Choose atleast two elements from array such that their GCD is 1 and cost is minimum

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Given two integer arrays arr[] and cost[] where cost[i] is the cost of choosing arr[i]. The task is to choose a subset with at least two elements such that the GCD of all the elements from the subset is 1 and the cost of choosing those elements is as minimum as possible then print the minimum cost.

Examples:

Input: arr[] = {5, 10, 12, 1}, cost[] = {2, 1, 2, 6}
Output: 4
{5, 12} is the required subset with cost = 2 + 2 = 4

Input: arr[] = {50, 100, 150, 200, 300}, cost[] = {2, 3, 4, 5, 6}
Output: -1
No subset possible with gcd = 1

Approach: Add GCD of any two elements to a map, now for every element arr[i] calculate its gcd with all the gcd values found so far (saved in the map) and update map[gcd] = min(map[gcd], map[gcd] + cost[i]). If in the end, map doesn’t contain any entry for gcd = 1 then print -1 else print the stored minimum cost.

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 cost required
int getMinCost(int arr[], int n, int cost[])
{
 
    // Map to store <gcd, cost> pair where
    // cost is the cost to get the current gcd
    map<int, int> mp;
    mp.clear();
    mp[0] = 0;
 
    for (int i = 0; i < n; i++) {
        for (auto it : mp) {
            int gcd = __gcd(arr[i], it.first);
 
            // If current gcd value already exists in map
            if (mp.count(gcd) == 1)
 
                // Update the minimum cost
                // to get the current gcd
                mp[gcd] = min(mp[gcd], it.second + cost[i]);
 
            else
                mp[gcd] = it.second + cost[i];
        }
    }
 
    // If there can be no sub-set such that
    // the gcd of all the elements is 1
    if (mp[1] == 0)
        return -1;
    else
        return mp[1];
}
 
// Driver code
int main()
{
    int arr[] = { 5, 10, 12, 1 };
    int cost[] = { 2, 1, 2, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << getMinCost(arr, n, cost);
    return 0;
}

Java

// Java implementation of the approach
 
import java.util.*;
import java.util.concurrent.ConcurrentHashMap;
 
class GFG{
  
// Function to return the minimum cost required
static int getMinCost(int arr[], int n, int cost[])
{
  
    // Map to store <gcd, cost> pair where
    // cost is the cost to get the current gcd
    Map<Integer,Integer> mp = new ConcurrentHashMap<Integer,Integer>();
    mp.clear();
    mp.put(0, 0);
  
    for (int i = 0; i < n; i++) {
        for (Map.Entry<Integer,Integer> it : mp.entrySet()){
            int gcd = __gcd(arr[i], it.getKey());
  
            // If current gcd value already exists in map
            if (mp.containsKey(gcd))
  
                // Update the minimum cost
                // to get the current gcd
                mp.put(gcd, Math.min(mp.get(gcd), it.getValue() + cost[i]));
  
            else
                mp.put(gcd,it.getValue() + cost[i]);
        }
    }
  
    // If there can be no sub-set such that
    // the gcd of all the elements is 1
    if (!mp.containsKey(1))
        return -1;
    else
        return mp.get(1);
}
static int __gcd(int a, int b)  
{  
    return b == 0? a:__gcd(b, a % b);     
} 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 5, 10, 12, 1 };
    int cost[] = { 2, 1, 2, 6 };
    int n = arr.length;
  
    System.out.print(getMinCost(arr, n, cost));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
from math import gcd as __gcd
 
# Function to return the minimum cost required
def getMinCost(arr, n, cost):
 
    # Map to store <gcd, cost> pair where
    # cost is the cost to get the current gcd
    mp = dict()
    mp[0] = 0
 
    for i in range(n):
        for it in list(mp):
            gcd = __gcd(arr[i], it)
 
            # If current gcd value 
            # already exists in map
            if (gcd in mp):
 
                # Update the minimum cost
                # to get the current gcd
                mp[gcd] = min(mp[gcd], 
                              mp[it] + cost[i])
 
            else:
                mp[gcd] = mp[it] + cost[i]
 
    # If there can be no sub-set such that
    # the gcd of all the elements is 1
    if (mp[1] == 0):
        return -1
    else:
        return mp[1]
 
# Driver code
arr = [ 5, 10, 12, 1]
cost = [ 2, 1, 2, 6]
n = len(arr)
 
print(getMinCost(arr, n, cost))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach 
using System;
using System.Collections.Generic;
using System.Linq;
class GFG
{
  static int __gcd(int a, int b)   
  {   
    return b == 0? a:__gcd(b, a % b);      
  }
 
  // Function to return the minimum cost required 
  static int getMinCost(int[] arr, int n, int[] cost) 
  {
 
    // Map to store <gcd, cost> pair where 
    // cost is the cost to get the current gcd 
    Dictionary<int, int> mp = new Dictionary<int, int>();
    mp.Add(0, 0);
    for (int i = 0; i < n; i++)
    {
 
      foreach (int it in mp.Keys.ToList())
      {
        int gcd = __gcd(arr[i], it);
 
        // If current gcd value already exists in map 
        if(mp.ContainsKey(gcd))
        {
 
          // Update the minimum cost 
          // to get the current gcd 
          mp[gcd] = Math.Min(mp[gcd], mp[it] + cost[i]);
        }
        else
        {
          mp.Add(gcd, mp[it] + cost[i]);
        }
      }  
    }
 
    // If there can be no sub-set such that 
    // the gcd of all the elements is 1 
    if (mp[1] == 0)
    {
      return -1;
    }
    else
    {
      return mp[1]; 
    }        
  }
 
  // Driver code
  static public void Main ()
  {
    int[] arr = { 5, 10, 12, 1 };
    int[] cost = { 2, 1, 2, 6 };
    int n = arr.Length; 
    Console.WriteLine(getMinCost(arr, n, cost));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript

<script>
 
// JavaScript implementation of the approach
 
// Function to return the minimum cost required
function getMinCost(arr,n,cost)
{
    // Map to store <gcd, cost> pair where
    // cost is the cost to get the current gcd
    let mp = new Map();
     
    mp.set(0, 0);
   
    for (let i = 0; i < n; i++) {
        for (let [key, value] of mp.entries()){
            let gcd = __gcd(arr[i], key);
   
            // If current gcd value already exists in map
            if (mp.has(gcd))
   
                // Update the minimum cost
                // to get the current gcd
                mp.set(gcd, Math.min(mp.get(gcd), value + cost[i]));
   
            else
                mp.set(gcd,value + cost[i]);
        }
    }
   
    // If there can be no sub-set such that
    // the gcd of all the elements is 1
    if (!mp.has(1))
        return -1;
    else
        return mp.get(1);
}
 
function __gcd(a,b)
{
    return b == 0? a:__gcd(b, a % b);
}
 
// Driver code
let arr=[5, 10, 12, 1 ];
let cost=[2, 1, 2, 6 ];
let n = arr.length;
document.write(getMinCost(arr, n, cost));
 
 
// This code is contributed by unknown2108
 
</script>

Output

Complexity Analysis:

Time Complexity: O(n² log(n)) where n is the number of elements.
Auxiliary Space: O(n)

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