Dada una array arr[] , la tarea es verificar si es posible ordenar la array dada usando cualquier cantidad de operaciones donde en cada operación, dos elementos arr[i] y arr[j] pueden intercambiarse si GCD de arr[ i] y arr[j] es 1 .
Ejemplo:
Entrada: a = {3, 2, 1}
Salida: Posible
explicación: La array dada se puede ordenar intercambiando arr[0] y arr[2], es decir, 3 y 1 como mcd(3, 1) = 1.Entrada: arr[] = {10, 15, 6, 2, 9}
Salida: No es posible
Explicación: No es posible ordenar la array dada usando ninguna secuencia de operaciones válidas.Entrada: arr[] = {1, 9, 3, 7, 2, 4}
Salida: Posible
Enfoque: el problema dado se puede resolver con la ayuda de la recursividad y el retroceso . Cree una función recursiva para iterar sobre todas las inversiones de la array dada , intercambie los elementos invertidos uno a la vez si su GCD es 1 y llame recursivamente a la array restante. En cualquier momento, si la array está ordenada, lo que se puede verificar con la función is_sorted() , devuelve verdadero ; de lo contrario, devuelve falso .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Recursive function to check if it is
// possible to sort the given array by
// swapping elements with their GCD = 1
bool isPossible(vector<int> arr)
{
// If the given array is sorted
if (is_sorted(arr.begin(), arr.end())) {
return true;
}
// Stores if it is possible to sort
// the given array
bool res = false;
// Iterate for all possible pairs of
// Inversions
for (int i = 0; i < arr.size(); i++) {
for (int j = i + 1; j < arr.size(); j++) {
// If for current inversion,
// GCD of both elements is 1,
// GCD of both the indices
if (arr[i] > arr[j]
&& __gcd(arr[i], arr[j]) == 1) {
swap(arr[i], arr[j]);
// Recursive Call for the
// remaining array
res = (res | isPossible(arr));
// Backtrack
swap(arr[i], arr[j]);
}
}
}
// Return Answer
return res;
}
// Driver Code
int main()
{
vector<int> arr = { 1, 9, 3, 7, 2, 4 };
if (isPossible(arr))
cout << "Possible";
else
cout << "Not Possible";
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Recursive function to check if it is
// possible to sort the given array by
// swapping elements with their GCD = 1
static boolean isPossible(int[] arr)
{
// If the given array is sorted
if (is_sorted(arr)) {
return true;
}
// Stores if it is possible to sort
// the given array
boolean res = false;
// Iterate for all possible pairs of
// Inversions
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
// If for current inversion,
// GCD of both elements is 1,
// GCD of both the indices
if (arr[i] > arr[j]
&& __gcd(arr[i], arr[j]) == 1) {
arr = swap(arr,i,j);
// Recursive Call for the
// remaining array
res = (res | isPossible(arr));
// Backtrack
arr = swap(arr,i,j);
}
}
}
// Return Answer
return res;
}
static int __gcd(int a, int b)
{
return b == 0? a:__gcd(b, a % b);
}
static int[] swap(int []arr, int i, int j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
private static boolean is_sorted(int[] a) {
int i;
for(i = 0; i < a.length - 1 && a[i] < a[i+1]; i++){}
return (i == a.length - 1);
}
public static void main(String[] args)
{
int[] arr = { 1, 9, 3, 7, 2, 4 };
if (isPossible(arr))
System.out.print("Possible");
else
System.out.print("Not Possible");
}
}
// This code is contributed by shikhasingrajput
Python3
# Python code for the above approach
# Recursive function to return gcd of a and b
def __gcd(a, b):
# Everything divides 0
if (a == 0):
return b
if (b == 0):
return a
# base case
if (a == b):
return a
# a is greater
if (a > b):
return __gcd(a - b, b)
return __gcd(a, b - a)
# Recursive function to check if it is
# possible to sort the given array by
# swapping elements with their GCD = 1
def isPossible(arr):
# If the given array is sorted
brr = sorted(arr)
if (arr == brr):
return True
# Stores if it is possible to sort
# the given array
res = False
# Iterate for all possible pairs of
# Inversions
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
# If for current inversion,
# GCD of both elements is 1,
# GCD of both the indices
if (arr[i] > arr[j] and __gcd(arr[i], arr[j]) == 1):
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
# Recursive Call for the
# remaining array
res = (res | isPossible(arr))
# Backtrack
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
# Return Answer
return res
# Driver Code
arr = [1, 9, 3, 7, 2, 4]
if (isPossible(arr)):
print("Possible")
else:
print("Not Possible")
# This code is contributed by Saurabh Jaiswal
C#
// C# program for the above approach
using System;
public class GFG{
// Recursive function to check if it is
// possible to sort the given array by
// swapping elements with their GCD = 1
static bool isPossible(int[] arr)
{
// If the given array is sorted
if (is_sorted(arr)) {
return true;
}
// Stores if it is possible to sort
// the given array
bool res = false;
// Iterate for all possible pairs of
// Inversions
for (int i = 0; i < arr.Length; i++) {
for (int j = i + 1; j < arr.Length; j++) {
// If for current inversion,
// GCD of both elements is 1,
// GCD of both the indices
if (arr[i] > arr[j]
&& __gcd(arr[i], arr[j]) == 1) {
arr = swap(arr,i,j);
// Recursive Call for the
// remaining array
res = (res | isPossible(arr));
// Backtrack
arr = swap(arr,i,j);
}
}
}
// Return Answer
return res;
}
static int __gcd(int a, int b)
{
return b == 0? a:__gcd(b, a % b);
}
static int[] swap(int []arr, int i, int j){
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
return arr;
}
private static bool is_sorted(int[] a) {
int i;
for(i = 0; i < a.Length - 1 && a[i] < a[i+1]; i++){}
return (i == a.Length - 1);
}
public static void Main(String[] args)
{
int[] arr = { 1, 9, 3, 7, 2, 4 };
if (isPossible(arr))
Console.Write("Possible");
else
Console.Write("Not Possible");
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript code for the above approach
// Recursive function to return gcd of a and b
function __gcd(a, b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return __gcd(a - b, b);
return __gcd(a, b - a);
}
// Recursive function to check if it is
// possible to sort the given array by
// swapping elements with their GCD = 1
function isPossible(arr)
{
// If the given array is sorted
brr = arr.sort(function (a, b) { return a - b })
if (arr == brr) {
return true;
}
// Stores if it is possible to sort
// the given array
let res = false;
// Iterate for all possible pairs of
// Inversions
for (let i = 0; i < arr.length; i++)
{
for (let j = i + 1; j < arr.length; j++)
{
// If for current inversion,
// GCD of both elements is 1,
// GCD of both the indices
if (arr[i] > arr[j]
&& __gcd(arr[i], arr[j]) == 1) {
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
// Recursive Call for the
// remaining array
res = (res | isPossible(arr));
// Backtrack
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
}
// Return Answer
return res;
}
// Driver Code
let arr = [1, 9, 3, 7, 2, 4];
if (isPossible(arr))
document.write("Possible")
else
document.write("Not Possible");
// This code is contributed by Potta Lokesh
</script>
Producción
Possible
Complejidad temporal: O(N 2 N!)
Espacio auxiliar: O(1)