Given a sequence of N elements, only three operations can be performed on any element at most one time. The operations are:

- Add one to the element.
- Subtract one from the element.
- Leave the element unchanged.

Perform any one of the operations on all elements in the array. The task is to find the minimum number of operations(addition and subtraction) that can be performed on the sequence, in order to convert it into a Geometric Progression. If it is not possible to generate a GP by performing the above operations, print -1.

**Examples**:

Input: a[] = {1, 1, 4, 7, 15, 33}

Output: The minimum number of operations are 4.

Steps:

- Keep a
_{1}unchanged- Add one to a
_{2}.- Keep a
_{3}unchanged- Subtract one from a
_{4}.- Subtract one from a
_{5}.- Add one to a
_{6}.The resultant sequence is {1, 2, 4, 8, 16, 32}

Input: a[] = {20, 15, 20, 15}

Output: -1

**Approach** The key observation to be made here is that any Geometric Progression is uniquely determined by only its first two elements (Since the ratio between each of the next pairs has to be the same as the ratio between this pair, consisting of the first two elements). Since only **3*3** permutations are possible. The possible combination of operations are (+1, +1), (+1, 0), (+1, -1), (-1, +1), (-1, 0), (-1, -1), (0, +1), (0, 0) and (0, -1). Using brute force all these **9** permutations and checking if they form a GP in linear time will give us the answer. The minimum of the operations which result in combinations which are in GP will be the answer.

Below is the implementation of the above approach:

// C++ program to find minimum number // of operations to convert a given // sequence to an Geometric Progression #include <bits/stdc++.h> using namespace std; // Function to print the GP series void construct(int n, pair<double, double> ans_pair) { // Check for possibility if (ans_pair.first == -1) { cout << "Not possible"; return; } double a1 = ans_pair.first; double a2 = ans_pair.second; double r = a2 / a1; cout << "The resultant sequence is:\n"; for (int i = 1; i <= n; i++) { double ai = a1 * pow(r, i - 1); cout << ai << " "; } } // Function for getting the Arithmetic Progression void findMinimumOperations(double* a, int n) { int ans = INT_MAX; // The array c describes all the given set of // possible operations. int c[] = { -1, 0, 1 }; // Size of c int possiblities = 3; // candidate answer int pos1 = -1, pos2 = -1; // loop through all the permutations of the first two // elements. for (int i = 0; i < possiblities; i++) { for (int j = 0; j < possiblities; j++) { // a1 and a2 are the candidate first two elements // of the possible GP. double a1 = a[1] + c[i]; double a2 = a[2] + c[j]; // temp stores the current answer, including the // modification of the first two elements. int temp = abs(a1 - a[1]) + abs(a2 - a[2]); if (a1 == 0 || a2 == 0) continue; // common ratio of the possible GP double r = a2 / a1; // To check if the chosen set is valid, and id yes // find the number of operations it takes. for (int pos = 3; pos <= n; pos++) { // ai is value of a[i] according to the assumed // first two elements a1, a2 // ith element of an GP = a1*((a2-a1)^(i-1)) double ai = a1 * pow(r, pos - 1); // Check for the "proposed" element to be only // differing by one if (a[pos] == ai) { continue; } else if (a[pos] + 1 == ai || a[pos] - 1 == ai) { temp++; } else { temp = INT_MAX; // set the temporary ans break; // to infinity and break } } // update answer if (temp < ans) { ans = temp; pos1 = a1; pos2 = a2; } } } if (ans == -1) { cout << "-1"; return; } cout << "Minimum Number of Operations are " << ans << "\n"; pair<double, double> ans_pair = { pos1, pos2 }; // Calling function to print the sequence construct(n, ans_pair); } // Driver Code int main() { // array is 1-indexed, with a[0] = 0 // for the sake of simplicity double a[] = { 0, 7, 20, 49, 125 }; int n = sizeof(a) / sizeof(a[0]); // Function to print the minimum operations // and the sequence of elements findMinimumOperations(a, n - 1); return 0; }

**Output:**

Minimum Number of Operations are 2 The resultant sequence is: 8 20 50 125

**Time Complexity** : O(9*N)

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