# Removing a number from array to make it Geometric Progression

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++ program to find the element removing which // complete array becomes a GP #include 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; }

 // Java program to find the element removing which // complete array becomes a GP import java.util.*;   class GFG {       final static double EPS = (double) 1e-6;       // Utitilty method to compare two double values     static boolean fEqual(double a, double b) {         return (Math.abs(a - b) < EPS);     }       // Utility method to check, after deleting arr[ignore],     // remaining array is GP or not     static boolean 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     static 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 podouble, we know that elements from arr[0]                 * to arr[i-1] are in GP. Soarr[i] is the element                 * removing which may make GP. We check if removing                 * arr[i] actually makes it GP or not.                 */                 double[] temp = new double[N - i + 2];                 int k = 0;                 for (int j = i - 2; j < N; j++) {                     temp[k++] = arr[j];                 }                 return (isGP(temp, N - i + 2, 2)) ? i : -1;             }         }           return -1;     }       // Driver Code     public static void main(String[] args) {           double arr[] = { 2, 4, 8, 30, 16 };         int N = arr.length;           int index = makeGPbyRemovingOneElement(arr, N);         if (index == -1)             System.out.println("Not possible");         else             System.out.println("Remove " + arr[index] +                         " to get geometric progression");     } }   // This code is contributed by // sanjeev2552

 # Python program to find the element removing which # complete array becomes a GP   EPS = 1e-7   # Utitilty method to compare two double values def fEqual(a, b):     return abs(a - b) < EPS   # Utility method to check, after deleting arr[ignore], # remaining array is GP or not def isGP(arr, N, ignore):     last = -1     ratio = -1       for i in range(N):           # 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 = arr[i] / last                       # if ratio is not constant return false                 elif not fEqual(ratio, arr[i] / last):                     return False             last = arr[i]       return True   # Method return value removing which array becomes GP def makeGPbyRemovingOneElement(arr, 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     ratio = arr[1] / arr[0]       for i in range(2, N):         if not fEqual(ratio, 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 i if isGP(arr[i - 2:], N - i + 2, 2) else -1       return -1   # Driver Code if __name__ == "__main__":     arr = [2, 4, 8, 30, 16]     N = len(arr)       index = makeGPbyRemovingOneElement(arr, N)       if index == -1:         print("Not Possible")     else:         print("Remove %d to get geometric progression" % arr[index])   # This code is contributed by # sanjeev2552

 // C# program to find the element // removing which complete array // becomes a GP using System;    class GFG{    static double EPS = (double)1e-6;   // Utitilty method to compare two // double values static bool fEqual(double a, double b) {     return (Math.Abs(a - b) < EPS); }   // Utility method to check, after // deleting arr[ignore], remaining // array is GP or not static 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 static 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 podouble, we know that             // elements from arr[0] to arr[i-1]             // are in GP. Soarr[i] is the element             // removing which may make GP. We check             // if removing arr[i] actually makes             // it GP or not.             double[] temp = new double[N - i + 2];             int k = 0;             for(int j = i - 2; j < N; j++)             {                 temp[k++] = arr[j];             }             return (isGP(temp, N - i + 2, 2)) ? i : -1;         }     }     return -1; }   // Driver Code public static void Main(string[] args) {           double []arr = { 2, 4, 8, 30, 16 };     int N = arr.Length;       int index = makeGPbyRemovingOneElement(arr, N);     if (index == -1)         Console.Write("Not possible");     else         Console.Write("Remove " + arr[index] +                       " to get geometric progression"); } }   // This code is contributed by rutvik_56

Output:

Remove 30 to get geometric progression

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