Enhance Merge Sort using C#

Software Development
#1

Approach:
Suppose the number of inversions in the left half and right half of the array (let be inv1 and inv2); what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions that need to be counted during the merge step. Therefore, to get the total number of inversions that needs to be added are the number of inversions in the left subarray, right subarray, and merge().

  • How to get the number of inversions in merge()?
    In merge process, let i is used for indexing left sub-array and j for right sub-array. At any step in merge(), if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j]

  • Algorithm:

    1. The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.
    2. Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for the first half, and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].
    3. Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, the number of inversion in the second half and the number of inversions by merging the two.
    4. The base case of recursion is when there is only one element in the given half.
    5. Print the answer

  • Implementation:

// C# implementation of counting the

// inversion using merge sort

using System;

public class Test {

/* This method sorts the input array and returns the

number of inversions in the array */

static int mergeSort( int [] arr, int array_size)

{

int [] temp = new int [array_size];

return _mergeSort(arr, temp, 0, array_size - 1);

}

/* An auxiliary recursive method that sorts the input

array and returns the number of inversions in the

array. */

static int _mergeSort( int [] arr, int [] temp, int left,

int right)

{

int mid, inv_count = 0;

if (right > left) {

/* Divide the array into two parts and call

_mergeSortAndCountInv() for each of the parts */

mid = (right + left) / 2;

/* Inversion count will be the sum of inversions

in left-part, right-part

and number of inversions in merging */

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

inv_count

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

/*Merge the two parts*/

inv_count

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

}

return inv_count;

}

/* This method merges two sorted arrays and returns

inversion count in the arrays.*/

static int merge( int [] arr, int [] temp, int left,

int mid, int right)

{

int i, j, k;

int inv_count = 0;

i = left; /* i is index for left subarray*/

j = mid; /* j is index for right subarray*/

k = left; /* k is index for resultant merged

subarray*/

while ((i <= mid - 1) && (j <= right)) {

if (arr[i] <= arr[j]) {

temp[k++] = arr[i++];

}

else {

temp[k++] = arr[j++];

/*this is tricky -- see above

* explanation/diagram for merge()*/

inv_count = inv_count + (mid - i);

}

}

/* Copy the remaining elements of left subarray

(if there are any) to temp*/

while (i <= mid - 1)

temp[k++] = arr[i++];

/* Copy the remaining elements of right subarray

(if there are any) to temp*/

while (j <= right)

temp[k++] = arr[j++];

/*Copy back the merged elements to original array*/

for (i = left; i <= right; i++)

arr[i] = temp[i];

return inv_count;

}

// Driver method to test the above function

public static void Main()

{

int [] arr = new int [] { 1, 20, 6, 4, 5 };

Console.Write( "Number of inversions are "

+ mergeSort(arr, 5));

}

}

Output:

Number of inversions are 5

Complexity Analysis:

  • Time Complexity: O(n log n), The algorithm used is divide and conquer, So in each level, one full array traversal is needed, and there are log n levels, so the time complexity is O(n log n).
  • Space Complexity**:** O(n), Temporary array.