Rearrange array to make it non-decreasing by swapping pairs having GCD equal to minimum array element

Given an array, arr[] consisting of N positive integers, the task is to make the array non-decreasing by swapping pairs (arr[i], arr[j]) such that i != j (1 ≤ i, j ≤ n) and GCD(arr[i], arr[j]) is equal to the minimum element present in the array.

Examples:

Input: arr[] = {4, 3, 6, 6, 2, 9}
Output: Yes
Explanation: 
Minimum array element = 2. 
Swap arr[0] and arr[2], since gcd(4, 6) = 2. Array modifies to {6, 3, 4, 6, 2, 9}. 
Swap arr[0] and arr[4], since gcd(6, 2) = 2. 
Therefore, the modified array {2, 3, 4, 6, 6, 9} is non-decreasing.

Input: arr[] = {7, 5, 2, 2, 4}
Output: No

Approach:

1. Firstly, traverse the array to find the minimum element. Store all these elements in another array and sort that array.
2. For every array element, check if it is at the correct position or not. If found to be true, proceed to next element. Otherwise, check if it is divisible by the smallest element of the array as only these elements will have GCD with other elements equal to the minimum element of the array.
3. If any array element is not at its correct position and that element is not divisible by the minimum element of the array, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if the array
// can be made non-decreasing
bool check(int a[], int n)
{
    int b[n];
    int minElement = INT_MAX;

    // Iterate till N
    for(int i = 0; i < n; i++)
    {

        // Find the minimum element
        b[i] = a[i];
        minElement = min(minElement, a[i]);
    }

    // Sort the array
    sort(b, b + n);
    int k = 1;

    // Iterate till N
    for(int i = 0; i < n; i++)
    {

        // Check if the element is
        // at its correct position
        if (a[i] != b[i] &&
            a[i] % minElement != 0)
        {
            k = 0;
            break;
        }
    }

    // Return the answer
    return k == 1 ? true : false;
}

// Driver Code
int main()
{
    int a[] = { 4, 3, 6, 6, 2, 9 };

    int n = sizeof(a) / sizeof(a[0]);

    // Print the answer
    if (check(a, n) == true)
        cout << "Yes \n";
    else
        cout<<"No \n";

    return 0;
}

// This code is contributed by akhilsaini

Java

// Java program for above approach

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

class GFG {

    // Function to check if the array
    // can be made non-decreasing
    public static boolean check(int[] a, int n)
    {
        int[] b = new int[n];
        int minElement = Integer.MAX_VALUE;

        // Iterate till N
        for (int i = 0; i < n; i++) {
            // Find the minimum element
            b[i] = a[i];
            minElement
                = Math.min(minElement, a[i]);
        }

        // Sort the array
        Arrays.sort(b);
        int k = 1;

        // Iterate till N
        for (int i = 0; i < n; i++) {
            // Check if the element is
            // at its correct position
            if (a[i] != b[i]
                && a[i] % minElement != 0) {
                k = 0;
                break;
            }
        }

        // Return the answer
        return k == 1 ? true : false;
    }

    // Driver Code
    public static void main(String[] args)
    {

        int[] a = { 4, 3, 6, 6, 2, 9 };

        int n = a.length;

        // Print the answer
        if (check(a, n) == true) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}

Python3

# Python3 program for above approach
import sys

# Function to check if the array
# can be made non-decreasing
def check(a, n):

    b = [None] * n
    minElement = sys.maxsize

    # Iterate till N
    for i in range(0, n):

        # Find the minimum element
        b[i] = a[i]
        minElement = min(minElement, a[i])

    # Sort the array
    b.sort()
    k = 1

    # Iterate till N
    for i in range(0, n):

        # Check if the element is
        # at its correct position
        if ((a[i] != b[i]) and
            (a[i] % minElement != 0)):
            k = 0
            break

    # Return the answer
    if k == 1:
        return True
    else:
        return False

# Driver Code
if __name__ == "__main__":

    a = [ 4, 3, 6, 6, 2, 9 ]

    n = len(a)

    # Print the answer
    if check(a, n) == True:
        print("Yes")
    else:
        print("No")

# This code is contributed by akhilsaini

C#

// C# program for above approach
using System;

class GFG{

// Function to check if the array
// can be made non-decreasing
static bool check(int[] a, int n)
{
    int[] b = new int[n];
    int minElement = int.MaxValue;

    // Iterate till N
    for(int i = 0; i < n; i++)
    {

        // Find the minimum element
        b[i] = a[i];
        minElement = Math.Min(minElement, a[i]);
    }

    // Sort the array
    Array.Sort(b);
    int k = 1;

    // Iterate till N
    for(int i = 0; i < n; i++)
    {

        // Check if the element is
        // at its correct position
        if (a[i] != b[i] &&
            a[i] % minElement != 0)
        {
            k = 0;
            break;
        }
    }

    // Return the answer
    return k == 1 ? true : false;
}

// Driver Code
static public void Main()
{
    int[] a = { 4, 3, 6, 6, 2, 9 };

    int n = a.Length;

    // Print the answer
    if (check(a, n) == true)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by akhilsaini

Javascript

<script>

// Javascript program to implement
// the above approach

   // Function to check if the array
    // can be made non-decreasing
    function check(a, n)
    {
        let b = new Array(n).fill(0);
        let minElement = Number.MAX_VALUE;

        // Iterate till N
        for (let i = 0; i < n; i++) {
            // Find the minimum element
            b[i] = a[i];
            minElement
                = Math.min(minElement, a[i]);
        }

        // Sort the array
        b.sort();
        let k = 1;

        // Iterate till N
        for (let i = 0; i < n; i++) {
            // Check if the element is
            // at its correct position
            if (a[i] != b[i]
                && a[i] % minElement != 0) {
                k = 0;
                break;
            }
        }

        // Return the answer
        return k == 1 ? true : false;
    }

// Driver Code

    let a = [ 4, 3, 6, 6, 2, 9 ];

        let n = a.length;

        // Print the answer
        if (check(a, n) == true) {
            document.write("Yes");
        }
        else {
           document.write("No");
        }

 // This code is contributed by avijitmondal1998.
</script>

Output:

Yes

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