# Position of an element after stable sort

Given an array of integers which may contain duplicate elements, an element of this array is given to us, we need to tell the final position of this element in the array, if a stable sort algorithm is applied.

Examples:

```Input  : arr[] = [3, 4, 3, 5, 2, 3, 4, 3, 1, 5], index = 5
Output : 4
Element initial index – 5 (third 3)
After sorting array by stable sorting algorithm, we get
array as shown below
[1(8), 2(4), 3(0), 3(2), 3(5), 3(7), 4(1), 4(6), 5(3), 5(9)]
with their initial indices shown in parentheses next to them,
Element's index after sorting = 4
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

One easy way to solve this problem is to use any stable sorting algorithm like Insertion Sort, Merge Sort etc and then get the new index of given element but we can solve this problem without sorting the array.
As position of an element in a sorted array is decided by only those elements which are smaller than given element. We count all array elements smaller than given element and for those elements which are equal to given element, elements occurring before given elements’ index will be included in count of smaller elements this will insure the stability of the result’s index.
Simple code to implement above approach is implemented below:

## C/C++

```// C++ program to get index of array element in
// sorted array
#include <bits/stdc++.h>
using namespace std;

//  Method returns the position of arr[idx] after
// performing stable-sort on array
int getIndexInSortedArray(int arr[], int n, int idx)
{
/*  Count of elements smaller than current
element plus the equal element occurring
before given index*/
int result = 0;
for (int i = 0; i < n; i++)
{
// If element is smaller then increase
// the smaller count
if (arr[i] < arr[idx])
result++;

// If element is equal then increase count
// only if it occurs before
if (arr[i] == arr[idx] && i < idx)
result++;
}
return result;
}

//  Driver code to test above methods
int main()
{
int arr[] = {3, 4, 3, 5, 2, 3, 4, 3, 1, 5};
int n = sizeof(arr) / sizeof(arr[0]);

int idxOfEle = 5;
cout << getIndexInSortedArray(arr, n, idxOfEle);

return 0;
}
```

## Java

```// Java program to get index of array
// element in sorted array

class ArrayIndex
{
//  Method returns the position of
// arr[idx] after performing stable-sort
// on array
static int getIndexInSortedArray(int arr[],
int n, int idx)
{
/*  Count of elements smaller than
current element plus the equal element
occurring before given index*/
int result = 0;
for (int i = 0; i < n; i++)
{
// If element is smaller then
// increase the smaller count
if (arr[i] < arr[idx])
result++;

// If element is equal then increase
// count only if it occurs before
if (arr[i] == arr[idx] && i < idx)
result++;
}
return result;
}

//  Driver code to test above methods
public static void main(String[] args)
{
int arr[] = {3, 4, 3, 5, 2, 3, 4, 3, 1, 5};
int n = arr.length;

int idxOfEle = 5;
System.out.println(getIndexInSortedArray(arr,
n, idxOfEle));

}
}

// This code is contributed by Raghav sharma
```

## Python

```# Python program to get index of array element in
# sorted array
# Method returns the position of arr[idx] after
# performing stable-sort on array

def getIndexInSortedArray(arr, n, idx):
# Count of elements smaller than current
# element plus the equal element occurring
# before given index
result = 0
for i in range(n):
# If element is smaller then increase
# the smaller count
if (arr[i] < arr[idx]):
result += 1

# If element is equal then increase count
# only if it occurs before
if (arr[i] == arr[idx] and i < idx):
result += 1
return result;

#  Driver code to test above methods
arr = [3, 4, 3, 5, 2, 3, 4, 3, 1, 5]
n = len(arr)

idxOfEle = 5
print getIndexInSortedArray(arr, n, idxOfEle)

# Contributed by: Afzal Ansari

```

Output:

```4
```

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