Removing a number from array to make it Geometric Progression

3

Given an array arr[] of N positive elements, the task is to find whether it is possible to convert this array into Geometric Progression (GP) by removing at-most one element. If yes, then find index of the number removing which converts the array into a geometric progression.
Special Cases :
1) If whole array is already in GP, then return any index.
2) If it is not possible to convert array into GP, then print “Not possible”.

Examples:

Input  : arr[] = [2, 4, 8, 24, 16, 32]
Output : 3
Number to remove is arr[3], i.e., 24
After removing 24 array will be [2, 4, 8, 16, 32] which 
is a GP with starting value 2 and common ratio 2.

Input  : arr[] = [2, 8, 30, 60]
Output : Not Possible

We can solve this problem by handling some special cases and then finding the pivot element, removing which makes array a GP. First we will check that our pivot element is first or second element, if not then the multiplier between them will be common ration of our GP, if yes then we found our solution.
Once we get common ratio of GP, we can check array element with this ratio, if this ratio violates at some index, then we skip this element and check from next index whether it is a continuation of previous GP or not.
In below code a method isGP is implemented which checks array to be GP after removing element at index ‘index’. This method is written for special case handling of first, second and last element.
Please see below code for better understanding.

//  C/C++ program to find the element removing which
// complete array becomes a GP
#include <bits/stdc++.h>
using namespace std;
#define EPS 1e-6

//  Utitilty method to compare two double values
bool fEqual(double a, double b)
{
    return (abs(a - b) < EPS);
}

//  Utility method to check, after deleting arr[ignore],
//  remaining array is GP or not
bool isGP(double arr[], int N, int ignore)
{
    double last = -1;
    double ratio = -1;

    for (int i = 0; i < N; i++)
    {
        //  check ratio only if i is not ignore
        if (i != ignore)
        {
            //  last will be -1 first time
            if (last != -1)
            {
                //  ratio will be -1 at first time
                if (ratio == -1)
                    ratio = (double)arr[i] / last;

                //  if ratio is not constant return false
                else if (!fEqual(ratio, (double)arr[i] / last))
                    return false;
            }
            last = arr[i];
        }
    }
    return true;
}

//  method return value removing which array becomes GP
int makeGPbyRemovingOneElement(double arr[], int N)
{
    /*  solving special cases separately */
    //  Try removing first element
    if (isGP(arr, N, 0))
        return 0;

    //  Try removing second element
    if (isGP(arr, N, 1))
        return 1;

    //  Try removing last element
    if (isGP(arr, N, N-1))
        return (N-1);

    /*  now we know that first and second element will be
        part of our GP so getting constant ratio of our GP */
    double ratio = (double)arr[1]/arr[0];
    for (int i = 2; i < N; i++)
    {
        if (!fEqual(ratio, (double)arr[i]/arr[i-1]))
        {
             /* At this point, we know that elements from arr[0]
               to arr[i-1] are in GP. So arr[i] is the element
               removing which may make GP.  We check if removing
               arr[i] actually makes it GP or not. */
            return (isGP(arr+i-2, N-i+2, 2))? i : -1;
         }
    }

    return -1;
}

//  Driver code to test above method
int main()
{
    double arr[] = {2, 4, 8, 30, 16};
    int N = sizeof(arr) / sizeof(arr[0]);

    int index = makeGPbyRemovingOneElement(arr, N);
    if (index == -1)
        cout << "Not possible\n";
    else
        cout << "Remove " << arr[index]
             << " to get geometric progression\n";

    return 0;
}

Output:

Remove 30 to get geometric progression

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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