Minimum De-arrangements present in array of AP (Arithmetic Progression)

Last Updated : 13 Jun, 2022

Given an array of n-elements. Given array is a permutation of some Arithmetic Progression. Find the minimum number of De-arrangements present in that array so as to make that array an Arithmetic progression.

Examples:

Input : arr[] = [8, 6, 10 ,4, 2]
Output : Minimum De-arrangement = 3
Explanation : arr[] = [10, 8, 6, 4, 2] is permutation 
which forms an AP and has minimum de-arrangements.

Input : arr[] = [5, 10, 15, 25, 20]
Output : Minimum De-arrangement = 2
Explanation : arr[] = [5, 10, 15, 20, 25] is permutation
which forms an AP and has minimum de-arrangements.

As per property of Arithmetic Progression our sequence will be either in increasing or decreasing manner. Also, we know that reverse of any Arithmetic Progression also form another Arithmetic Progression. So, we create a copy of original array and then once sort our given array in increase order and find total count of mismatch again after that we will reverse our sorted array and found new count of mismatch. Comparing both the counts of mismatch we can find the minimum number of de-arrangements.

C++

// CPP for counting minimum de-arrangements present
// in an array.
#include<bits/stdc++.h>
using namespace std;

// function to count Dearrangement
int countDe (int arr[], int n)
{
    // create a copy of original array
    vector <int> v (arr, arr+n);

    // sort the array
    sort(arr, arr+n);
    
    // traverse sorted array for counting mismatches
    int count1 = 0;
    for (int i=0; i<n; i++)   
        if (arr[i] != v[i])
            count1++;        
    
    // reverse the sorted array
    reverse(arr,arr+n);    

    // traverse reverse sorted array for counting 
    // mismatches
    int count2 = 0;
    for (int i=0; i<n; i++)
        if (arr[i] != v[i])       
            count2++;

    // return minimum mismatch count
    return (min (count1, count2));
}

// driver program
int main()
{
    int arr[] = {5, 9, 21, 17, 13};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Minimum Dearrangement = " << countDe(arr, n);
    return 0;
}

Java

// Java code for counting minimum 
// de-arrangements present in an array.
import java.util.*;
import java.lang.*;
import java.util.Arrays;

public class GeeksforGeeks{
    
    // function to count Dearrangement
    public static int countDe(int arr[], int n){
        int v[] = new int[n];
        
        // create a copy of original array
        for(int i = 0; i < n; i++)
            v[i] = arr[i]; 
            
        // sort the array
        Arrays.sort(arr);
    
        // traverse sorted array for 
        // counting mismatches
        int count1 = 0;
        for (int i = 0; i < n; i++) 
            if (arr[i] != v[i])
            count1++;     
    
        // reverse the sorted array
        Collections.reverse(Arrays.asList(arr));

        // traverse reverse sorted array 
        // for counting mismatches
        int count2 = 0;
        for (int i = 0; i < n; i++)
            if (arr[i] != v[i])     
                count2++;

        // return minimum mismatch count
        return (Math.min (count1, count2));
    }

    // driver code
    public static void main(String argc[]){
        int arr[] = {5, 9, 21, 17, 13};
        int n = 5;
        System.out.println("Minimum Dearrangement = "+
                            countDe(arr, n));
    }
}

/*This code is contributed by Sagar Shukla.*/

Python3

# Python3 code for counting minimum 
# de-arrangements present in an array.

# function to count Dearrangement
def countDe(arr, n):

        i = 0

        # create a copy of 
        # original array
        v = arr.copy()
            
        # sort the array
        arr.sort()
    
        # traverse sorted array for 
        # counting mismatches
        count1 = 0
        i = 0
        while( i < n ): 
            if (arr[i] != v[i]):
                count1 = count1 + 1
            i = i + 1
    
        # reverse the sorted array
        arr.sort(reverse=True)

        # traverse reverse sorted array 
        # for counting mismatches
        count2 = 0
        i = 0
        while( i < n ):
            if (arr[i] != v[i]):     
                count2 = count2 + 1
            i = i + 1

        # return minimum mismatch count
        return (min (count1, count2))

# Driven code
arr = [5, 9, 21, 17, 13]
n = 5
print ("Minimum Dearrangement =",countDe(arr, n))

# This code is contributed by "rishabh_jain".

// C# code for counting 
// minimum de-arrangements 
// present in an array.
using System;

class GFG
{

// function to count
// Dearrangement
public static int countDe(int[] arr, 
                          int n)
{
    int[] v = new int[n];
    
    // create a copy
    // of original array
    for(int i = 0; i < n; i++)
        v[i] = arr[i]; 
        
    // sort the array
    Array.Sort(arr);

    // traverse sorted array for 
    // counting mismatches
    int count1 = 0;
    for (int i = 0; i < n; i++) 
        if (arr[i] != v[i])
        count1++; 

    // reverse the sorted array
    Array.Reverse(arr);

    // traverse reverse sorted array 
    // for counting mismatches
    int count2 = 0;
    for (int i = 0; i < n; i++)
        if (arr[i] != v[i]) 
            count2++;

    // return minimum 
    // mismatch count
    return (Math.Min (count1, count2));
}

// Driver code
public static void Main()
{
    int[] arr = new int[]{5, 9, 21, 17, 13};
    int n = 5;
    Console.WriteLine("Minimum Dearrangement = " + 
                                 countDe(arr, n));
}
}

// This code is contributed by mits

PHP

<?php
// PHP for counting minimum de-arrangements 
// present in an array.

// function to count Dearrangement
function countDe ($arr, $n)
{
    // create a copy of original array
    $v = $arr;

    // sort the array
    sort($arr);
    
    // traverse sorted array for 
    // counting mismatches
    $count1 = 0;
    for ($i = 0; $i < $n; $i++) 
        if ($arr[$i] != $v[$i])
            $count1++; 
    
    // reverse the sorted array
    rsort($arr); 

    // traverse reverse sorted array 
    // for counting mismatches
    $count2 = 0;
    for ($i = 0; $i < $n; $i++)
        if ($arr[$i] != $v[$i]) 
            $count2++;

    // return minimum mismatch count
    return (min($count1, $count2));
}

// Driver Code
$arr = array(5, 9, 21, 17, 13);
$n = count($arr);
echo "Minimum Dearrangement = " . 
               countDe($arr, $n);

// This code is contributed by mits
?>

JavaScript

<script>

// JavaScript code for counting minimum 
// de-arrangements present in an array

// Function to count Dearrangement
function countDe(arr, n)
{
    let v = [];
      
    // Create a copy of original array
    for(let i = 0; i < n; i++)
        v[i] = arr[i]; 
          
    // Sort the array
    arr.sort();
  
    // Traverse sorted array for 
    // counting mismatches
    let count1 = 0;
    for(let i = 0; i < n; i++) 
        if (arr[i] != v[i])
            count1++;     
  
    // Reverse the sorted array
    arr.reverse();

    // Traverse reverse sorted array 
    // for counting mismatches
    let count2 = 0;
    for(let i = 0; i < n; i++)
        if (arr[i] != v[i])     
            count2++;

    // Return minimum mismatch count
    return (Math.min (count1, count2));
}

// Driver Code
let arr = [ 5, 9, 21, 17, 13 ];
let n = 5;

document.write("Minimum Dearrangement = " +
               countDe(arr, n));
               
// This code is contributed by sanjoy_62

</script>

Output:

Minimum Dearrangement = 2

Time Complexity: O(nlogn).

Auxiliary Space: O(n)

Change one element in the given array to make it an Arithmetic Progression

Shivam.Pradhan

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Minimum De-arrangements present in array of AP (Arithmetic Progression)

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