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.


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.


  • 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++ program to find the minimum number
// of elements to be changed in the array
// to make it a Lucas Sequence
#include <bits/stdc++.h>
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
    if (n >= 2)
    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())
    // return the remaining number of
    // elemnets 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[0]);
    cout << lucasArray(arr, n);
    return 0;




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