Minimum number of operations to convert a given sequence into a Geometric Progression

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

  1. Add one to the element.
  2. Subtract one from the element.
  3. 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:

  1. Keep a1 unchanged
  2. Add one to a2.
  3. Keep a3 unchanged
  4. Subtract one from a4.
  5. Subtract one from a5.
  6. Add one to a6.

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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