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

• Difficulty Level : Hard
• Last Updated : 10 Mar, 2022

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 couldnâ€™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 0â€™s

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``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

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to find elements of vector``static` `void` `Print(Vector 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 ans = ``new` `Vector();` `    ``// 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`

## Python3

 `# 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#

 `// 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`

## Javascript

 ``

Output:

`8 2 4 6 8`

Time complexity: O(NlogN)

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