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Minimum deletions required to make GCD of the array equal to 1

  • Difficulty Level : Easy
  • Last Updated : 21 Apr, 2021

Given an array arr[] of N integers, the task is to find the minimum deletions required to make the GCD of the resulting array elements equal to 1. If it is impossible then print -1.
Examples: 
 

Input: arr[] = {2, 4, 6, 3} 
Output:
It is clear that GCD(2, 4, 6, 3) = 1 
So, we do not need to delete any elements.
Input: arr[] = {8, 14, 16, 26} 
Output: -1 
No matter how many elements get deleted, the gcd will never be 1. 
 

 

Approach: If the GCD of the initial array is 1 then we do not need to delete any of the elements and the result will be 0. If GCD is not 1 then the GCD can never be 1 no matter what elements we delete. Let’s say gcd be the gcd of the elements of the array and we delete k elements. Now, N – k elements are left and they still have gcd as their factor. So, it is impossible to get GCD of the array elements equal to 1.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum
// deletions required
int MinDeletion(int a[], int n)
{
 
    // To store the GCD of the array
    int gcd = 0;
    for (int i = 0; i < n; i++)
        gcd = __gcd(gcd, a[i]);
 
    // GCD cannot be 1
    if (gcd > 1)
        return -1;
 
    // GCD of the elements is already 1
    else
        return 0;
}
 
// Driver code
int main()
{
    int a[] = { 3, 6, 12, 81, 9 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << MinDeletion(a, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
    // Recursive function to return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0)
        return b;
        if (b == 0)
        return a;
         
        // base case
        if (a == b)
            return a;
         
        // a is greater
        if (a > b)
            return __gcd(a-b, b);
        return __gcd(a, b-a);
    }
 
    // Function to return the minimum
    // deletions required
    static int MinDeletion(int a[], int n)
    {
     
        // To store the GCD of the array
        int gcd = 0;
        for (int i = 0; i < n; i++)
            gcd = __gcd(gcd, a[i]);
     
        // GCD cannot be 1
        if (gcd > 1)
            return -1;
     
        // GCD of the elements is already 1
        else
            return 0;
    }
 
 
    // Driver code
    public static void main (String[] args)
    {
        int a[] = { 3, 6, 12, 81, 9 };
        int n = a.length;
 
        System.out.print(MinDeletion(a, n));
    }
}
 
// This code is contributed by anuj_67..

Python3




# Python3 implementation of the approach
from math import gcd
 
# Function to return the minimum
# deletions required
def MinDeletion(a, n) :
 
    # To store the GCD of the array
    __gcd = 0;
    for i in range(n) :
        __gcd = gcd(__gcd, a[i]);
 
    # GCD cannot be 1
    if (__gcd > 1) :
        return -1;
 
    # GCD of the elements is already 1
    else :
        return 0;
 
# Driver code
if __name__ == "__main__" :
 
    a = [ 3, 6, 12, 81, 9 ];
    n = len(a)
 
    print(MinDeletion(a, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Recursive function to return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;
         
        // base case
        if (a == b)
            return a;
         
        // a is greater
        if (a > b)
            return __gcd(a-b, b);
        return __gcd(a, b-a);
    }
 
    // Function to return the minimum
    // deletions required
    static int MinDeletion(int []a, int n)
    {
     
        // To store the GCD of the array
        int gcd = 0;
        for (int i = 0; i < n; i++)
            gcd = __gcd(gcd, a[i]);
     
        // GCD cannot be 1
        if (gcd > 1)
            return -1;
     
        // GCD of the elements is already 1
        else
            return 0;
    }
 
 
    // Driver code
    public static void Main ()
    {
        int []a = { 3, 6, 12, 81, 9 };
        int n = a.Length;
 
        Console.WriteLine(MinDeletion(a, n));
    }
}
 
// This code is contributed by anuj_67..

Javascript




<script>
// javascript implementation of the approach
 
    // Recursive function to return gcd of a and b
    function __gcd(a , b) {
        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;
 
        // base case
        if (a == b)
            return a;
 
        // a is greater
        if (a > b)
            return __gcd(a - b, b);
        return __gcd(a, b - a);
    }
 
    // Function to return the minimum
    // deletions required
    function MinDeletion(a , n) {
 
        // To store the GCD of the array
        var gcd = 0;
        for (i = 0; i < n; i++)
            gcd = __gcd(gcd, a[i]);
 
        // GCD cannot be 1
        if (gcd > 1)
            return -1;
 
        // GCD of the elements is already 1
        else
            return 0;
    }
 
    // Driver code
     
        var a = [ 3, 6, 12, 81, 9 ];
        var n = a.length;
 
        document.write(MinDeletion(a, n));
 
// This code is contributed by aashish1995
</script>
Output: 



-1

 

Time Complexity: O(N)
 

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