Given three integers A, B, and C, the task is to find out a positive integer K (if exists) such that if k is multiplied by any of the given integers A, B, or C they form G.P.(Geometric Progression) in the given order. If there exists no K print -1.
Examples:
Input: A = 1, B = 9, C = 27
Output: 3
Explanation: If A is multiplied by 3 then the integers 3, 9, and 27 forms a G.P. with a common ratio of 3.Input: A = 3, B = 2, C = 12
Output: 3
Explanation: If B is multiplied by 3 then the integers 3, 6, and 12 form a G.P. with a common ratio of 2.Input: A = 2, B = 6, C = 8
Output: -1
Explanation: There exists no positive integer K
Approach: Implement the idea below to solve the problem:
Three numbers A, B, and C form a G.P. if and only if C / B = B / A = r. ( where r is the common ratio ). For three integers we have 3 cases:
- Case 1: A is multiplied by K, Now the integers become A*K, B, and C. Using the G.P. property we have C / B = B / A*K => K = ( B*B ) / ( A*C )
- Case 2: B is multiplied by K, Now the integers become A, B*K, and C. Using the G.P. property we have C / B*k = B*K / A => K = sqrt(A*C)/ B
- Case 3: C is multiplied by K, Now the integers become A, B, and C*K. Using the G.P. property we have C / B = B / A*K => K = ( B*B ) / ( A*C )
If any of the above 3 three cases yield a positive integer K then the answer(s) is/are found else return -1.
Follow these steps to solve the above problem:
- Check the cases mentioned above.
- If any of the cases return true, Print respectively K value for that case.
- Otherwise, return -1.
Below is the implementation of the above approach:
C++
// C++ code for the given approach#include <cmath>#include <iostream>using namespace std;
// Utility function to check if given a number is a// positive integerbool ifposInt(double num)
{ int a = (int)num;
if (num - a > 0) {
return false;
}
else {
return true;
}
}// Function to find any integer M if existsvoid findk(int A, int B, int C)
{ double k1 = (double)((B * B) / (A * C));
double k2 = (double)(sqrt(C * A) / B);
double k3 = (double)((B * B) / (A * C));
// Checks if possible values k are positive
// integers or not
if (k1 > 1 && ifposInt(k1)) {
cout << (int)k1 << endl;
}
else if (k2 > 1 && ifposInt(k2)) {
cout << (int)k2 << endl;
}
else if (k3 > 1 && ifposInt(k3)) {
cout << (int)k3 << endl;
}
else {
cout << "-1" << endl;
}
}int main()
{ int A = 3;
int B = 2;
int C = 12;
// Function call
findk(A, B, C);
return 0;
}// This code is contributed by lokeshmvs21. |
Java
// Java code for the given approachimport java.io.*;
class GFG {
// Utility function to check if given a number is a
// positive integer
static boolean ifposInt(double num)
{
int a = (int)num;
if (num - a > 0) {
return false;
}
else {
return true;
}
}
// Function to find any integer M if exists
static void findk(int A, int B, int C)
{
double k1 = (double)((B * B) / (A * C));
double k2 = (double)(Math.sqrt(C * A) / B);
double k3 = (double)((B * B) / (A * C));
// Checks if possible values k are positive
// integers or not
if (k1 > 1 && ifposInt(k1)) {
System.out.println((int)k1);
}
else if (k2 > 1 && ifposInt(k2)) {
System.out.println((int)k2);
}
else if (k3 > 1 && ifposInt(k3)) {
System.out.println((int)k3);
}
else {
System.out.println("-1");
}
}
public static void main(String[] args)
{
int A = 3;
int B = 2;
int C = 12;
// Function call
findk(A, B, C);
}
}// This code is contributed by lokesh. |
Python3
# Python3 code for the given approachimport math
# Utility function to check if given a# number is a positive integerdef ifposInt(num):
a = int(num)
if num - a > 0:
return False
else:
return True
# Function to find any integer# M if existsdef findk(A, B, C):
k1 = (B * B) / (A * C)
k2 = math.sqrt(C * A) / B
k3 = (B * B) / (A * C)
# Checks if possible values k are
# positive integers or not
if k1 > 1 and ifposInt(k1):
k1 = int(k1)
print(k1)
elif k2 > 1 and ifposInt(k2):
k2 = int(k2)
print(k2)
elif k3 > 1 and ifposInt(k3):
k3 = int(k3)
print(k3)
else:
print(-1)
# Driver codeA = 3
B = 2
C = 12
#Function callfindk(A, B, C)#This code is contributed by nikhilsainiofficial546 |
C#
// C# code for the given approachusing System;
class MainClass {
// Utility function to check if given a
// number is a positive integer
static bool ifposInt(double num)
{
int a = (int)num;
if (num - a > 0)
return false;
else
return true;
}
// Function to find any integer
// M if exists
static void findk(int A, int B, int C)
{
double k1 = (double)((B * B) / (A * C));
double k2 = (double)(Math.Sqrt(C * A) / B);
double k3 = (double)((B * B) / (A * C));
// Checks if possible values k are
// positive integers or not
if (k1 > 1 && ifposInt(k1))
Console.WriteLine(k1);
else if (k2 > 1 && ifposInt(k2))
Console.WriteLine(k2);
else if (k3 > 1 && ifposInt(k3))
Console.WriteLine(k3);
else
Console.WriteLine("-1");
}
//Drive Code
public static void Main(string[] args)
{
int A = 3;
int B = 2;
int C = 12;
// Function call
findk(A, B, C);
}
}// This code is contributed by nikhilsainiofficial546 |
Javascript
// Javascript code for the given approach// Utility function to check if given a number is a// positive integerfunction ifposInt(num)
{ let a = parseInt(num);
if (num - a > 0) {
return false;
}
else {
return true;
}
}// Function to find any integer M if existsfunction findk( A, B, C)
{ let k1 = ((B * B) / (A * C));
let k2 = (Math.sqrt(C * A) / B);
let k3 = ((B * B) / (A * C));
// Checks if possible values k are positive
// integers or not
if (k1 > 1 && ifposInt(k1)) {
console.log(parseInt(k1));
}
else if (k2 > 1 && ifposInt(k2)) {
console.log(parseInt(k2));
}
else if (k3 > 1 && ifposInt(k3)) {
console.log(parseInt(k3));
}
else {
console.log("-1");
}
} let A = 3;
let B = 2;
let C = 12;
// Function call
findk(A, B, C);
// This code is contributed by poojaagarwal2.
|
Output
3
Time Complexity: O(1)
Auxiliary Space: O(1)