Minimize Array sum by replacing an element with GCD

Given an array A[] of the length N. The task is to find the minimum sum of the array (i.e. A₁ + A₂ + … + A_N) by performing the following operation on the given array any number (0 or more) of times:

• Take any two indices i and j such that 0 ? i, j < N and swap either A_i or A_j with gcd(A_i, A_j).

Examples:

Input: A[] = {3, 3, 9} 
Output: 9
Explanation: Choose i = 0 and j = 2 and replace A?₃ with gcd(3, 9) = 3.
Now array A[]={3, 3, 3} and sum = 9 which is minimum possible sum of given array.

Input: A[] = {7, 14}
Output: 14

Approach 1:

The idea is to find the GCD of each pair of elements in the array and replace the larger element with the GCD. This will reduce the sum of the array as much as possible.

Algorithm:

• Calculate the GCD of each pair of elements in the array.
• Replace the larger element with the GCD.
• Repeat the above steps until all the elements in the array are equal.
• Calculate the sum of the array.
  

C++

#include <bits/stdc++.h>
using namespace std;

// Function to calculate GCD of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to calculate the minimum sum of the array
int minSum(int arr[], int n)
{
    // Iterate over all pairs of elements in the array
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // Calculate GCD of arr[i] and arr[j]
            int g = gcd(arr[i], arr[j]);

            // Replace arr[i] and arr[j] with GCD
            arr[i] = g;
            arr[j] = g;
        }
    }

    // Calculate sum of the array
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];

    return sum;
}

// Driver code
int main()
{
    int arr[] = { 3, 3, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << minSum(arr, n);

    return 0;
}

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

public class GFG {

    // Function to find gcd of two numbers
    public static int gcd(int a, int b)
    {
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

 // Function to find the minimum
// sum of array
    public static int minSum(int arr[], int n)
{
    // Iterate over all pairs of elements in the array
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // Calculate GCD of arr[i] and arr[j]
            int g = gcd(arr[i], arr[j]);

            // Replace arr[i] and arr[j] with GCD
            arr[i] = g;
            arr[j] = g;
        }
    }

    // Calculate sum of the array
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];

    return sum;
}

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 3, 3, 9 };
        int n = arr.length;

        // Function call
        System.out.println(minSum(arr, n));
    }
}

Python3

def gcd(a: int, b: int) -> int:
    if a == 0:
        return b
    return gcd(b % a, a)

# Function to calculate the minimum sum of the array
def min_sum(arr, n):

    # Iterate over all pairs of elements in the array
    for i in range(n):
        for j in range(i + 1, n):

            # Calculate GCD of arr[i] and arr[j]
            g = gcd(arr[i], arr[j])

            # Replace arr[i] and arr[j] with GCD
            arr[i] = g
            arr[j] = g

    # Calculate sum of the array
    s = 0
    for i in range(n):
        s += arr[i]

    return s

# Driver code
if __name__ == "__main__":
    arr = [3, 3, 9]
    n = len(arr)

    print(min_sum(arr, n))

    # This code is contributed by divya_p123.

C#

using System;

public class Program {
    // Function to find gcd of two numbers
    public static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // Function to find the minimum sum of array
    public static int minSum(int[] arr, int n)
    {
        // Iterate over all pairs of elements in the array
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                // Calculate GCD of arr[i] and arr[j]
                int g = gcd(arr[i], arr[j]);

                // Replace arr[i] and arr[j] with GCD
                arr[i] = g;
                arr[j] = g;
            }
        }

        // Calculate sum of the array
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];

        return sum;
    }

    static void Main(string[] args)
    {
        int[] arr = { 3, 3, 9 };
        int n = arr.Length;

        // Function call
        Console.WriteLine(minSum(arr, n));
    }
}

Javascript

<script>
// JavaScript code to implement the approach


    // Function to find gcd of two numbers
    function gcd(a,  b)
    {
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

 // Function to find the minimum
// sum of array
   function minSum( arr,n)
{
    // Iterate over all pairs of elements in the array
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            // Calculate GCD of arr[i] and arr[j]
           let g = gcd(arr[i], arr[j]);

            // Replace arr[i] and arr[j] with GCD
            arr[i] = g;
            arr[j] = g;
        }
    }

    // Calculate sum of the array
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += arr[i];

    return sum;
}

    // Driver code

        let arr = [3, 3, 9];
        let n = arr.length;

        // Function call
        console.log(minSum(arr, n));
    </script>

Output

9

Time Complexity: O(N²)
Auxiliary Space: O(1)

Approach 2: To solve the problem follow the below idea:

Let G = gcd(A₁, A₂, . . ., A_N). The answer is simply N*G. because every element can be converted to be the same as G.

Follow the below steps to solve the problem:

• Calculate the GCD of all the array elements.
• Multiply the result by N to get the minimum sum of the array.

Below is the implementation of the above approach.  

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find gcd of two numbers
int gcd(int p, int q) { return q == 0 ? p : gcd(q, p % q); }

// Function to find the minimum
// sum of array
int minSum(int A[], int N)
{

  int min = INT_MAX;
  for (int i = 0; i < N - 1; i++) {
    int b = gcd(A[i], A[i + 1]);
    if (b < min)
      min = b;
  }
  return min * N;
}

// Driver Code
int main()
{
  int A[] = { 3, 3, 9 };
  int N = sizeof(A) / sizeof(A[0]);;

  // Function call
  cout << minSum(A, N) << endl;

  return 0;
}

// This code is contributed by aarohirai2616.

Java

// Java code to implement the approach

import java.io.*;
import java.util.*;

public class GFG {

    // Function to find gcd of two numbers
    public static int gcd(int p, int q)
    {
        if (q == 0) {
            return p;
        }
        return gcd(q, p % q);
    }

    // Function to find the minimum
    // sum of array
    public static int minSum(int A[], int N)
    {
       int min = Integer.MAX_VALUE;
        for (int i = 0; i < N - 1; i++) {
            int b = gcd(A[i], A[i + 1]);
            if (b < min)
                min = b;
        }
        return min * N;
    }

    // Driver code
    public static void main(String[] args)
    {
        int A[] = { 3, 3, 9 };
        int N = A.length;

        // Function call
        System.out.println(minSum(A, N));
    }
}

Python3

# Python code for the above approach
import sys

# Function to find gcd of two numbers
def gcd(a,b):

    # Everything divides 0
    if (b == 0):
         return a
    return gcd(b, a%b)

# Function to find the minimum
# sum of array
def minSum(A, N) :

    min = sys.maxsize

    for i in range(0, N-1, 1):
        b = gcd(A[i], A[i + 1])
        if (b < min) :
            min = b

    return min * N


# Driver Code

A = [ 3, 3, 9 ]
N = len(A)

# Function call
print(minSum(A, N))

# This code is contributed by code_hunt.

C#

// C# code to implement the approach
using System;
class GFG
{

  // Function to find gcd of two numbers
  public static int gcd(int p, int q)
  {
    if (q == 0) {
      return p;
    }
    return gcd(q, p % q);
  }

  // Function to find the minimum
  // sum of array
  public static int minSum(int[] A, int N)
  {
    int min = Int32.MaxValue;
    for (int i = 0; i < N - 1; i++) {
      int b = gcd(A[i], A[i + 1]);
      if (b < min)
        min = b;
    }
    return min * N;
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int[] A = { 3, 3, 9 };
    int N = A.Length;

    // Function call
    Console.WriteLine(minSum(A, N));
  }
}

// This code is contributed by sanjoy_62.

Javascript

<script>
// Javascript code to implement the approach

// Function to find gcd of two numbers
function gcd(a, b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to find the minimum
// sum of array
function minSum(A,N)
{
    let min = Number.MAX_SAFE_INTEGER;
        for (let i = 0; i < N - 1; i++) {
            let b = gcd(A[i], A[i + 1]);
            if (b < min)
                min = b;
        }
        return min * N;
}

// Driver code
        let A = [ 3, 3, 9 ];
        let N = A.length;

        // Function call
        console.log(minSum(A, N));

// This code is contributed by aarohirai2616.
</script>

Output

9

Time Complexity: O(N) 
Auxiliary Space: O(1)