Dada una array arr[] de enteros no negativos. Podemos realizar una operación de intercambio en cualquiera de los dos elementos adyacentes de la array. Encuentre la cantidad mínima de intercambios necesarios para ordenar la array en orden ascendente.
Ejemplos:
Input : arr[] = {3, 2, 1}
Output : 3
We need to do following swaps
(3, 2), (3, 1) and (1, 2)
Input : arr[] = {1, 20, 6, 4, 5}
Output : 5
Hay una solución interesante a este problema. Se puede resolver utilizando el hecho de que el número de intercambios necesarios es igual al número de inversiones . Entonces, básicamente, necesitamos contar las inversiones en array .
El hecho se puede establecer utilizando las siguientes observaciones:
- Una array ordenada no tiene inversiones.
- Un intercambio adyacente puede reducir una inversión. Hacer x intercambios adyacentes puede reducir x inversiones en una array.
Implementación:
C++
// C++ program to count number of swaps required
// to sort an array when only swapping of adjacent
// elements is allowed.
#include <bits/stdc++.h>
/* This function merges two sorted arrays and returns inversion
count in the arrays.*/
int merge(int arr[], int temp[], int left, int mid, int right)
{
int inv_count = 0;
int i = left; /* i is index for left subarray*/
int j = mid; /* j is index for right subarray*/
int k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right))
{
if (arr[i] <= arr[j])
temp[k++] = arr[i++];
else
{
temp[k++] = arr[j++];
/* this is tricky -- see above explanation/
diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for (i=left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
/* An auxiliary recursive function that sorts the input
array and returns the number of inversions in the
array. */
int _mergeSort(int arr[], int temp[], int left, int right)
{
int mid, inv_count = 0;
if (right > left)
{
/* Divide the array into two parts and call
_mergeSortAndCountInv() for each of the parts */
mid = (right + left)/2;
/* Inversion count will be sum of inversions in
left-part, right-part and number of inversions
in merging */
inv_count = _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid+1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid+1, right);
}
return inv_count;
}
/* This function sorts the input array and returns the
number of inversions in the array */
int countSwaps(int arr[], int n)
{
int temp[n];
return _mergeSort(arr, temp, 0, n - 1);
}
/* Driver program to test above functions */
int main(int argv, char** args)
{
int arr[] = {1, 20, 6, 4, 5};
int n = sizeof(arr)/sizeof(arr[0]);
printf("Number of swaps is %d \n", countSwaps(arr, n));
return 0;
}
Java
// Java program to count number of
// swaps required to sort an array
// when only swapping of adjacent
// elements is allowed.
import java.io.*;
class GFG {
// This function merges two sorted
// arrays and returns inversion
// count in the arrays.
static int merge(int arr[], int temp[],
int left, int mid, int right)
{
int inv_count = 0;
/* i is index for left subarray*/
int i = left;
/* j is index for right subarray*/
int j = mid;
/* k is index for resultant merged subarray*/
int k = left;
while ((i <= mid - 1) && (j <= right))
{
if (arr[i] <= arr[j])
temp[k++] = arr[i++];
else
{
temp[k++] = arr[j++];
/* this is tricky -- see above /
explanation diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left
subarray (if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right
subarray (if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements
to original array*/
for (i=left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
// An auxiliary recursive function that
// sorts the input array and returns
// the number of inversions in the array.
static int _mergeSort(int arr[], int temp[],
int left, int right)
{
int mid, inv_count = 0;
if (right > left)
{
// Divide the array into two parts and
// call _mergeSortAndCountInv() for
// each of the parts
mid = (right + left)/2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count = _mergeSort(arr, temp,
left, mid);
inv_count += _mergeSort(arr, temp,
mid+1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp,
left, mid+1, right);
}
return inv_count;
}
// This function sorts the input
// array and returns the number
// of inversions in the array
static int countSwaps(int arr[], int n)
{
int temp[] = new int[n];
return _mergeSort(arr, temp, 0, n - 1);
}
// Driver Code
public static void main (String[] args)
{
int arr[] = {1, 20, 6, 4, 5};
int n = arr.length;
System.out.println("Number of swaps is "
+ countSwaps(arr, n));
}
}
// This code is contributed by vt_m
Python3
# python 3 program to count number of swaps required
# to sort an array when only swapping of adjacent
# elements is allowed.
#include <bits/stdc++.h>
#This function merges two sorted arrays and returns inversion count in the arrays.*/
def merge(arr, temp, left, mid, right):
inv_count = 0
i = left #i is index for left subarray*/
j = mid #j is index for right subarray*/
k = left #k is index for resultant merged subarray*/
while ((i <= mid - 1) and (j <= right)):
if (arr[i] <= arr[j]):
temp[k] = arr[i]
k += 1
i += 1
else:
temp[k] = arr[j]
k += 1
j += 1
#this is tricky -- see above explanation/
# diagram for merge()*/
inv_count = inv_count + (mid - i)
#Copy the remaining elements of left subarray
# (if there are any) to temp*/
while (i <= mid - 1):
temp[k] = arr[i]
k += 1
i += 1
#Copy the remaining elements of right subarray
# (if there are any) to temp*/
while (j <= right):
temp[k] = arr[j]
k += 1
j += 1
# Copy back the merged elements to original array*/
for i in range(left,right+1,1):
arr[i] = temp[i]
return inv_count
#An auxiliary recursive function that sorts the input
# array and returns the number of inversions in the
# array. */
def _mergeSort(arr, temp, left, right):
inv_count = 0
if (right > left):
# Divide the array into two parts and call
#_mergeSortAndCountInv()
# for each of the parts */
mid = int((right + left)/2)
#Inversion count will be sum of inversions in
# left-part, right-part and number of inversions
# in merging */
inv_count = _mergeSort(arr, temp, left, mid)
inv_count += _mergeSort(arr, temp, mid+1, right)
# Merge the two parts*/
inv_count += merge(arr, temp, left, mid+1, right)
return inv_count
#This function sorts the input array and returns the
#number of inversions in the array */
def countSwaps(arr, n):
temp = [0 for i in range(n)]
return _mergeSort(arr, temp, 0, n - 1)
# Driver program to test above functions */
if __name__ == '__main__':
arr = [1, 20, 6, 4, 5]
n = len(arr)
print("Number of swaps is",countSwaps(arr, n))
# This code is contributed by
# Surendra_Gangwar
C#
// C# program to count number of
// swaps required to sort an array
// when only swapping of adjacent
// elements is allowed.
using System;
class GFG
{
// This function merges two
// sorted arrays and returns
// inversion count in the arrays.
static int merge(int []arr, int []temp,
int left, int mid,
int right)
{
int inv_count = 0;
/* i is index for
left subarray*/
int i = left;
/* j is index for
right subarray*/
int j = mid;
/* k is index for resultant
merged subarray*/
int k = left;
while ((i <= mid - 1) &&
(j <= right))
{
if (arr[i] <= arr[j])
temp[k++] = arr[i++];
else
{
temp[k++] = arr[j++];
/* this is tricky -- see above /
explanation diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements
of left subarray (if there are
any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements
of right subarray (if there are
any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged
elements to original array*/
for (i=left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
// An auxiliary recursive function
// that sorts the input array and
// returns the number of inversions
// in the array.
static int _mergeSort(int []arr, int []temp,
int left, int right)
{
int mid, inv_count = 0;
if (right > left)
{
// Divide the array into two parts
// and call _mergeSortAndCountInv()
// for each of the parts
mid = (right + left) / 2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count = _mergeSort(arr, temp,
left, mid);
inv_count += _mergeSort(arr, temp,
mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp,
left, mid + 1, right);
}
return inv_count;
}
// This function sorts the input
// array and returns the number
// of inversions in the array
static int countSwaps(int []arr, int n)
{
int []temp = new int[n];
return _mergeSort(arr, temp, 0, n - 1);
}
// Driver Code
public static void Main ()
{
int []arr = {1, 20, 6, 4, 5};
int n = arr.Length;
Console.Write("Number of swaps is " +
countSwaps(arr, n));
}
}
// This code is contributed by nitin mittal.
PHP
<?php
// PHP program to count number of swaps required
// to sort an array when only swapping of adjacent
// elements is allowed.
/* This function merges two sorted arrays and returns inversion
count in the arrays.*/
function merge(&$arr, &$temp, $left, $mid, $right)
{
$inv_count = 0;
$i = $left; /* i is index for left subarray*/
$j = $mid; /* j is index for right subarray*/
$k = $left; /* k is index for resultant merged subarray*/
while (($i <= $mid - 1) && ($j <= $right))
{
if ($arr[$i] <= $arr[$j])
$temp[$k++] = $arr[$i++];
else
{
$temp[$k++] = $arr[$j++];
/* this is tricky -- see above explanation/
diagram for merge()*/
$inv_count = $inv_count + ($mid - $i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while ($i <= $mid - 1)
$temp[$k++] = $arr[$i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while ($j <= $right)
$temp[$k++] = $arr[$j++];
/*Copy back the merged elements to original array*/
for ($i=$left; $i <= $right; $i++)
$arr[$i] = $temp[$i];
return $inv_count;
}
/* An auxiliary recursive function that sorts the input
array and returns the number of inversions in the
array. */
function _mergeSort(&$arr, &$temp, $left, $right)
{
$inv_count = 0;
if ($right > $left)
{
/* Divide the array into two parts and call
_mergeSortAndCountInv() for each of the parts */
$mid = intval(($right + $left)/2);
/* Inversion count will be sum of inversions in
left-part, right-part and number of inversions
in merging */
$inv_count = _mergeSort($arr, $temp, $left, $mid);
$inv_count += _mergeSort($arr, $temp, $mid+1, $right);
/*Merge the two parts*/
$inv_count += merge($arr, $temp, $left, $mid+1, $right);
}
return $inv_count;
}
/* This function sorts the input array and returns the
number of inversions in the array */
function countSwaps(&$arr, $n)
{
$temp = array_fill(0,$n,NULL);
return _mergeSort($arr, $temp, 0, $n - 1);
}
/* Driver program to test above functions */
$arr = array(1, 20, 6, 4, 5);
$n = sizeof($arr)/sizeof($arr[0]);
echo "Number of swaps is ". countSwaps($arr, $n);
return 0;
?>
Javascript
<script>
// Javascript program to count number of
// swaps required to sort an array
// when only swapping of adjacent
// elements is allowed.
// This function merges two sorted
// arrays and returns inversion
// count in the arrays.
function merge(arr,temp,left,mid,right)
{
let inv_count = 0;
/* i is index for left subarray*/
let i = left;
/* j is index for right subarray*/
let j = mid;
/* k is index for resultant merged subarray*/
let k = left;
while ((i <= mid - 1) && (j <= right))
{
if (arr[i] <= arr[j])
temp[k++] = arr[i++];
else
{
temp[k++] = arr[j++];
/* this is tricky -- see above /
explanation diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left
subarray (if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right
subarray (if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements
to original array*/
for (i=left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
// An auxiliary recursive function that
// sorts the input array and returns
// the number of inversions in the array.
function _mergeSort(arr,temp,left,right)
{
let mid, inv_count = 0;
if (right > left)
{
// Divide the array into two parts and
// call _mergeSortAndCountInv() for
// each of the parts
mid = Math.floor((right + left)/2);
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count = _mergeSort(arr, temp,
left, mid);
inv_count += _mergeSort(arr, temp,
mid+1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp,
left, mid+1, right);
}
return inv_count;
}
// This function sorts the input
// array and returns the number
// of inversions in the array
function countSwaps(arr,n)
{
let temp = new Array(n);
for(let i = 0; i < n; i++)
{
temp[i] = 0;
}
return _mergeSort(arr, temp, 0, n - 1);
}
// Driver Code
let arr=[1, 20, 6, 4, 5];
let n = arr.length;
document.write("Number of swaps is "
+ countSwaps(arr, n));
// This code is contributed by rag2127
</script>
Producción
Number of swaps is 5
Complejidad de tiempo: O (n Log n)
Publicación relacionada:
Número mínimo de intercambios necesarios para ordenar una array
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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