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

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: 2
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: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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; ` ` `  `    ``// 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 required ` `int` `recamanArray(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Set to store first n Recaman numbers ` `    ``unordered_set<``int``> 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 code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 7, 11, 20, 4, 2, 1, 8, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << recamanArray(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `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 required ` `static` `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 code ` `public` `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 approach ` `def` `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 required ` `def` `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 code ` `arr ``=` `[``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 approach  ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG  ` `{  ` ` `  `static` `int` `recamanGenerator(``int` `[]arr, ``int` `n)  ` `{  ` `    ``// First term of the sequence is always 0  ` `    ``arr = 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 required  ` `static` `int` `recamanArray(``int` `[]arr, ``int` `n)  ` `{  ` `     `  `    ``// Set to store first n Recaman numbers  ` `    ``HashSet<``int``> s=``new` `HashSet<``int``>();  ` ` `  `    ``// 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 code  ` `static` `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 `

Output:

```3
```

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