Dada una array de n elementos distintos. Compruebe si la array dada es una array ordenada k o no. Una array ordenada k es una array en la que cada elemento está como máximo a k distancias de su posición de destino en la array ordenada.
Por ejemplo, consideremos k es 2, un elemento en el índice 7 en la array ordenada, puede estar en los índices 5, 6, 7, 8, 9 en la array dada.
Ejemplos:
Input : arr[] = {3, 2, 1, 5, 6, 4}, k = 2
Output : Yes
Every element is at most 2 distance away
from its target position in the sorted array.
Input : arr[] = {13, 8, 10, 7, 15, 14, 12}, k = 3
Output : No
13 is more than k = 3 distance away
from its target position in the sorted array.
Copie elementos de la array original arr[] en una array auxiliar aux[] .
Ordenar aux[] . Ahora, para cada elemento en el índice i en arr[] , encuentre su índice j en aux[] usando la búsqueda binaria. Si para cualquier elemento k < abs(ij) , entonces arr[] no es un arreglo k ordenado. De lo contrario, es una array ordenada k . Aquí los abdominales son el valor absoluto.
Implementación:
C++
// C++ implementation to check whether the given array
// is a k sorted array or not
#include <bits/stdc++.h>
using namespace std;
// function to find index of element 'x' in sorted 'arr'
// uses binary search technique
int binarySearch(int arr[], int low, int high, int x)
{
while (low <= high)
{
int mid = (low + high) / 2;
if (arr[mid] == x)
return mid;
else if (arr[mid] > x)
high = mid - 1;
else
low = mid + 1;
}
}
// function to check whether the given array is
// a 'k' sorted array or not
string isKSortedArray(int arr[], int n, int k)
{
// auxiliary array 'aux'
int aux[n];
// copy elements of 'arr' to 'aux'
for (int i = 0; i<n; i++)
aux[i] = arr[i];
// sort 'aux'
sort(aux, aux + n);
// for every element of 'arr' at index 'i',
// find its index 'j' in 'aux'
for (int i = 0; i<n; i++)
{
// index of arr[i] in sorted array 'aux'
int j = binarySearch(aux, 0, n-1, arr[i]);
// if abs(i-j) > k, then that element is
// not at-most k distance away from its
// target position. Thus, 'arr' is not a
// k sorted array
if (abs(i - j) > k)
return "No";
}
// 'arr' is a k sorted array
return "Yes";
}
// Driver program to test above
int main()
{
int arr[] = {3, 2, 1, 5, 6, 4};
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
cout << "Is it a k sorted array?: "
<< isKSortedArray(arr, n, k);
return 0;
}
Java
// Java implementation to check whether the given array
// is a k sorted array or not
import java.util.Arrays;
class Test
{
// Method to check whether the given array is
// a 'k' sorted array or not
static String isKSortedArray(int arr[], int n, int k)
{
// auxiliary array 'aux'
int aux[] = new int[n];
// copy elements of 'arr' to 'aux'
for (int i = 0; i<n; i++)
aux[i] = arr[i];
// sort 'aux'
Arrays.sort(aux);
// for every element of 'arr' at index 'i',
// find its index 'j' in 'aux'
for (int i = 0; i<n; i++)
{
// index of arr[i] in sorted array 'aux'
int j = Arrays.binarySearch(aux,arr[i]);
// if abs(i-j) > k, then that element is
// not at-most k distance away from its
// target position. Thus, 'arr' is not a
// k sorted array
if (Math.abs(i - j) > k)
return "No";
}
// 'arr' is a k sorted array
return "Yes";
}
// Driver method
public static void main(String args[])
{
int arr[] = {3, 2, 1, 5, 6, 4};
int k = 2;
System.out.println("Is it a k sorted array ?: " +
isKSortedArray(arr, arr.length, k));
}
}
Python3
# Python 3 implementation to check
# whether the given array is a k
# sorted array or not
# function to find index of element
# 'x' in sorted 'arr' uses binary
# search technique
def binarySearch(arr, low, high, x):
while (low <= high):
mid = int((low + high) / 2)
if (arr[mid] == x):
return mid
elif(arr[mid] > x):
high = mid - 1
else:
low = mid + 1
# function to check whether the given
# array is a 'k' sorted array or not
def isKSortedArray(arr, n, k):
# auxiliary array 'aux'
aux = [0 for i in range(n)]
# copy elements of 'arr' to 'aux'
for i in range(0, n, 1):
aux[i] = arr[i]
# sort 'aux'
aux.sort(reverse = False)
# for every element of 'arr' at
# index 'i', find its index 'j' in 'aux'
for i in range(0, n, 1):
# index of arr[i] in sorted
# array 'aux'
j = binarySearch(aux, 0, n - 1, arr[i])
# if abs(i-j) > k, then that element is
# not at-most k distance away from its
# target position. Thus, 'arr' is not a
# k sorted array
if (abs(i - j) > k):
return "No"
# 'arr' is a k sorted array
return "Yes"
# Driver Code
if __name__ == '__main__':
arr = [3, 2, 1, 5, 6, 4]
n = len(arr)
k = 2
print("Is it a k sorted array?:",
isKSortedArray(arr, n, k))
# This code is contributed by
# Shashank_Sharma
C#
// C# implementation to check
// whether the given array is a
// k sorted array or not
using System;
using System.Collections;
class GFG {
// Method to check whether the given
// array is a 'k' sorted array or not
static String isKSortedArray(int []arr, int n, int k)
{
// auxiliary array 'aux'
int []aux = new int[n];
// copy elements of 'arr' to 'aux'
for (int i = 0; i<n; i++)
aux[i] = arr[i];
// sort 'aux'
Array.Sort(aux);
// for every element of 'arr' at index
// 'i', find its index 'j' in 'aux'
for (int i = 0; i<n; i++)
{
// index of arr[i] in sorted array 'aux'
int j = Array.BinarySearch(aux,arr[i]);
// if abs(i-j) > k, then that element is
// not at-most k distance away from its
// target position. Thus, 'arr' is not a
// k sorted array
if (Math.Abs(i - j) > k)
return "No";
}
// 'arr' is a k sorted array
return "Yes";
}
// Driver method
public static void Main()
{
int []arr = {3, 2, 1, 5, 6, 4};
int k = 2;
Console.WriteLine("Is it a k sorted array ?: " +
isKSortedArray(arr, arr.Length, k));
}
}
// This code is contributed by Sam007
Javascript
<script>
// Javascript implementation to check whether the given array
// is a k sorted array or not
// function to find index of element 'x' in sorted 'arr'
// uses binary search technique
function binarySearch(arr, low, high, x)
{
while (low <= high)
{
var mid = parseInt((low + high) / 2);
if (arr[mid] == x)
return mid;
else if (arr[mid] > x)
high = mid - 1;
else
low = mid + 1;
}
}
// function to check whether the given array is
// a 'k' sorted array or not
function isKSortedArray(arr, n, k)
{
// auxiliary array 'aux'
var aux = Array(n);
// copy elements of 'arr' to 'aux'
for (var i = 0; i<n; i++)
aux[i] = arr[i];
// sort 'aux'
aux.sort((a,b)=> a-b)
// for every element of 'arr' at index 'i',
// find its index 'j' in 'aux'
for (var i = 0; i<n; i++)
{
// index of arr[i] in sorted array 'aux'
var j = binarySearch(aux, 0, n-1, arr[i]);
// if abs(i-j) > k, then that element is
// not at-most k distance away from its
// target position. Thus, 'arr' is not a
// k sorted array
if (Math.abs(i - j) > k)
return "No";
}
// 'arr' is a k sorted array
return "Yes";
}
// Driver program to test above
var arr = [3, 2, 1, 5, 6, 4];
var n = arr.length;
var k = 2;
document.write( "Is it a k sorted array?: "
+ isKSortedArray(arr, n, k));
</script>
Producción
Is it a k sorted array?: Yes
Complejidad temporal: O(nlogn)
Espacio auxiliar: O(n)
Otro enfoque puede ser almacenar los índices correspondientes de los elementos en la array auxiliar . Luego simplemente verifique si abs (i – aux[i].segundo) <= k, devuelva “No” si la condición no se cumple. Es un poco más rápido que el enfoque mencionado anteriormente, ya que no tenemos que realizar una búsqueda binaria para verificar la distancia desde el índice original, aunque la «notación O» seguirá siendo la misma.
Implementación:
C++
#include <bits/stdc++.h>
using namespace std;
string isKSortedArray(int arr[], int n, int k)
{
// creating an array to store value, index of the original array
vector<pair<int, int>> aux;
for(int i=0;i<n;i++){
aux.push_back({arr[i], i}); // pushing the elements and index of arr to aux
}
// sorting the aux array
sort(aux.begin(), aux.end());
// for every element, check if the absolute value of (currIndex-originalIndex) <= k
// if not, then return "NO"
for(auto i=0;i<n;i++){
if(abs(i-aux[i].second)>k) return "No";
}
// If all elements satisfy the condition, the loop will terminate and
// "Yes" will be returned.
return "Yes";
}
int main() {
int arr[] = {3, 2, 1, 5, 6, 4}; // input array
int n = sizeof(arr)/sizeof(int); // number of elements in array(arr)
int k = 2; // value to check is array is "k" sorted
cout<<isKSortedArray(arr, n, k); // prints "Yes" since the input array is k-sorted
return 0;
}
Python3
# Python code for the same approach def isKSortedArray(arr, n, k): # creating an array to store value, index of the original array aux = [] for i in range(n): aux.append([arr[i], i]) # pushing the elements and index of arr to aux # sorting the aux array aux.sort() # for every element, check if the absolute value of (currIndex-originalIndex) <= k # if not, then return "NO" for i in range(n): if(abs(i-aux[i][1])>k): return "No" # If all elements satisfy the condition, the loop will terminate and # "Yes" will be returned. return "Yes" # driver code arr = [3, 2, 1, 5, 6, 4] # input array n = len(arr) # number of elements in array(arr) k = 2 # value to check is array is "k" sorted print(isKSortedArray(arr, n, k)) # prints "Yes" since the input array is k-sorted # This code is contributed by shinjanpatra
Producción
Yes
Complejidad de tiempo: O(nlogn)
Complejidad de espacio: O(n)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA