Dada una array arr[] que consta de una permutación de enteros [1, N] , derivada de reorganizar el orden ordenado [1, N] , la tarea es encontrar el número mínimo de pasos después de los cuales el orden ordenado [1, N] se repite, repitiendo el mismo proceso mediante el cual se obtiene arr[] de la secuencia ordenada en cada paso.
Ejemplos:
Entrada: arr[ ] = {3, 6, 5, 4, 1, 2}
Salida: 6
Explicación:
Permutación creciente: {1, 2, 3, 4, 5, 6}
Paso 1: arr[] = {3, 6, 5, 4, 1, 2} (array dada)
Paso 2: arr[] = {5, 2, 1, 4, 3, 6}
Paso 3: arr[] = {1, 6, 3, 4, 5, 2}
Paso 4: arr[] = {3, 2, 5, 4, 1, 6}
Paso 5: arr[] = {5, 6, 1, 4, 3, 2}
Paso 6: arr[] = {1, 2, 3, 4, 5, 6} (Permutación creciente)
Por lo tanto, el número total de pasos necesarios es 6.
Entrada: arr[ ] = [5, 1, 4, 3, 2]
Salida: 6
Enfoque:
este problema se puede resolver simplemente utilizando el concepto de direccionamiento directo . Siga los pasos que se indican a continuación para resolver el problema:
- Inicialice una array dat[] para el direccionamiento directo.
- Iterar sobre [1, N] y calcular la diferencia del índice actual de cada elemento a partir de su índice en la secuencia ordenada.
- Calcule el LCM de la array dat[] .
- Ahora, imprima el LCM obtenido como los pasos mínimos requeridos para obtener el orden clasificado.
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find
// GCD of two numbers
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the
// LCM of array elements
int findlcm(int arr[], int n)
{
// Initialize result
int ans = 1;
for (int i = 1; i <= n; i++)
ans = (((arr[i] * ans))
/ (gcd(arr[i], ans)));
return ans;
}
// Function to find minimum steps
// required to obtain sorted sequence
void minimumSteps(int arr[], int n)
{
// Initialize dat[] array for
// Direct Address Table.
int i, dat[n + 1];
for (i = 1; i <= n; i++)
dat[arr[i - 1]] = i;
int b[n + 1], j = 0, c;
// Calculating steps required
// for each element to reach
// its sorted position
for (i = 1; i <= n; i++) {
c = 1;
j = dat[i];
while (j != i) {
c++;
j = dat[j];
}
b[i] = c;
}
// Calculate LCM of the array
cout << findlcm(b, n);
}
// Driver Code
int main()
{
int arr[] = { 5, 1, 4, 3, 2, 7, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
minimumSteps(arr, N);
return 0;
}
Java
// Java program to implement
// the above approach
class GFG{
// Function to find
// GCD of two numbers
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the
// LCM of array elements
static int findlcm(int arr[], int n)
{
// Initialize result
int ans = 1;
for(int i = 1; i <= n; i++)
ans = (((arr[i] * ans)) /
(gcd(arr[i], ans)));
return ans;
}
// Function to find minimum steps
// required to obtain sorted sequence
static void minimumSteps(int arr[], int n)
{
// Initialize dat[] array for
// Direct Address Table.
int i;
int dat[] = new int[n + 1];
for(i = 1; i <= n; i++)
dat[arr[i - 1]] = i;
int b[] = new int[n + 1];
int j = 0, c;
// Calculating steps required
// for each element to reach
// its sorted position
for(i = 1; i <= n; i++)
{
c = 1;
j = dat[i];
while (j != i)
{
c++;
j = dat[j];
}
b[i] = c;
}
// Calculate LCM of the array
System.out.println(findlcm(b, n));
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 5, 1, 4, 3, 2, 7, 6 };
int N = arr.length;
minimumSteps(arr, N);
}
}
// This code is contributed by rutvik_56
Python3
# Python3 program to implement # the above approach # Function to find # GCD of two numbers def gcd(a, b): if(b == 0): return a return gcd(b, a % b) # Function to calculate the # LCM of array elements def findlcm(arr, n): # Initialize result ans = 1 for i in range(1, n + 1): ans = ((arr[i] * ans) // (gcd(arr[i], ans))) return ans # Function to find minimum steps # required to obtain sorted sequence def minimumSteps(arr, n): # Initialize dat[] array for # Direct Address Table. dat = [0] * (n + 1) for i in range(1, n + 1): dat[arr[i - 1]] = i b = [0] * (n + 1) j = 0 # Calculating steps required # for each element to reach # its sorted position for i in range(1, n + 1): c = 1 j = dat[i] while(j != i): c += 1 j = dat[j] b[i] = c # Calculate LCM of the array print(findlcm(b, n)) # Driver Code arr = [ 5, 1, 4, 3, 2, 7, 6 ] N = len(arr) minimumSteps(arr, N) # This code is contributed by Shivam Singh
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to find
// GCD of two numbers
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the
// LCM of array elements
static int findlcm(int []arr, int n)
{
// Initialize result
int ans = 1;
for(int i = 1; i <= n; i++)
ans = (((arr[i] * ans)) /
(gcd(arr[i], ans)));
return ans;
}
// Function to find minimum steps
// required to obtain sorted sequence
static void minimumSteps(int []arr, int n)
{
// Initialize dat[] array for
// Direct Address Table.
int i;
int []dat = new int[n + 1];
for(i = 1; i <= n; i++)
dat[arr[i - 1]] = i;
int []b = new int[n + 1];
int j = 0, c;
// Calculating steps required
// for each element to reach
// its sorted position
for(i = 1; i <= n; i++)
{
c = 1;
j = dat[i];
while (j != i)
{
c++;
j = dat[j];
}
b[i] = c;
}
// Calculate LCM of the array
Console.WriteLine(findlcm(b, n));
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 5, 1, 4, 3, 2, 7, 6 };
int N = arr.Length;
minimumSteps(arr, N);
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// JavaScript program for the above approach
// Function to find
// GCD of two numbers
function gcd(a, b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the
// LCM of array elements
function findlcm(arr, n)
{
// Initialize result
let ans = 1;
for(let i = 1; i <= n; i++)
ans = (((arr[i] * ans)) /
(gcd(arr[i], ans)));
return ans;
}
// Function to find minimum steps
// required to obtain sorted sequence
function minimumSteps(arr, n)
{
// Initialize dat[] array for
// Direct Address Table.
let i;
let dat = Array.from({length: n+1}, (_, i) => 0);
for(i = 1; i <= n; i++)
dat[arr[i - 1]] = i;
let b = Array.from({length: n+1}, (_, i) => 0);
let j = 0, c;
// Calculating steps required
// for each element to reach
// its sorted position
for(i = 1; i <= n; i++)
{
c = 1;
j = dat[i];
while (j != i)
{
c++;
j = dat[j];
}
b[i] = c;
}
// Calculate LCM of the array
document.write(findlcm(b, n));
}
// Driver Code
let arr = [ 5, 1, 4, 3, 2, 7, 6 ];
let N = arr.length;
minimumSteps(arr, N);
</script>
Producción:
6
Complejidad temporal: O(NlogN)
Espacio auxiliar: O(N)