Check if GCD of Array can be made greater than 1 by replacing pairs with their products
Last Updated :
09 Feb, 2022
Given three integers L, R, and K. Consider an array arr[] consisting of all the elements from L to R, the task is to check whether the GCD of the array can be made greater than 1 using at most K operations. An operation is defined below:
- Choose any two numbers from the array
- Remove them from the array
- Insert their product back into the array
Examples:
Input: L = 4, R = 10, K = 3
Output: true
Explanation: Array will be arr[] = {4, 5, 6, 7, 8, 9, 10}
Choose arr[0], arr[1]: arr[] = {20, 6, 7, 8, 9, 10}
Choose arr[1], arr[2]: arr[] = {20, 42, 8, 9, 10}
Choose arr[2], arr[3]: arr[] = {20, 42, 72, 10}
GCD of the formed array = 2
Input: L = 3, R = 5, K = 1
Output: false
Explanation: Array will be arr[] = {3, 4, 5}]
Operation on arr[0], arr[1]: arr[] = {12, 5}, GCD = 1, or
Operation on arr[1], arr[2]: arr[] = {3, 20}, GCD = 1, or
Operation on arr[0], arr[2]: arr[] = {4, 15}, GCD = 1
Approach: The task can be solved by converting all the odd array elements to even so that the overall GCD of the array becomes even i.e greater than 1. To check if it is possible or not follow the cases below:
- Case 1: if L = R = 1 then GCD will always be 1, return false
- Case 2: if L = R (and L≠1) then GCD = L, return true
- Case 3: if K is greater equal to the number of odds between range L and R then return true
- If any of the above cases didn't imply then return false.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find that GCD of array is
// greater than 1 or
// not after at most K operations
bool gcdOfArray(int L, int R, int K)
{
// Finding number of integers
// between L and R
int range = (R - L + 1);
int even = 0;
int odd = 0;
// Finding number of odd and even integers
// in the given range
if (range % 2 == 0) {
even = range / 2;
odd = range - even;
}
else {
if (L % 2 != 0 || R % 2 != 0) {
odd = (range / 2) + 1;
even = range - odd;
}
else {
odd = range / 2;
even = range - odd;
}
}
// Case 1
if (L == R && L == 1)
return false;
// Case 2
if (L == R)
return true;
// Case 3
if (K >= odd)
return true;
// Otherwise not possible
else
return false;
}
// Driver Code
int main()
{
int L = 4;
int R = 10;
int K = 3;
bool isPossible = gcdOfArray(L, R, K);
if (isPossible)
cout << "true" << endl;
else
cout << "false" << endl;
return 0;
}
// JAVA program for the above approach
import java.util.*;
class GFG
{
// Function to find that GCD of array is
// greater than 1 or
// not after at most K operations
public static boolean gcdOfArray(int L, int R, int K)
{
// Finding number of integers
// between L and R
int range = (R - L + 1);
int even = 0;
int odd = 0;
// Finding number of odd and even integers
// in the given range
if (range % 2 == 0) {
even = range / 2;
odd = range - even;
}
else {
if (L % 2 != 0 || R % 2 != 0) {
odd = (range / 2) + 1;
even = range - odd;
}
else {
odd = range / 2;
even = range - odd;
}
}
// Case 1
if (L == R && L == 1)
return false;
// Case 2
if (L == R)
return true;
// Case 3
if (K >= odd)
return true;
// Otherwise not possible
else
return false;
}
// Driver Code
public static void main(String[] args)
{
int L = 4;
int R = 10;
int K = 3;
boolean isPossible = gcdOfArray(L, R, K);
if (isPossible)
System.out.println("true");
else
System.out.println("false");
}
}
// This code is contributed by Taranpreet
# Python program for the above approach
# Function to find that GCD of array is
# greater than 1 or
# not after at most K operations
def gcdOfArray(L, R, K):
# Finding number of integers
# between L and R
range = (R - L + 1)
even = 0
odd = 0
# Finding number of odd and even integers
# in the given range
if (range % 2 == 0):
even = range // 2
odd = range - even
else:
if (L % 2 != 0 or R % 2 != 0):
odd = (range // 2) + 1
even = range - odd
else:
odd = range // 2
even = range - odd
# Case 1
if (L == R and L == 1):
return False
# Case 2
if (L == R):
return True
# Case 3
if (K >= odd):
return True
# Otherwise not possible
else:
return False
# Driver Code
L = 4
R = 10
K = 3
isPossible = gcdOfArray(L, R, K)
if (isPossible):
print("true")
else:
print("false")
# This code is contributed by gfgking
// C# program for the above approach
using System;
public class GFG{
// Function to find that GCD of array is
// greater than 1 or
// not after at most K operations
public static bool gcdOfArray(int L, int R, int K)
{
// Finding number of integers
// between L and R
int range = (R - L + 1);
int even = 0;
int odd = 0;
// Finding number of odd and even integers
// in the given range
if (range % 2 == 0) {
even = range / 2;
odd = range - even;
}
else {
if (L % 2 != 0 || R % 2 != 0) {
odd = (range / 2) + 1;
even = range - odd;
}
else {
odd = range / 2;
even = range - odd;
}
}
// Case 1
if (L == R && L == 1)
return false;
// Case 2
if (L == R)
return true;
// Case 3
if (K >= odd)
return true;
// Otherwise not possible
else
return false;
}
// Driver Code
static public void Main (){
int L = 4;
int R = 10;
int K = 3;
bool isPossible = gcdOfArray(L, R, K);
if (isPossible)
Console.WriteLine("true");
else
Console.WriteLine("false");
}
}
// This code is contributed by Shubham Singh
<script>
// JavaScript program for the above approach
// Function to find that GCD of array is
// greater than 1 or
// not after at most K operations
const gcdOfArray = (L, R, K) => {
// Finding number of integers
// between L and R
let range = (R - L + 1);
let even = 0;
let odd = 0;
// Finding number of odd and even integers
// in the given range
if (range % 2 == 0) {
even = parseInt(range / 2);
odd = range - even;
}
else {
if (L % 2 != 0 || R % 2 != 0) {
odd = parseInt(range / 2) + 1;
even = range - odd;
}
else {
odd = parseInt(range / 2);
even = range - odd;
}
}
// Case 1
if (L == R && L == 1)
return false;
// Case 2
if (L == R)
return true;
// Case 3
if (K >= odd)
return true;
// Otherwise not possible
else
return false;
}
// Driver Code
let L = 4;
let R = 10;
let K = 3;
let isPossible = gcdOfArray(L, R, K);
if (isPossible)
document.write("true");
else
document.write("false");
// This code is contributed by rakeshsahni
</script>
Time Complexity: O(1)
Auxiliary Space: O(1)
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