Given two Geometric Progressions (a1, r1) and (a2, r2) where (x, y) represents GP with initial term x and common ratio y and an integer N, the task is to find the count of the distinct integers that belong to the first N terms of at least one of the given geometric progressions.
Examples:
Input: N = 5, a1 = 3, r1 = 2, a2 = 6, r2 = 2
Output: 6
Explanation: The first 5 terms of the given geometric progressions are {3, 6, 12, 24, 48} and {6, 12, 24, 48, 96} respectively. Hence, the total count of distinct integers in the GP is 6.Input: N = 5, a1 = 3, r1 = 2, a2 = 2, r2 = 3
Output: 9
Explanation: The first 5 terms of the given geometric progressions are {3, 6, 12, 24, 48} and {2, 6, 18, 54, 162} respectively. Hence, the total count of distinct integers in the GP is 9.
Approach: The given problem can be solved by the observation that the total count of distinct integers can be calculated by generating the first N terms of both the Geometric Progressions and removing the duplicates terms. This can be achieved by the use of the set data structure. Firstly, generate all the N terms of the 1st GP and insert the terms into a set S. Similarly, insert the terms of the 2nd GP into the set S. The size of the set S is the required answer.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of distinct
// integers that belong to the first N
// terms of at least one of them is GP
int UniqueGeometricTerms(int N, int a1,
int r1, int a2,
int r2)
{
// Stores the integers that occur in
// GPs in a set data-structure
set<int> S;
// Stores the current integer of
// the first GP
long long p1 = a1;
// Iterate first N terms of first GP
for (int i = 0; i < N; i++) {
// Insert the ith term of GP in S
S.insert(p1);
p1 = (long long)(p1 * r1);
}
// Stores the current integer
// of the second GP
long long p2 = a2;
// Iterate first N terms of second GP
for (int i = 0; i < N; i++) {
S.insert(p2);
p2 = (long long)(p2 * r2);
}
// Return Answer
return S.size();
}
// Driver Code
int main()
{
int N = 5;
int a1 = 3, r1 = 2, a2 = 2, r2 = 3;
cout << UniqueGeometricTerms(
N, a1, r1, a2, r2);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the count of distinct
// integers that belong to the first N
// terms of at least one of them is GP
static int UniqueGeometricTerms(int N, int a1,
int r1, int a2,
int r2)
{
// Stores the integers that occur in
// GPs in a set data-structure
HashSet<Integer> S=new HashSet<Integer>();
// Stores the current integer of
// the first GP
int p1 = a1;
// Iterate first N terms of first GP
for (int i = 0; i < N; i++) {
// Insert the ith term of GP in S
S.add(p1);
p1 = (p1 * r1);
}
// Stores the current integer
// of the second GP
int p2 = a2;
// Iterate first N terms of second GP
for (int i = 0; i < N; i++) {
S.add(p2);
p2 = (p2 * r2);
}
// Return Answer
return S.size();
}
// Driver Code
public static void main(String[] args)
{
int N = 5;
int a1 = 3, r1 = 2, a2 = 2, r2 = 3;
System.out.print(UniqueGeometricTerms(
N, a1, r1, a2, r2));
}
}
// This code is contributed by shikhasingrajput
# Python 3 program for the above approach
# Function to find the count of distinct
# integers that belong to the first N
# terms of at least one of them is GP
def UniqueGeometricTerms(N, a1, r1, a2, r2):
# Stores the integers that occur in
# GPs in a set data-structure
S = set()
# Stores the current integer of
# the first GP
p1 = a1
# Iterate first N terms of first GP
for i in range(N):
# Insert the ith term of GP in S
S.add(p1)
p1 = (p1 * r1)
# Stores the current integer
# of the second GP
p2 = a2
# Iterate first N terms of second GP
for i in range(N):
S.add(p2)
p2 = (p2 * r2)
# Return Answer
return len(S)
# Driver Code
if __name__ == '__main__':
N = 5
a1 = 3
r1 = 2
a2 = 2
r2 = 3
print(UniqueGeometricTerms(N, a1, r1, a2, r2))
# This code is contributed by SURENDRA_GANGWAR.
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to find the count of distinct
// integers that belong to the first N
// terms of at least one of them is GP
static int UniqueGeometricTerms(int N, int a1,
int r1, int a2,
int r2)
{
// Stores the integers that occur in
// GPs in a set data-structure
HashSet<int> S=new HashSet<int>();
// Stores the current integer of
// the first GP
int p1 = a1;
// Iterate first N terms of first GP
for (int i = 0; i < N; i++) {
// Insert the ith term of GP in S
S.Add(p1);
p1 = (p1 * r1);
}
// Stores the current integer
// of the second GP
int p2 = a2;
// Iterate first N terms of second GP
for (int i = 0; i < N; i++) {
S.Add(p2);
p2 = (p2 * r2);
}
// Return Answer
return S.Count;
}
// Driver Code
public static void Main(string[] args)
{
int N = 5;
int a1 = 3, r1 = 2, a2 = 2, r2 = 3;
Console.Write(UniqueGeometricTerms(
N, a1, r1, a2, r2));
}
}
// This code is contributed by AnkThon
<script>
// JavaScript Program to implement
// the above approach
// Function to find the count of distinct
// integers that belong to the first N
// terms of at least one of them is GP
function UniqueGeometricTerms(N, a1,
r1, a2,
r2)
{
// Stores the integers that occur in
// GPs in a set data-structure
let S = new Set();
// Stores the current integer of
// the first GP
let p1 = a1;
// Iterate first N terms of first GP
for (let i = 0; i < N; i++) {
// Insert the ith term of GP in S
S.add(p1);
p1 = (p1 * r1);
}
// Stores the current integer
// of the second GP
let p2 = a2;
// Iterate first N terms of second GP
for (let i = 0; i < N; i++) {
S.add(p2);
p2 = (p2 * r2);
}
// Return Answer
return S.size;
}
// Driver Code
let N = 5;
let a1 = 3, r1 = 2, a2 = 2, r2 = 3;
document.write(UniqueGeometricTerms(
N, a1, r1, a2, r2));
// This code is contributed by Potta Lokesh
</script>
Output:
9
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
