Counting Inversion by piggybacking on Merge Sort with Python and C++

Abhishek Dubey — Tue, 15 Jun 2021 05:07:01 +0000

Solution to the problem.

Solution in Python

import os


def main():
 array = []
 with open("IntegerArray.txt") as file:
      array = [line.strip() for line in file]
 mergesort(array)


def merge(a,b):
    c = []
    count = 0
    while len(a) != 0 and len(b) != 0:
        if a[0] < b[0]:
            c.append(a[0])
            a.remove(a[0])
            count += 1
        else:
            c.append(b[0])
            b.remove(b[0])
            count += 1
    if len(a) == 0:
        c += b
    else:
        c += a
    # print("\n\n\n\n"+str(count)+"\n\n\n\n")
    return count


def mergesort(x):
    if len(x) == 0 or len(x) == 1:
        return x
    else:
        middle = len(x)/2
        middle = int(middle)

        a = mergesort(x[:middle])
        b = mergesort(x[middle:])
        return merge(a,b)


if __name__ == '__main__':
 main()

Solution of the same problem in C++.

#include "bits/stdc++.h"
#include "iostream"
#include "fstream"
using namespace std;

long long  _mergeSort(long long arr[], long long temp[], long long left, long long right);
long long merge(long long arr[], long long temp[], long long left, long long mid, long long right);

long long mergeSort(long long arr[], long long array_size)
{
    long long *temp = (long long *)malloc(sizeof(long long)*array_size);
    return _mergeSort(arr, temp, 0, array_size - 1);
}

long long _mergeSort(long long arr[], long long temp[], long long left, long long right)
{
  long long mid, inv_count = 0;
  if (right > left)
  {
    mid = (right + left)/2;

    inv_count  = _mergeSort(arr, temp, left, mid);
    inv_count += _mergeSort(arr, temp, mid+1, right);

    inv_count += merge(arr, temp, left, mid+1, right);
  }
  return inv_count;
}


long long merge(long long arr[], long long temp[], long long left, long long mid, long long right)
{
  long long i, j, k;
  long long inv_count = 0;

  i = left;
  j = mid;
  k = left;
  while ((i <= mid - 1) && (j <= right))
  {
    if (arr[i] <= arr[j])
    {
      temp[k++] = arr[i++];
    }
    else
    {
      temp[k++] = arr[j++];
      inv_count = inv_count + (mid - i);
    }
  }


  while (i <= mid - 1)
    temp[k++] = arr[i++];

  while (j <= right)
    temp[k++] = arr[j++];

  for (i=left; i <= right; i++)
    arr[i] = temp[i];

  return inv_count;
}


int main()
{
    long long arr[100000];
    ifstream file("..PATH_TO_FILE/file.txt");
        if(file.is_open())
        {

            for(long long i = 0; i < 100000; ++i)
            {
                file >> arr[i];
             }
        }

        std::cout << mergeSort(arr, 100000);
  return 0;
}

DEV Community: Abhishek Dubey

Counting Inversion by piggybacking on Merge Sort with Python and C++

Solution to the problem.