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# Minimum array elements to be changed to make Recaman’s sequence

• Last Updated : 12 May, 2021

Given an array arr[] of N elements. The task is to find the minimum number of elements to be changed in the array such that the array contains first N Recaman’s Sequence terms. Note that Recaman terms may be present in any order in the array.
First few terms of Recaman’s Sequence are:

0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, …..

Examples:

Input: arr[] = {44, 0, 2, 3, 9}
Output:
N = 5 and first 5 Recaman Numbers are 0, 1, 3, 6 and 2
44 and 9 must be replaced with 6 and 1
Hence 2 changes are required.
Input: arr[] = {0, 33, 3, 1}
Output:

Approach:

• Insert first N Recaman’s Sequence terms in a set.
• Traverse the array from left to right and check if array element is present in the set.
• If current element is present in the set that remove it from the set.
• Minimum changes required is the size of the final reduced set.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; int recamanGenerator(int arr[], int n){    // First term of the sequence is always 0    arr[0] = 0;     // Fill remaining terms using recursive    // formula    for (int i = 1; i <= n; i++) {        int temp = arr[i - 1] - i;        int j;         for (j = 0; j < i; j++) {             // If arr[i-1] - i is negative or            // already exists            if ((arr[j] == temp) || temp < 0) {                temp = arr[i - 1] + i;                break;            }        }         arr[i] = temp;    }} // Function that returns minimum changes requiredint recamanArray(int arr[], int n){     // Set to store first n Recaman numbers    unordered_set s;     // Generate and store    // first n Recaman numbers    int recaman[n];    recamanGenerator(recaman, n);     // Insert first n Recaman numbers to set    for (int i = 0; i < n; i++)        s.insert(recaman[i]);     for (int i = 0; i < n; i++) {         // If current element of the array        // is present in the set        auto it = s.find(arr[i]);        if (it != s.end())            s.erase(it);    }     // Return the remaining number of    // elements in the set    return s.size();} // Driver codeint main(){     int arr[] = { 7, 11, 20, 4, 2, 1, 8, 6 };    int n = sizeof(arr) / sizeof(arr[0]);     cout << recamanArray(arr, n);     return 0;}

## Java

 // Java implementation of the approachimport java.util.*; class GFG{ static int recamanGenerator(int arr[], int n){    // First term of the sequence is always 0    arr[0] = 0;     // Fill remaining terms using recursive    // formula    for (int i = 1; i <= n; i++)    {        int temp = arr[i - 1] - i;        int j;         for (j = 0; j < i; j++)        {             // If arr[i-1] - i is negative or            // already exists            if ((arr[j] == temp) || temp < 0)            {                temp = arr[i - 1] + i;                break;            }        }         arr[i] = temp;    }    return 0;} // Function that returns minimum changes requiredstatic int recamanArray(int arr[], int n){          // Set to store first n Recaman numbers    Set s=new HashSet();     // Generate and store    // first n Recaman numbers    int recaman[]=new int[n+1];    recamanGenerator(recaman, n);     // Insert first n Recaman numbers to set    for (int i = 0; i < n; i++)        s.add(recaman[i]);     for (int i = 0; i < n; i++)    {        // If current element of the array        // is present in the set        if (s.contains(arr[i]))            s.remove(arr[i]);    }     // Return the remaining number of    // elements in the set    return s.size();} // Driver codepublic static void main(String args[]){     int arr[] = { 7, 11, 20, 4, 2, 1, 8, 6 };    int n = arr.length;     System.out.print( recamanArray(arr, n));}} // This code is contributed by Arnab Kundu

## Python3

 # Python3 implementation of the approachdef recamanGenerator(arr, n):         # First term of the sequence    # is always 0    arr[0] = 0     # Fill remaining terms using    # recursive formula    for i in range(1, n):        temp = arr[i - 1] - i        j = 0         for j in range(i):             # If arr[i-1] - i is negative or            # already exists            if ((arr[j] == temp) or temp < 0):                temp = arr[i - 1] + i                break         arr[i] = temp # Function that returns minimum# changes requireddef recamanArray(arr, n):     # Set to store first n Recaman numbers    s = dict()     # Generate and store    # first n Recaman numbers    recaman = [0 for i in range(n)]    recamanGenerator(recaman, n)     # Insert first n Recaman numbers to set    for i in range(n):        s[recaman[i]] = s.get(recaman[i], 0) + 1     for i in range(n):         # If current element of the array        # is present in the set        if arr[i] in s.keys():            del s[arr[i]]     # Return the remaining number of    # elements in the set    return len(s) # Driver codearr = [7, 11, 20, 4, 2, 1, 8, 6 ]n = len(arr) print(recamanArray(arr, n)) # This code is contributed# by mohit kumar

## C#

 // C# implementation of the approachusing System;using System.Collections.Generic; class GFG{ static int recamanGenerator(int []arr, int n){    // First term of the sequence is always 0    arr[0] = 0;     // Fill remaining terms using recursive    // formula    for (int i = 1; i <= n; i++)    {        int temp = arr[i - 1] - i;        int j;         for (j = 0; j < i; j++)        {             // If arr[i-1] - i is negative or            // already exists            if ((arr[j] == temp) || temp < 0)            {                temp = arr[i - 1] + i;                break;            }        }         arr[i] = temp;    }    return 0;} // Function that returns minimum changes requiredstatic int recamanArray(int []arr, int n){         // Set to store first n Recaman numbers    HashSet s=new HashSet();     // Generate and store    // first n Recaman numbers    int[] recaman=new int[n+1];    recamanGenerator(recaman, n);     // Insert first n Recaman numbers to set    for (int i = 0; i < n; i++)        s.Add(recaman[i]);     for (int i = 0; i < n; i++)    {        // If current element of the array        // is present in the set        if (s.Contains(arr[i]))            s.Remove(arr[i]);    }     // Return the remaining number of    // elements in the set    return s.Count;} // Driver codestatic void Main(){     int []arr = { 7, 11, 20, 4, 2, 1, 8, 6 };    int n = arr.Length;     Console.Write( recamanArray(arr, n));}} // This code is contributed by mits

## Javascript


Output:
3

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