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
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++ 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 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 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 |
// C# program to get index of array // element in sorted array using System;
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()
{
int [] arr = { 3, 4, 3, 5, 2, 3, 4, 3, 1, 5 };
int n = arr.Length;
int idxOfEle = 5;
Console.WriteLine(getIndexInSortedArray(arr, n,
idxOfEle));
}
} // This code is contributed by vt_m |
<?php // PHP program to get index of // array element in sorted array // Method returns the position of // arr[idx] after performing // stable-sort on array function getIndexInSortedArray( $arr , $n , $idx )
{ /* Count of elements smaller
than current element plus
the equal element occurring
before given index */
$result = 0;
for ( $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 ] and $i < $idx )
$result ++;
}
return $result ;
} // Driver Code
$arr = array (3, 4, 3, 5, 2, 3, 4, 3, 1, 5);
$n = count ( $arr );
$idxOfEle = 5;
echo getIndexInSortedArray( $arr , $n ,
$idxOfEle );
// This code is contributed by anuj_67. ?> |
<script> // JavaScript program to get index of array // element in sorted array // Method returns the position of
// arr[idx] after performing stable-sort
// on array
function getIndexInSortedArray(arr,
n, idx)
{
/* Count of elements smaller than
current element plus the equal element
occurring before given index*/
let result = 0;
for (let 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 let arr = [ 3, 4, 3, 5, 2, 3, 4, 3, 1, 5 ];
let n = arr.length;
let idxOfEle = 5;
document.write(getIndexInSortedArray(arr,
n, idxOfEle));
// This code is contributed by code_hunt. </script> |
4
Time Complexity: O(n) where n is the size of the array.
Auxiliary Space: O(1)