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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find elements of vectorvoid Print(vector<int>& ans){ for (auto i : ans) cout << i << " ";}// Function to find the lexicographically largest// permutation that satisfies the given conditionvoid 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 codeint 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 approachimport java.util.*;class GFG{// Function to find elements of vectorstatic void Print(Vector<Integer> ans){ for (Integer i : ans) System.out.print(i + " ");}// Function to find the lexicographically largest// permutation that satisfies the given conditionstatic 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 codepublic 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 approachfrom math import gcd# Function to find elements of vectordef Print(ans): for i in range(len(ans)): print(ans[i], end = " ")# Function to find the lexicographically # largest permutation that satisfies # the given conditiondef 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 codeif __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 approachusing System;using System.Collections.Generic; class GFG{// Function to find elements of vectorstatic void Print(List<int> ans){ foreach (int i in ans) Console.Write(i + " ");}// Function to find the lexicographically largest// permutation that satisfies the given conditionstatic 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 codepublic 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
<script>// JavaScript implementation of the approach// Function to find elements of vectorfunction Print(ans) { for (let i of ans) document.write(i + " ");}// Function to find the lexicographically largest// permutation that satisfies the given conditionfunction 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 codelet a = [4, 6, 2, 8, 8];let n = a.length;Permutation(a, n);// This code is contributed by _saurabh_jaiswal</script> |
Output:
8 2 4 6 8
Time complexity: O(NlogN)
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