Maximize sum of GCD of two subsets of given Array
Last Updated :
13 Dec, 2022
Given an array arr[] of positive integers of size N, the task is to divide the array into two non-empty subsets X and Y in such a way that the sum of their GCD turns out to be the maximum possible
Examples:
Input: N = 3, arr[] = {1, 2, 9}
Output: 10
Explanation: We can divide the array as
X = (1, 2) with GCD = 1
Y = (9) with GCD = 9
Maximum sum comes as 10
Input: N = 4, arr[] = {7, 21, 27, 28}
Output: 34
Explanation: Possible divisions :
X= 7, 21, 28 with GCD = 7 and Y = 27 with GCD =27. Sum = 34
X = 7, 21, 27 with GCD = 1 and Y = 28 with GCD =28. Sum = 29
Use the first way for the maximum sum possible i.e., 34.
Approach: The problem can be solved based on the following idea:
To get the maximum GCD sum, keep the highest one in one subset (say X) and the others in another (say Y).
But it can give raise to a case where the second maximum unique element causes the GCD of Y = 1 but the maximum element when considered in Y and the second max in X then the GCD sum may be greater. (Happens when the maximum element when kept with elements of Y subset the GCD is greater than 1 but when 2nd largest was in Y it was 1).
We don't need to consider for 3rd largest because if both first and second largest are divisible by the GCD then there difference is greater than the GCD[say a] and 3rd largest + a < 1 + largest as largest > 3rd largest+a
- So for gaining max gcd sum we will take gcd of smallest n-2 unique elements.
- Find including which one gives the greater sum.
Follow the below steps to understand better:
- First sort the array arr[].
- Traverse arr from i = 0 to i = N-1 and Find all the unique elements.
- If there was only one element then division in subsets is not possible.
- If only one unique element but more than once, then the same element will be present in both subsets.
- If there are only two unique elements after avoiding repetitions then return their sum.
- Take gcd of all elements except the last two:
- Check which is better to keep alone last or second last element.
- Return the maximum sum.
Below is the code for the above implementation:
C++
// C++ code for above approach
#include <bits/stdc++.h>
using namespace std;
// Function for finding maximum value sum of
// GCD of Subsets
int SubMaxGCD(int N, vector<int>& A)
{
// If only one element
if (A.size() == 1)
return -1;
// Sort the given array
sort(A.begin(), A.end());
vector<int> v;
// Running till last-1 element
for (int i = 0; i < N - 1; ++i) {
// Avoiding repetitions
// as GCD remains same
if (A[i] == A[i + 1])
continue;
else
v.push_back(A[i]);
}
// Pushing the last element
// It avoids the case of
// all the same element
v.push_back(A[N - 1]);
if (v.size() == 1) {
// Returning if v.size is 1
// i.e. all elements are same
return v[0] * 2;
}
// Corner case if there are only two
// elements after avoiding repetitions
if (v.size() == 2) {
return v[0] + v[1];
}
int n = v.size();
int g = v[0];
// Taking gcd of all elements
// except last two elements
for (int i = 1; i < n - 2; ++i) {
g = __gcd(g, v[i]);
}
// Check which is better to keep alone
// last or second last element
int gcd_a = __gcd(g, v[n - 2]) + v[n - 1];
int gcd_b = __gcd(g, v[n - 1]) + v[n - 2];
// Return the maximum sum.
if (gcd_a >= gcd_b)
return gcd_a;
else
return gcd_b;
}
// Driver code
int main()
{
int N = 3;
vector<int> arr = { 1, 2, 9 };
// Function call
cout << SubMaxGCD(N, arr);
return 0;
}
// Java code for above approach
import java.io.*;
import java.util.*;
class GFG {
public static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function for finding maximum value sum of
// GCD of Subsets
public static int SubMaxGCD(int N, int A[])
{
// If only one element
if (A.length == 1)
return -1;
// Sort the given array
Arrays.sort(A);
ArrayList<Integer> v = new ArrayList<>();
// Running till last-1 element
for (int i = 0; i < N - 1; ++i) {
// Avoiding repetitions
// as GCD remains same
if (A[i] == A[i + 1])
continue;
else
v.add(A[i]);
}
// Pushing the last element
// It avoids the case of
// all the same element
v.add(A[N - 1]);
if (v.size() == 1) {
// Returning if v.size is 1
// i.e. all elements are same
return v.get(0) * 2;
}
// Corner case if there are only two
// elements after avoiding repetitions
if (v.size() == 2) {
return v.get(0) + v.get(1);
}
int n = v.size();
int g = v.get(0);
// Taking gcd of all elements
// except last two elements
for (int i = 1; i < n - 2; ++i) {
g = gcd(g, v.get(i));
}
// Check which is better to keep alone
// last or second last element
int gcd_a = gcd(g, v.get(n - 2)) + v.get(n - 1);
int gcd_b = gcd(g, v.get(n - 1)) + v.get(n - 2);
// Return the maximum sum.
if (gcd_a >= gcd_b)
return gcd_a;
else
return gcd_b;
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
int arr[] = { 1, 2, 9 };
// Function call
System.out.print(SubMaxGCD(N, arr));
}
}
// This code is contributed by Rohit Pradhan
# python3 code for above approach
def gcd( a, b) :
if (b == 0) :
return a
return gcd(b, a % b)
# Function for finding maximum value sum of
# GCD of Subsets
def SubMaxGCD( N, A) :
# If only one element
if len(A) == 1 :
return -1
# Sort the given array
A.sort()
v=[]
# Running till last-1 element
for i in range(N-1) :
# Avoiding repetitions
# as GCD remains same
if A[i] == A[i + 1] :
continue
else :
v.append(A[i])
# Pushing the last element
# It avoids the case of
# all the same element
v.append(A[N - 1])
if len(v) == 1 :
# Returning if v.size is 1
# i.e. all elements are same
return v[0] * 2
# Corner case if there are only two
# elements after avoiding repetitions
if len(v) == 2 :
return v[0] + v[1]
n = len(v)
g = v[0]
# Taking gcd of all elements
# except last two elements
for i in range(N-2):
g = gcd(g, v[i])
# Check which is better to keep alone
# last or second last element
gcd_a = gcd(g, v[n - 2]) + v[n - 1]
gcd_b = gcd(g, v[n - 1]) + v[n - 2]
# Return the maximum sum.
if gcd_a >= gcd_b :
return gcd_a
else :
return gcd_b
# Driver Code
if __name__ == "__main__":
N = 3
arr = [ 1, 2, 9 ]
# Function call
print(SubMaxGCD(N, arr))
# This code is contributed by jana_sayantan.
// C# code for above approach
using System;
using System.Collections;
class GFG {
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function for finding maximum value sum of
// GCD of Subsets
static int SubMaxGCD(int N, int[] A)
{
// If only one element
if (A.Length == 1)
return -1;
// Sort the given array
Array.Sort(A);
ArrayList v = new ArrayList();
// Running till last-1 element
for (int i = 0; i < N - 1; ++i) {
// Avoiding repetitions
// as GCD remains same
if (A[i] == A[i + 1])
continue;
else
v.Add(A[i]);
}
// Pushing the last element
// It avoids the case of
// all the same element
v.Add(A[N - 1]);
if (v.Count == 1) {
// Returning if v.size is 1
// i.e. all elements are same
return (int)v[0] * 2;
}
// Corner case if there are only two
// elements after avoiding repetitions
if (v.Count == 2) {
return (int)v[0] + (int)v[1];
}
int n = v.Count;
int g = (int)v[0];
// Taking gcd of all elements
// except last two elements
for (int i = 1; i < n - 2; ++i) {
g = gcd(g, (int)v[i]);
}
// Check which is better to keep alone
// last or second last element
int gcd_a = gcd(g, (int)v[n - 2]) + (int)v[n - 1];
int gcd_b = gcd(g, (int)v[n - 1]) + (int)v[n - 2];
// Return the maximum sum.
if (gcd_a >= gcd_b)
return gcd_a;
else
return gcd_b;
}
// Driver Code
public static void Main()
{
int N = 3;
int[] arr = { 1, 2, 9 };
// Function call
Console.Write(SubMaxGCD(N, arr));
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// JavaScript code for above approach
// Function for GCD
const __gcd = (a, b) => {
if (b == 0) return a;
return __gcd(b, a % b);
}
// Function for finding maximum value sum of
// GCD of Subsets
const SubMaxGCD = (N, A) => {
// If only one element
if (A.length == 1)
return -1;
// Sort the given array
A.sort();
let v = [];
// Running till last-1 element
for (let i = 0; i < N - 1; ++i) {
// Avoiding repetitions
// as GCD remains same
if (A[i] == A[i + 1])
continue;
else
v.push(A[i]);
}
// Pushing the last element
// It avoids the case of
// all the same element
v.push(A[N - 1]);
if (v.length == 1) {
// Returning if v.size is 1
// i.e. all elements are same
return v[0] * 2;
}
// Corner case if there are only two
// elements after avoiding repetitions
if (v.length == 2) {
return v[0] + v[1];
}
let n = v.length;
let g = v[0];
// Taking gcd of all elements
// except last two elements
for (let i = 1; i < n - 2; ++i) {
g = __gcd(g, v[i]);
}
// Check which is better to keep alone
// last or second last element
let gcd_a = __gcd(g, v[n - 2]) + v[n - 1];
let gcd_b = __gcd(g, v[n - 1]) + v[n - 2];
// Return the maximum sum.
if (gcd_a >= gcd_b)
return gcd_a;
else
return gcd_b;
}
// Driver code
let N = 3;
let arr = [1, 2, 9];
// Function call
document.write(SubMaxGCD(N, arr));
// This code is contributed by rakeshsahni.
</script>
Time Complexity: O(N * logN)
Auxiliary Space: O(N)
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