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 = 4Input: 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 requiredint 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 codeint 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 approachimport java.util.*;import java.util.concurrent.ConcurrentHashMap;class GFG{ // Function to return the minimum cost requiredstatic 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 codepublic 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 approachfrom math import gcd as __gcd# Function to return the minimum cost requireddef 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 codearr = [ 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 requiredfunction 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 codelet 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
4
Complexity Analysis:
- Time Complexity: O(n2 log(n)) where n is the number of elements.
- Auxiliary Space: O(n)
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