# Minimum array elements to be changed to make it a Lucas Sequence

Given an array with N distinct elements. The task is to find the minimum number of elements to be changed in the array such that, the array contains first N Lucas Sequence terms.

Note: Lucas terms may be present in any order in the array.

Examples:

Input : arr[] = {29, 1, 3, 4, 5, 11, 18, 2}
Output : 1
5 must be changed to 7, to get first N(8) terms of Lucas Sequence.
Hence, 1 change is required

Input : arr[] = {4, 2, 3, 1}
Output : 0
All elements are already first N(4) terms in Lucas sequence.

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

Approach:

• Insert first N(size of input array) Lucas Sequence terms in a set.
• Traverse array from left to right and check if array element is present in the set.
• If it is present that remove it from the set.
• Minimum changes required is the size of the final remaining set.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the minimum number ` `// of elements to be changed in the array ` `// to make it a Lucas Sequence ` `#include ` `using` `namespace` `std; ` ` `  `// Function that finds minimum changes to ` `// be made in the array ` `int` `lucasArray(``int` `arr[], ``int` `n) ` `{ ` `    ``set<``int``> s; ` ` `  `    ``// a and b are first two ` `    ``// lucas numbers ` `    ``int` `a = 2, b = 1; ` `    ``int` `c; ` ` `  `    ``// insert first n lucas elements to set ` `    ``s.insert(a); ` `    ``if` `(n >= 2) ` `        ``s.insert(b); ` ` `  `    ``for` `(``int` `i = 0; i < n - 2; i++) { ` `        ``s.insert(a + b); ` `        ``c = a + b; ` `        ``a = b; ` `        ``b = c; ` `    ``} ` ` `  `    ``set<``int``>::iterator it; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// if lucas element is present in array, ` `        ``// remove it from set ` `        ``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, 22, 4, 2, 1, 8, 9 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << lucasArray(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the minimum number ` `// of elements to be changed in the array ` `// to make it a Lucas Sequence ` `import` `java.util.HashSet; ` `import` `java.util.Set; ` ` `  `class` `GfG ` `{ ` ` `  `    ``// Function that finds minimum changes ` `    ``// to be made in the array ` `    ``static` `int` `lucasArray(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``HashSet s = ``new` `HashSet<>(); ` `     `  `        ``// a and b are first two lucas numbers ` `        ``int` `a = ``2``, b = ``1``, c; ` `     `  `        ``// insert first n lucas elements to set ` `        ``s.add(a); ` `        ``if` `(n >= ``2``) ` `            ``s.add(b); ` `     `  `        ``for` `(``int` `i = ``0``; i < n - ``2``; i++)  ` `        ``{ ` `            ``s.add(a + b); ` `            ``c = a + b; ` `            ``a = b; ` `            ``b = c; ` `        ``} ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``// if lucas element is present in array, ` `            ``// remove it from set ` `            ``if` `(s.contains(arr[i])) ` `                ``s.remove(arr[i]); ` `        ``} ` `     `  `        ``// return the remaining number of ` `        ``// elemnets in the set ` `        ``return` `s.size(); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `         `  `        ``int` `arr[] = { ``7``, ``11``, ``22``, ``4``, ``2``, ``1``, ``8``, ``9` `}; ` `        ``int` `n = arr.length; ` `     `  `        ``System.out.println(lucasArray(arr, n)); ` `    ``} ` `} ` `     `  `// This code is contributed by Rituraj Jain `

## Python3

 `# Python 3 program to find the minimum number ` `# of elements to be changed in the array ` `# to make it a Lucas Sequence ` ` `  `# Function that finds minimum changes to ` `# be made in the array ` `def` `lucasArray(arr, n): ` `    ``s ``=` `set``() ` ` `  `    ``# a and b are first two ` `    ``# lucas numbers ` `    ``a ``=` `2` `    ``b ``=` `1` ` `  `    ``# insert first n lucas elements to set ` `    ``s.add(a) ` `    ``if` `(n >``=` `2``): ` `        ``s.add(b) ` ` `  `    ``for` `i ``in` `range``(n ``-` `2``): ` `        ``s.add(a ``+` `b) ` `        ``c ``=` `a ``+` `b ` `        ``a ``=` `b ` `        ``b ``=` `c ` ` `  `    ``for` `i ``in` `range``(n): ` `         `  `        ``# if lucas element is present in array, ` `        ``# remove it from set ` `        ``if` `(arr[i] ``in` `s): ` `            ``s.remove(arr[i]) ` ` `  `    ``# return the remaining number of ` `    ``# elemnets in the set ` `    ``return` `len``(s) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``7``, ``11``, ``22``, ``4``, ``2``, ``1``, ``8``, ``9``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``print``(lucasArray(arr, n)) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# program to find the minimum number ` `// of elements to be changed in the array ` `// to make it a Lucas Sequence ` `using` `System; ` `using` `System.Collections.Generic;  ` `     `  `class` `GFG ` `{ ` ` `  `    ``// Function that finds minimum changes ` `    ``// to be made in the array ` `    ``static` `int` `lucasArray(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``HashSet<``int``> s = ``new` `HashSet<``int``>(); ` `     `  `        ``// a and b are first two lucas numbers ` `        ``int` `a = 2, b = 1, c; ` `     `  `        ``// insert first n lucas elements to set ` `        ``s.Add(a); ` `        ``if` `(n >= 2) ` `            ``s.Add(b); ` `     `  `        ``for` `(``int` `i = 0; i < n - 2; i++)  ` `        ``{ ` `            ``s.Add(a + b); ` `            ``c = a + b; ` `            ``a = b; ` `            ``b = c; ` `        ``} ` `     `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``// if lucas element is present in array, ` `            ``// remove it from set ` `            ``if` `(s.Contains(arr[i])) ` `                ``s.Remove(arr[i]); ` `        ``} ` `     `  `        ``// return the remaining number of ` `        ``// elemnets in the set ` `        ``return` `s.Count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `         `  `        ``int` `[]arr = { 7, 11, 22, 4, 2, 1, 8, 9 }; ` `        ``int` `n = arr.Length; ` `     `  `        ``Console.WriteLine(lucasArray(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992  `

Output:

```3
```

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