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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum cost required ` `int` `getMinCost(``int` `arr[], ``int` `n, ``int` `cost[]) ` `{ ` ` `  `    ``// Map to store pair where ` `    ``// cost is the cost to get the current gcd ` `    ``map<``int``, ``int``> mp; ` `    ``mp.clear(); ` `    ``mp = 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 == 0) ` `        ``return` `-1; ` `    ``else` `        ``return` `mp; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 5, 10, 12, 1 }; ` `    ``int` `cost[] = { 2, 1, 2, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << getMinCost(arr, n, cost); ` `    ``return` `0; ` `} `

 `# 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 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 `

Output:
```4
```

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Improved By : mohit kumar 29

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