Skip to content
Related Articles

Related Articles

Improve Article
Count of elements which are not at the correct position
  • Last Updated : 04 Jun, 2021

Given an array arr[] of N elements and the task is to count the number of elements from this array which are not at the correct position. An element is said to be in an incorrect position if its position changes in the array when the array is sorted.
Examples: 
 

Input: arr[] = {1, 2, 6, 2, 4, 5} 
Output:
Array in the sorted form will be {1, 2, 2, 4, 5, 6}
Input: arr[] = {1, 2, 3, 4} 
Output:
All the elements are already sorted. 
 

 

Approach: First copy the array elements in another array say B[] then sort the given array. Start traversing the array and for every element if arr[i] != B[i] then it is the element which was not at the right position in the given array.
Below is the implementation of the above approach: 
 

C++






// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// elements which are not in
// the correct position when sorted
int cntElements(int arr[], int n)
{
 
    // To store a copy of the
    // original array
    int copy_arr[n];
 
    // Copy the elements of the given
    // array to the new array
    for (int i = 0; i < n; i++)
        copy_arr[i] = arr[i];
 
    // To store the required count
    int count = 0;
 
    // Sort the original array
    sort(arr, arr + n);
    for (int i = 0; i < n; i++) {
 
        // If current element was not
        // at the right position
        if (arr[i] != copy_arr[i]) {
            count++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 6, 2, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << cntElements(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
    // Function to return the count of
    // elements which are not in
    // the correct position when sorted
    static int cntElements(int arr[], int n)
    {
     
        // To store a copy of the
        // original array
        int copy_arr[] = new int[n];
     
        // Copy the elements of the given
        // array to the new array
        for (int i = 0; i < n; i++)
            copy_arr[i] = arr[i];
     
        // To store the required count
        int count = 0;
     
        // Sort the original array
        Arrays.sort(arr);
         
        for (int i = 0; i < n; i++)
        {
     
            // If current element was not
            // at the right position
            if (arr[i] != copy_arr[i])
            {
                count++;
            }
        }
        return count;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 1, 2, 6, 2, 4, 5 };
        int n = arr.length;
     
        System.out.println(cntElements(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach
 
# Function to return the count of
# elements which are not in
# the correct position when sorted
def cntElements(arr, n) :
 
    # To store a copy of the
    # original array
    copy_arr = [0] * n
 
    # Copy the elements of the given
    # array to the new array
    for i in range(n):
        copy_arr[i] = arr[i]
 
    # To store the required count
    count = 0
 
    # Sort the original array
    arr.sort()
    for i in range(n):
 
        # If current element was not
        # at the right position
        if (arr[i] != copy_arr[i]) :
            count += 1
             
    return count
     
# Driver code
arr = [ 1, 2, 6, 2, 4, 5 ]
n = len(arr)
 
print(cntElements(arr, n))
 
# This code is contributed by
# divyamohan123

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return the count of
    // elements which are not in
    // the correct position when sorted
    static int cntElements(int [] arr, int n)
    {
     
        // To store a copy of the
        // original array
        int [] copy_arr = new int[n];
     
        // Copy the elements of the given
        // array to the new array
        for (int i = 0; i < n; i++)
            copy_arr[i] = arr[i];
     
        // To store the required count
        int count = 0;
     
        // Sort the original array
        Array.Sort(arr);
         
        for (int i = 0; i < n; i++)
        {
     
            // If current element was not
            // at the right position
            if (arr[i] != copy_arr[i])
            {
                count++;
            }
        }
        return count;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int [] arr = { 1, 2, 6, 2, 4, 5 };
        int n = arr.Length;
     
        Console.WriteLine(cntElements(arr, n));
    }
}
 
// This code is contributed by Mohit kumar

Javascript




<script>
// Javascript implementation of the approach
 
 
// Function to return the count of
// elements which are not in
// the correct position when sorted
function cntElements(arr, n) {
 
    // To store a copy of the
    // original array
    let copy_arr = new Array(n);
 
    // Copy the elements of the given
    // array to the new array
    for (let i = 0; i < n; i++)
        copy_arr[i] = arr[i];
 
    // To store the required count
    let count = 0;
 
    // Sort the original array
    arr.sort((a, b) => a - b);
    for (let i = 0; i < n; i++) {
 
        // If current element was not
        // at the right position
        if (arr[i] != copy_arr[i]) {
            count++;
        }
    }
    return count;
}
 
// Driver code
 
let arr = [1, 2, 6, 2, 4, 5];
let n = arr.length;
 
document.write(cntElements(arr, n));
 
// This code is contributed by gfgking.
</script>
Output: 
4

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes




My Personal Notes arrow_drop_up
Recommended Articles
Page :