Minimum number of operations to convert a given sequence into a Geometric Progression (original) (raw)

Last Updated : 25 Apr, 2023

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++ `

// 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 possibilities = 3;

// candidate answer
int pos1 = -1, pos2 = -1;

// loop through all the permutations of the first two
// elements.
for (int i = 0; i < possibilities; i++) {
    for (int j = 0; j < possibilities; 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;

}

Java

// Java program to find minimum number // of operations to convert a given // sequence to an Geometric Progression import java.util.*;

class GFG {

static class pair { double first, second; public pair(double first, double second) { this.first = first; this.second = second; } } // Function to print the GP series static void construct(int n, pair ans_pair) { // Check for possibility if (ans_pair.first == -1) { System.out.print("Not possible"); return; } double a1 = ans_pair.first; double a2 = ans_pair.second; double r = a2 / a1;

System.out.print("The resultant sequence is:\n");
for (int i = 1; i <= n; i++)
{
    int ai = (int) (a1 * Math.pow(r, i - 1));
    System.out.print(ai + " ");
}

}

// Function for getting the Arithmetic Progression static void findMinimumOperations(double []a, int n) { int ans = Integer.MAX_VALUE;

// The array c describes all the given set of
// possible operations.
int c[] = { -1, 0, 1 };

// Size of c
int possibilities = 3;

// candidate answer
int pos1 = -1, pos2 = -1;

// loop through all the permutations of the first two
// elements.
for (int i = 0; i < possibilities; i++) 
{
    for (int j = 0; j < possibilities; 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 = (int) (Math.abs(a1 - a[1]) + Math.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 * Math.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 = Integer.MAX_VALUE; // set the temporary ans
                break; // to infinity and break
            }
        }

        // update answer
        if (temp < ans) 
        {
            ans = temp;
            pos1 = (int) a1;
            pos2 = (int) a2;
        }
    }
}
if (ans == -1) 
{
    System.out.print("-1");
    return;
}

System.out.print("Minimum Number of Operations are " + ans+ "\n");
pair ans_pair = new pair( pos1, pos2 );

// Calling function to print the sequence
construct(n, ans_pair);

}

// Driver Code public static void main(String[] args) {

// array is 1-indexed, with a[0] = 0
// for the sake of simplicity
double a[] = { 0, 7, 20, 49, 125 };

int n = a.length;

// Function to print the minimum operations
// and the sequence of elements
findMinimumOperations(a, n - 1);

} }

// This code is contributed by 29AjayKumar

Python3

Python program to find minimum number

of operations to convert a given

sequence to an Geometric Progression

from sys import maxsize as INT_MAX

Function to print the GP series

def construct(n: int, ans_pair: tuple):

# Check for possibility
if ans_pair[0] == -1:
    print("Not possible")
    return

a1 = ans_pair[0]
a2 = ans_pair[1]
r = a2 / a1

print("The resultant sequence is")
for i in range(1, n + 1):
    ai = a1 * pow(r, i - 1)
    print(int(ai), end=" ")

Function for getting the Arithmetic Progression

def findMinimumOperations(a: list, n: int): ans = INT_MAX

# The array c describes all the given set of
# possible operations.
c = [-1, 0, 1]

# Size of c
possibilities = 3

# candidate answer
pos1 = -1
pos2 = -1

# loop through all the permutations of the first two
# elements.
for i in range(possibilities):
    for j in range(possibilities):

        # a1 and a2 are the candidate first two elements
        # of the possible GP.
        a1 = a[1] + c[i]
        a2 = a[2] + c[j]

        # temp stores the current answer, including the
        # modification of the first two elements.
        temp = abs(a1 - a[1]) + abs(a2 - a[2])

        if a1 == 0 or a2 == 0:
            continue

        # common ratio of the possible GP
        r = a2 / a1

        # To check if the chosen set is valid, and id yes
        # find the number of operations it takes.
        for pos in range(3, n + 1):

            # 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))
            ai = a1 * pow(r, pos - 1)

            # Check for the "proposed" element to be only
            # differing by one
            if a[pos] == ai:
                continue
            elif a[pos] + 1 == ai or a[pos] - 1 == ai:
                temp += 1
            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:
    print("-1")
    return

print("Minimum number of Operations are", ans)
ans_pair = (pos1, pos2)

# Calling function to print the sequence
construct(n, ans_pair)

Driver Code

if name == "main":

# array is 1-indexed, with a[0] = 0
# for the sake of simplicity
a = [0, 7, 20, 49, 125]
n = len(a)

# Function to print the minimum operations
# and the sequence of elements
findMinimumOperations(a, n - 1)

This code is contributed by

sanjeev2552

C#

// C# program to find minimum number // of operations to convert a given // sequence to an Geometric Progression using System;

class GFG {

class pair { public double first, second; public pair(double first, double second) { this.first = first; this.second = second; } }

// Function to print the GP series static void construct(int n, pair ans_pair) { // Check for possibility if (ans_pair.first == -1) { Console.Write("Not possible"); return; } double a1 = ans_pair.first; double a2 = ans_pair.second; double r = a2 / a1;

Console.Write("The resultant sequence is:\n");
for (int i = 1; i <= n; i++)
{
    int ai = (int) (a1 * Math.Pow(r, i - 1));
    Console.Write(ai + " ");
}

}

// Function for getting the Arithmetic Progression static void findMinimumOperations(double []a, int n) { int ans = int.MaxValue;

// The array c describes all the given set of
// possible operations.
int []c = { -1, 0, 1 };

// Size of c
int possibilities = 3;

// candidate answer
int pos1 = -1, pos2 = -1;

// loop through all the permutations of the first two
// elements.
for (int i = 0; i < possibilities; i++) 
{
    for (int j = 0; j < possibilities; 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 = (int) (Math.Abs(a1 - a[1]) + Math.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 * Math.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.MaxValue; // set the temporary ans
                break; // to infinity and break
            }
        }

        // update answer
        if (temp < ans) 
        {
            ans = temp;
            pos1 = (int) a1;
            pos2 = (int) a2;
        }
    }
}
if (ans == -1) 
{
    Console.Write("-1");
    return;
}

Console.Write("Minimum Number of Operations are " + ans+ "\n");
pair ans_pair = new pair( pos1, pos2 );

// Calling function to print the sequence
construct(n, ans_pair);

}

// Driver Code public static void Main(String[] args) {

// array is 1-indexed, with a[0] = 0
// for the sake of simplicity
double []a = { 0, 7, 20, 49, 125 };

int n = a.Length;

// Function to print the minimum operations
// and the sequence of elements
findMinimumOperations(a, n - 1);

} }

// This code is contributed by Rajput-Ji

JavaScript

`

Output:

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

Time Complexity : O(9*N)

Space Complexity: O(1)