr = map(int, raw_input().rstrip().split())
result = mergeSort(arr)
print result