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: 0
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 requiredint 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 codeint 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 approachimport 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 approachusing 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> |
-1
Time Complexity: O(N)
Auxiliary Space: O(1)
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