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++ 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 arrangement
if (a[i] != a[i - 1] + __gcd(a[i - 1], a[i - 2])) {
flag = 1;
pos = i;
break ;
}
}
// If possible then check for lexicographically
// 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;
} |
// 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 arrangement
if (a[i] != a[i - 1 ] + __gcd(a[i - 1 ],
a[i - 2 ]))
{
flag = 1 ;
pos = i;
break ;
}
}
// If possible then check for lexicographically
// 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 |
# 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 arrangement
if (a[i] ! = a[i - 1 ] +
gcd(a[i - 1 ], a[i - 2 ])):
flag = 1
pos = i
break
# If possible then check for
# lexicographically 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 |
// 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 arrangement
if (a[i] != a[i - 1] + __gcd(a[i - 1],
a[i - 2]))
{
flag = 1;
pos = i;
break ;
}
}
// If possible then check for lexicographically
// 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 |
<script> // JavaScript implementation of the approach // Function to find elements of vector function Print(ans) {
for (let i of ans)
document.write(i + " " );
} // Function to find the lexicographically largest // permutation that satisfies the given condition function Permutation(a, n) {
let flag = 0, pos = 0;
// To store the required ans
let ans = new Array();
// Sort the array
a.sort((a, b) => a - b);
for (let i = 2; i < n; i++) {
// If need to make arrangement
if (a[i] != a[i - 1] + __gcd(a[i - 1],
a[i - 2])) {
flag = 1;
pos = i;
break ;
}
}
// If possible then check for lexicographically
// 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(a[n - 1]);
for (let i = 0; i < n - 1; i++)
ans.push(a[i]);
Print(ans);
return ;
}
// If no other arrangement is possible
else {
for (let i = 0; i < n; i++)
ans.push(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 (let 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.push(a[pos]);
for (let i = 0; i < n; i++)
if (i != pos)
ans.push(a[i]);
Print(ans);
return ;
}
}
}
ans.push(-1);
Print(ans);
} function __gcd(a, b) {
if (b == 0)
return a;
return __gcd(b, a % b);
} // Driver code let a = [4, 6, 2, 8, 8]; let n = a.length; Permutation(a, n); // This code is contributed by _saurabh_jaiswal </script> |
8 2 4 6 8
Time complexity: O(NlogN)