Lexicographically largest permutation of the array such that a[i] = a[i-1] + gcd(a[i-1], a[i-2])

Given an array arr[] of size N (N > 2). The task is to find lexicographically largest permutation of the array such that arr[i] = arr[i – 1] + gcd(arr[i – 1], arr[i – 2]). If it is not possible to find such arrangement then print -1.

Examples:

Input: arr[] = {4, 6, 2, 5, 3}
Output: 2 3 4 5 6
4 = 3 + gcd(2, 3)
5 = 4 + gcd(3, 4)
6 = 5 + gcd(4, 5)



Input: arr[] = {1, 6, 8}
Output: -1

Approach: If you are thinking about a solution that would involve sorting the array and then checking if the gcd condition holds. You are partly right, the numbers have to be in increasing sequence but except for one case where there could be a number that could appear at the start of the permutation. For Example, arr[] = {2, 4, 6, 8, 8} in this case, 8 can be placed at the starting of the array to get the permutation {8, 2, 4, 6, 8}.
Corner cases:

  • You couldn’t have more than two elements whose freq was more than 1.
  • If you had two zeros in the array, the only possible permutation possible was all 0’s

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find elements of vector
void Print(vector<int>& ans)
{
    for (auto i : ans)
        cout << i << " ";
}
  
// Function to find the lexicographically largest
// permutation that satisfies the given condition
void Permutation(int a[], int n)
{
    int flag = 0, pos;
  
    // To store the required ans
    vector<int> ans;
  
    // Sort the array
    sort(a, a + n);
  
    for (int i = 2; i < n; i++) {
  
        // If need to make arranagement
        if (a[i] != a[i - 1] + __gcd(a[i - 1], a[i - 2])) {
            flag = 1;
            pos = i;
            break;
        }
    }
  
    // If possible then check for lexographically
    // larger permutation (if any possible)
    if (flag == 0) {
  
        // If larger arrangement is possible
        if (a[1] == a[0] + __gcd(a[0], a[n - 1])) {
            ans.push_back(a[n - 1]);
            for (int i = 0; i < n - 1; i++)
                ans.push_back(a[i]);
  
            Print(ans);
            return;
        }
  
        // If no other arrangement is possible
        else {
            for (int i = 0; i < n; i++)
                ans.push_back(a[i]);
  
            Print(ans);
            return;
        }
    }
  
    // Need to re-arrange the array
    else {
  
        // If possible, place at first position
        if (a[1] == a[0] + __gcd(a[pos], a[0])) {
            flag = 0;
            for (int i = n - 1; i > pos + 2; i--) {
  
                // If even after one arrangement its impossible
                // to get the required array
                if (a[i] != a[i - 1] + __gcd(a[i - 1], a[i - 2])) {
                    flag = 1;
                    break;
                }
            }
  
            if (flag == 0 and pos < n - 1) {
  
                // If it is not possible to get
                // the required array
                if (a[pos + 1]
                    != a[pos - 1] + __gcd(a[pos - 1], a[pos - 2]))
                    flag = 1;
            }
  
            if (flag == 0 and pos < n - 2) {
  
                // If it is not possible to get
                // the required array
                if (a[pos + 2]
                    != a[pos + 1] + __gcd(a[pos - 1], a[pos + 1]))
                    flag = 1;
            }
  
            // If it is possible to get the answer
            if (flag == 0) {
                ans.push_back(a[pos]);
                for (int i = 0; i < n; i++)
                    if (i != pos)
                        ans.push_back(a[i]);
  
                Print(ans);
                return;
            }
        }
    }
  
    ans.push_back(-1);
    Print(ans);
}
  
// Driver code
int main()
{
    int a[] = { 4, 6, 2, 8, 8 };
    int n = sizeof(a) / sizeof(a[0]);
  
    Permutation(a, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to find elements of vector
static void Print(Vector<Integer> ans)
{
    for (Integer i : ans)
        System.out.print(i + " ");
}
  
// Function to find the lexicographically largest
// permutation that satisfies the given condition
static void Permutation(int a[], int n)
{
    int flag = 0, pos = 0;
  
    // To store the required ans
    Vector<Integer> ans = new Vector<Integer>();
  
    // Sort the array
    Arrays.sort(a);
  
    for (int i = 2; i < n; i++) 
    {
  
        // If need to make arranagement
        if (a[i] != a[i - 1] + __gcd(a[i - 1],
                                     a[i - 2]))
        {
            flag = 1;
            pos = i;
            break;
        }
    }
  
    // If possible then check for lexographically
    // larger permutation (if any possible)
    if (flag == 0)
    {
  
        // If larger arrangement is possible
        if (a[1] == a[0] + __gcd(a[0], 
                                 a[n - 1])) 
        {
            ans.add(a[n - 1]);
            for (int i = 0; i < n - 1; i++)
                ans.add(a[i]);
  
            Print(ans);
            return;
        }
  
        // If no other arrangement is possible
        else 
        {
            for (int i = 0; i < n; i++)
                ans.add(a[i]);
  
            Print(ans);
            return;
        }
    }
  
    // Need to re-arrange the array
    else 
    {
  
        // If possible, place at first position
        if (a[1] == a[0] + __gcd(a[pos], a[0])) 
        {
            flag = 0;
            for (int i = n - 1; i > pos + 2; i--)
            {
  
                // If even after one arrangement 
                // its impossible to get
                // the required array
                if (a[i] != a[i - 1] + __gcd(a[i - 1], 
                                             a[i - 2]))
                {
                    flag = 1;
                    break;
                }
            }
  
            if (flag == 0 & pos < n - 1
            {
  
                // If it is not possible to get
                // the required array
                if (a[pos + 1]
                    != a[pos - 1] + __gcd(a[pos - 1], 
                                          a[pos - 2]))
                    flag = 1;
            }
  
            if (flag == 0 & pos < n - 2
            {
  
                // If it is not possible to get
                // the required array
                if (a[pos + 2]
                    != a[pos + 1] + __gcd(a[pos - 1], 
                                          a[pos + 1]))
                    flag = 1;
            }
  
            // If it is possible to get the answer
            if (flag == 0)
            {
                ans.add(a[pos]);
                for (int i = 0; i < n; i++)
                    if (i != pos)
                        ans.add(a[i]);
  
                Print(ans);
                return;
            }
        }
    }
  
    ans.add(-1);
    Print(ans);
}
  
static int __gcd(int a, int b) 
    if (b == 0
        return a; 
    return __gcd(b, a % b);     
  
// Driver code
public static void main(String[] args)
{
    int a[] = { 4, 6, 2, 8, 8 };
    int n = a.length;
  
    Permutation(a, n);
    }
}
  
// This code is contributed 
// by PrinciRaj1992 

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Python3

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# Python 3 implementation of the approach
from math import gcd
  
# Function to find elements of vector
def Print(ans):
    for i in range(len(ans)):
        print(ans[i], end = " ")
  
# Function to find the lexicographically 
# largest permutation that satisfies 
# the given condition
def Permutation(a, n):
    flag = 0
  
    # To store the required ans
    ans = []
  
    # Sort the array
    a.sort(reverse = False)
  
    for i in range(2, n, 1):
          
        # If need to make arranagement
        if (a[i] != a[i - 1] + 
        gcd(a[i - 1], a[i - 2])):
            flag = 1
            pos = i
            break
  
    # If possible then check for 
    # lexographically larger 
    # permutation (if any possible)
    if (flag == 0):
          
        # If larger arrangement is possible
        if (a[1] == a[0] + 
        gcd(a[0], a[n - 1])):
            ans.append(a[n - 1])
            for i in range(n - 1):
                ans.append(a[i])
  
            Print(ans)
            return
  
        # If no other arrangement is possible
        else:
            for i in range(n):
                ans.append(a[i])
  
            Print(ans)
            return
  
    # Need to re-arrange the array
    else:
          
        # If possible, place at first position
        if (a[1] == a[0] + 
        gcd(a[pos], a[0])):
            flag = 0
            i = n - 1
            while(i > pos + 2):
                  
                # If even after one arrangement its 
                # impossible to get the required array
                if (a[i] != a[i - 1] + 
                gcd(a[i - 1], a[i - 2])):
                    flag = 1
                    break
  
                i -= 1
              
            if (flag == 0 and pos < n - 1):
                  
                # If it is not possible to get
                # the required array
                if (a[pos + 1] != a[pos - 1] + 
                gcd(a[pos - 1], a[pos - 2])):
                    flag = 1
  
            if (flag == 0 and pos < n - 2):
                  
                # If it is not possible to get
                # the required array
                if (a[pos + 2] != a[pos + 1] +
                gcd(a[pos - 1], a[pos + 1])):
                    flag = 1
  
            # If it is possible to get the answer
            if (flag == 0):
                ans.append(a[pos])
                for i in range(n):
                    if (i != pos):
                        ans.append(a[i])
  
                Print(ans)
                return
  
    ans.append(-1)
    Print(ans)
  
# Driver code
if __name__ == '__main__':
    a = [4, 6, 2, 8, 8]
    n = len(a)
  
    Permutation(a, n)
      
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic; 
  
class GFG
{
  
// Function to find elements of vector
static void Print(List<int> ans)
{
    foreach (int i in ans)
        Console.Write(i + " ");
}
  
// Function to find the lexicographically largest
// permutation that satisfies the given condition
static void Permutation(int []a, int n)
{
    int flag = 0, pos = 0;
  
    // To store the required ans
    List<int> ans = new List<int>();
  
    // Sort the array
    Array.Sort(a);
  
    for (int i = 2; i < n; i++) 
    {
  
        // If need to make arranagement
        if (a[i] != a[i - 1] + __gcd(a[i - 1],
                                     a[i - 2]))
        {
            flag = 1;
            pos = i;
            break;
        }
    }
  
    // If possible then check for lexographically
    // larger permutation (if any possible)
    if (flag == 0)
    {
  
        // If larger arrangement is possible
        if (a[1] == a[0] + __gcd(a[0], 
                                 a[n - 1])) 
        {
            ans.Add(a[n - 1]);
            for (int i = 0; i < n - 1; i++)
                ans.Add(a[i]);
  
            Print(ans);
            return;
        }
  
        // If no other arrangement is possible
        else
        {
            for (int i = 0; i < n; i++)
                ans.Add(a[i]);
  
            Print(ans);
            return;
        }
    }
  
    // Need to re-arrange the array
    else
    {
  
        // If possible, place at first position
        if (a[1] == a[0] + __gcd(a[pos], a[0])) 
        {
            flag = 0;
            for (int i = n - 1; i > pos + 2; i--)
            {
  
                // If even after one arrangement 
                // its impossible to get
                // the required array
                if (a[i] != a[i - 1] + __gcd(a[i - 1], 
                                             a[i - 2]))
                {
                    flag = 1;
                    break;
                }
            }
  
            if (flag == 0 & pos < n - 1) 
            {
  
                // If it is not possible to get
                // the required array
                if (a[pos + 1]
                    != a[pos - 1] + __gcd(a[pos - 1], 
                                          a[pos - 2]))
                    flag = 1;
            }
  
            if (flag == 0 & pos < n - 2) 
            {
  
                // If it is not possible to get
                // the required array
                if (a[pos + 2]
                    != a[pos + 1] + __gcd(a[pos - 1], 
                                          a[pos + 1]))
                    flag = 1;
            }
  
            // If it is possible to get the answer
            if (flag == 0)
            {
                ans.Add(a[pos]);
                for (int i = 0; i < n; i++)
                    if (i != pos)
                        ans.Add(a[i]);
  
                Print(ans);
                return;
            }
        }
    }
  
    ans.Add(-1);
    Print(ans);
}
  
static int __gcd(int a, int b) 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b);     
  
// Driver code
public static void Main(String[] args)
{
    int []a = { 4, 6, 2, 8, 8 };
    int n = a.Length;
  
    Permutation(a, n);
    }
}
  
// This code is contributed by Rajput-Ji

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Output:

8 2 4 6 8

Time complexity: O(NlogN)



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