Dada una array de 0 y 1, necesitamos escribir un programa para encontrar el número mínimo de intercambios necesarios para agrupar todos los 1 presentes en la array.
Ejemplos:
Input : arr[] = {1, 0, 1, 0, 1}
Output : 1
Explanation: Only 1 swap is required to
group all 1's together. Swapping index 1
and 4 will give arr[] = {1, 1, 1, 0, 0}
Input : arr[] = {1, 0, 1, 0, 1, 1}
Output : 1
Una solución simple es primero contar el número total de 1 en la array. Supongamos que este conteo es x, ahora necesitamos encontrar el subarreglo de longitud x de este arreglo con el número máximo de 1. Y los intercambios mínimos requeridos serán el número de 0 en el subarreglo de longitud x con el número máximo de 1.
Complejidad temporal: O(n 2 )
Una solución eficiente es optimizar la técnica de fuerza bruta para encontrar el subarreglo en el enfoque anterior utilizando el concepto de técnica de ventana deslizante . Podemos mantener un arreglo preCount para encontrar el número de 1 presentes en un subarreglo en una complejidad de tiempo O(1).
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to find minimum swaps
// required to group all 1's together
#include <iostream>
#include <limits.h>
using namespace std;
// Function to find minimum swaps
// to group all 1's together
int minSwaps(int arr[], int n) {
int noOfOnes = 0;
// find total number of all in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
int x = noOfOnes;
int maxOnes = INT_MIN;
// array to store number of 1's upto
// ith index
int preCompute[n] = {0};
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int noOfZeroes = x - maxOnes;
return noOfZeroes;
}
// Driver Code
int main() {
int a[] = {1, 0, 1, 0, 1};
int n = sizeof(a) / sizeof(a[0]);
cout << minSwaps(a, n);
return 0;
}
Java
// Java program to find minimum swaps
// required to group all 1's together
import java.io.*;
class GFG {
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int arr[], int n) {
int noOfOnes = 0;
// find total number of all in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
int x = noOfOnes;
int maxOnes = Integer.MIN_VALUE;
// array to store number of 1's upto
// ith index
int preCompute[] = new int[n];
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int noOfZeroes = x - maxOnes;
return noOfZeroes;
}
// Driver Code
public static void main (String[] args) {
int a[] = {1, 0, 1, 0, 1};
int n = a.length;
System.out.println( minSwaps(a, n));
}
}
// This code is contributed by vt_m.
Python3
# Python program to
# find minimum swaps
# required to group
# all 1's together
# Function to find minimum swaps
# to group all 1's together
def minSwaps(arr,n):
noOfOnes = 0
# find total number of
# all in the array
for i in range(n):
if (arr[i] == 1):
noOfOnes=noOfOnes+1
# length of subarray to check for
x = noOfOnes
maxOnes = -2147483648
# array to store number of 1's upto
# ith index
preCompute={}
# calculate number of 1's upto ith
# index and store in the
# array preCompute[]
if (arr[0] == 1):
preCompute[0] = 1
for i in range(1,n):
if (arr[i] == 1):
preCompute[i] = preCompute[i - 1] + 1
else:
preCompute[i] = preCompute[i - 1]
# using sliding window
# technique to find
# max number of ones in
# subarray of length x
for i in range(x-1,n):
if (i == (x - 1)):
noOfOnes = preCompute[i]
else:
noOfOnes = preCompute[i] - preCompute[i - x]
if (maxOnes < noOfOnes):
maxOnes = noOfOnes
# calculate number of zeros in subarray
# of length x with maximum number of 1's
noOfZeroes = x - maxOnes
return noOfZeroes
# Driver code
a = [1, 0, 1, 0, 1]
n = len(a)
print(minSwaps(a, n))
# This code is contributed
# by Anant Agarwal.
C#
// C# program to find minimum swaps
// required to group all 1's together
using System;
class GFG {
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int []arr, int n) {
int noOfOnes = 0;
// find total number of all in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
int x = noOfOnes;
int maxOnes = int.MinValue;
// array to store number of 1's upto
// ith index
int []preCompute = new int[n];
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int noOfZeroes = x - maxOnes;
return noOfZeroes;
}
// Driver Code
public static void Main ()
{
int []a = {1, 0, 1, 0, 1};
int n = a.Length;
Console.WriteLine( minSwaps(a, n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to find minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps($arr, $n)
{
$noOfOnes = 0;
// find total number of
// all in the array
for($i = 0; $i < $n; $i++)
{
if ($arr[$i] == 1)
$noOfOnes++;
}
// length of subarray
// to check for
$x = $noOfOnes;
$maxOnes = PHP_INT_MIN;
// array to store number of
// 1's upto ith index
$preCompute = array();
// calculate number of 1's
// upto ith index and store
// in the array preCompute[]
if ($arr[0] == 1)
$preCompute[0] = 1;
for($i = 1; $i < $n; $i++)
{
if ($arr[$i] == 1)
{
$preCompute[$i] = $preCompute[$i - 1] + 1;
}
else
$preCompute[$i] = $preCompute[$i - 1];
}
// using sliding window
// technique to find
// max number of ones in
// subarray of length x
for ( $i = $x - 1; $i < $n; $i++)
{
if ($i == ($x - 1))
$noOfOnes = $preCompute[$i];
else
$noOfOnes = $preCompute[$i] -
$preCompute[$i - $x];
if ($maxOnes < $noOfOnes)
$maxOnes = $noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
$noOfZeroes = $x - $maxOnes;
return $noOfZeroes;
}
// Driver Code
$a = array(1, 0, 1, 0, 10);
$n = count($a);
echo minSwaps($a, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// Javascript program to find minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps(arr, n) {
let noOfOnes = 0;
// find total number of all in the array
for (let i = 0; i < n; i++) {
if (arr[i] == 1)
noOfOnes++;
}
// length of subarray to check for
let x = noOfOnes;
let maxOnes = Number.MIN_VALUE;
// array to store number of 1's upto
// ith index
let preCompute = new Array(n);
preCompute.fill(0);
// calculate number of 1's upto ith
// index and store in the array preCompute[]
if (arr[0] == 1)
preCompute[0] = 1;
for (let i = 1; i < n; i++) {
if (arr[i] == 1) {
preCompute[i] = preCompute[i - 1] + 1;
} else
preCompute[i] = preCompute[i - 1];
}
// using sliding window technique to find
// max number of ones in subarray of length x
for (let i = x - 1; i < n; i++) {
if (i == (x - 1))
noOfOnes = preCompute[i];
else
noOfOnes = preCompute[i] - preCompute[i - x];
if (maxOnes < noOfOnes)
maxOnes = noOfOnes;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
let noOfZeroes = x - maxOnes;
return noOfZeroes;
}
let a = [1, 0, 1, 0, 1];
let n = a.length;
document.write( minSwaps(a, n));
</script>
Producción
1
Tiempo Complejidad : O(n)
Espacio Auxiliar : O(n)
Otro enfoque eficiente:
primero cuente el número total de 1 en la array. Suponga que este conteo es x, ahora encuentre el subarreglo de longitud x de este arreglo con el número máximo de 1 usando el concepto de técnica de deslizamiento de ventanas . Mantenga una variable para encontrar el número de 1 presentes en un subarreglo en O (1) espacio adicional y para cada subarreglo mantenga la variable maxOnes y, por último, devuelva el número de ceros (número de ceros = x – maxOnes).
C++
// C++ code for minimum swaps
// required to group all 1's together
#include <iostream>
#include <limits.h>
using namespace std;
// Function to find minimum swaps
// to group all 1's together
int minSwaps(int arr[], int n)
{
int numberOfOnes = 0;
// find total number of all 1's in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
int x = numberOfOnes;
int count_ones = 0, maxOnes;
// Find 1's for first subarray of length x
for(int i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = 1; i <= n-x; i++) {
// first remove leading element and check
// if it is equal to 1 then decrement the
// value of count_ones by 1
if (arr[i-1] == 1)
count_ones--;
// Now add trailing element and check
// if it is equal to 1 Then increment
// the value of count_ones by 1
if(arr[i+x-1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
// Driver Code
int main() {
int a[] = {0, 0, 1, 0, 1, 1, 0, 0, 1};
int n = sizeof(a) / sizeof(a[0]);
cout << minSwaps(a, n);
return 0;
}
Java
// java program to find largest number
// smaller than equal to n with m set
// bits then m-1 0 bits.
public class GFG {
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int arr[], int n)
{
int numberOfOnes = 0;
// find total number of all 1's
// in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
int x = numberOfOnes;
int count_ones = 0, maxOnes;
// Find 1's for first subarray
// of length x
for(int i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique
// to find max number of ones in
// subarray of length x
for (int i = 1; i <= n-x; i++) {
// first remove leading element
// and check if it is equal to
// 1 then decrement the
// value of count_ones by 1
if (arr[i - 1] == 1)
count_ones--;
// Now add trailing element
// and check if it is equal
// to 1 Then increment the
// value of count_ones by 1
if(arr[i + x - 1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in
// subarray of length x with
// maximum number of 1's
int numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
// Driver code
public static void main(String args[])
{
int a[] = new int[]{0, 0, 1, 0,
1, 1, 0, 0, 1};
int n = a.length;
System.out.println(minSwaps(a, n));
}
}
// This code is contributed by Sam007
Python3
# Python code for minimum # swaps required to group # all 1's together # Function to find minimum # swaps to group all 1's # together def minSwaps(arr, n) : numberOfOnes = 0 # find total number of # all 1's in the array for i in range(0, n) : if (arr[i] == 1) : numberOfOnes = numberOfOnes + 1 # length of subarray # to check for x = numberOfOnes count_ones = 0 maxOnes = 0 # Find 1's for first # subarray of length x for i in range(0, x) : if(arr[i] == 1) : count_ones = count_ones + 1 maxOnes = count_ones # using sliding window # technique to find # max number of ones in # subarray of length x for i in range(1, (n - x + 1)) : # first remove leading # element and check # if it is equal to 1 # then decrement the # value of count_ones by 1 if (arr[i - 1] == 1) : count_ones = count_ones - 1 # Now add trailing # element and check # if it is equal to 1 # Then increment # the value of count_ones by 1 if(arr[i + x - 1] == 1) : count_ones = count_ones + 1 if (maxOnes < count_ones) : maxOnes = count_ones # calculate number of # zeros in subarray # of length x with # maximum number of 1's numberOfZeroes = x - maxOnes return numberOfZeroes # Driver Code a = [0, 0, 1, 0, 1, 1, 0, 0, 1] n = 9 print (minSwaps(a, n)) # This code is contributed # by Manish Shaw(manishshaw1)
C#
// C# code for minimum swaps
// required to group all 1's together
using System;
class GFG{
// Function to find minimum swaps
// to group all 1's together
static int minSwaps(int []arr, int n)
{
int numberOfOnes = 0;
// find total number of all 1's in the array
for (int i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
int x = numberOfOnes;
int count_ones = 0, maxOnes;
// Find 1's for first subarray of length x
for(int i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique to find
// max number of ones in subarray of length x
for (int i = 1; i <= n-x; i++) {
// first remove leading element and check
// if it is equal to 1 then decrement the
// value of count_ones by 1
if (arr[i - 1] == 1)
count_ones--;
// Now add trailing element and check
// if it is equal to 1 Then increment
// the value of count_ones by 1
if(arr[i + x - 1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
int numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
// Driver Code
static public void Main ()
{
int []a = {0, 0, 1, 0, 1, 1, 0, 0, 1};
int n = a.Length;
Console.WriteLine(minSwaps(a, n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP code for minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps($arr, $n)
{
$numberOfOnes = 0;
// find total number of
// all 1's in the array
for ($i = 0; $i < $n; $i++)
{
if ($arr[$i] == 1)
$numberOfOnes++;
}
// length of subarray to check for
$x = $numberOfOnes;
$count_ones = 0;
$maxOnes;
// Find 1's for first
// subarray of length x
for($i = 0; $i < $x; $i++)
{
if($arr[$i] == 1)
$count_ones++;
}
$maxOnes = $count_ones;
// using sliding window
// technique to find
// max number of ones in
// subarray of length x
for ($i = 1; $i <= $n - $x; $i++)
{
// first remove leading
// element and check
// if it is equal to 1
// then decrement the
// value of count_ones by 1
if ($arr[$i - 1] === 1)
$count_ones--;
// Now add trailing
// element and check
// if it is equal to 1
// Then increment
// the value of count_ones by 1
if($arr[$i + $x - 1] === 1)
$count_ones++;
if ($maxOnes < $count_ones)
$maxOnes = $count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
$numberOfZeroes = $x - $maxOnes;
return $numberOfZeroes;
}
// Driver Code
$a = array(0, 0, 1, 0, 1, 1, 0, 0, 1);
$n = 9;
echo minSwaps($a, $n);
// This code is contributed by Anuj_67
?>
Javascript
<script>
// Javascript code for minimum swaps
// required to group all 1's together
// Function to find minimum swaps
// to group all 1's together
function minSwaps(arr, n)
{
let numberOfOnes = 0;
// find total number of all 1's in the array
for (let i = 0; i < n; i++) {
if (arr[i] == 1)
numberOfOnes++;
}
// length of subarray to check for
let x = numberOfOnes;
let count_ones = 0, maxOnes;
// Find 1's for first subarray of length x
for(let i = 0; i < x; i++){
if(arr[i] == 1)
count_ones++;
}
maxOnes = count_ones;
// using sliding window technique to find
// max number of ones in subarray of length x
for (let i = 1; i <= n-x; i++) {
// first remove leading element and check
// if it is equal to 1 then decrement the
// value of count_ones by 1
if (arr[i - 1] == 1)
count_ones--;
// Now add trailing element and check
// if it is equal to 1 Then increment
// the value of count_ones by 1
if(arr[i + x - 1] == 1)
count_ones++;
if (maxOnes < count_ones)
maxOnes = count_ones;
}
// calculate number of zeros in subarray
// of length x with maximum number of 1's
let numberOfZeroes = x - maxOnes;
return numberOfZeroes;
}
let a = [0, 0, 1, 0, 1, 1, 0, 0, 1];
let n = a.length;
document.write(minSwaps(a, n));
</script>
Producción
1
Complejidad de tiempo: O(n)
Espacio auxiliar: O(1)
Gracias al Sr. Gera por sugerir este enfoque.
Ventana corredera Versión fácil de entender
Algoritmo:
1. Almacene el número total de unos en una variable digamos contar. Este será el tamaño de la ventana.
2. Inicialice una variable para almacenar el número máximo de unos de todos los subconjuntos de conteo de tamaño y una variable para almacenar el conteo de unos en la ventana actual.
3. Ahora itere sobre la array y, tan pronto como alcance el tamaño de la ventana, compare el número de unos en esa ventana con el número máximo de unidades en todas las ventanas hasta el momento y actualice el número máximo de unidades si el número de unidades en la ventana actual es más. Si el primer elemento de la ventana es 1, disminuya el recuento actual.
4. La respuesta será el conteo total de unos – el conteo máximo de unos de todas las ventanas.
C++
#include <bits/stdc++.h>
using namespace std;
int minSwaps(int arr[], int n)
{
int totalCount = 0; // To store total number of ones
// count total no of ones
for (int i = 0; i < n; i++)
totalCount += arr[i];
int currCount = 0; // To store count of ones in current window
int maxCount = 0; // To store maximum count ones out of all windows
int i = 0; // start of window
int j = 0; // end of window
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e. total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count if first element of window is 1
i++; // slide window
}
j++;
}
return totalCount - maxCount; // return total no of ones in array - maximum count of ones out of all windows
}
// Driver Code
int main()
{
int a[] = {1, 0, 1, 0, 1, 1};
int n = sizeof(a) / sizeof(a[0]);
cout << minSwaps(a, n);
return 0;
}
Java
import java.io.*;
import java.util.*;
class GFG{
static int minSwaps(int[] arr, int n)
{
// To store total number of ones
int totalCount = 0;
// Count total no of ones
int i;
for(i = 0; i < n; i++)
totalCount += arr[i];
int currCount = 0; // To store count of ones in current window
int maxCount = 0; // To store maximum count ones out
// of all windows
// start of window
i = 0;
// end of window
int j = 0;
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e.
// total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = Math.max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count
// if first element of
// window is 1
// slide window
i++;
}
j++;
}
return totalCount - maxCount; // return total no of ones in array
// - maximum count of ones out of
// all windows
}
// Driver Code
public static void main(String args[])
{
int[] a = { 1, 0, 1, 0, 1, 1 };
int n = a.length;
System.out.println(minSwaps(a, n));
}
}
// This code is contributed by shivanisinghss2110
Python
def minSwaps(arr, n): # To store total number of ones totalCount = 0 # count total no of ones for i in range(0,n): totalCount += arr[i] currCount = 0 # To store count of ones in current window maxCount = 0 # To store maximum count ones out of all windows i = 0 # start of window j = 0 # end of window while (j < n): currCount += arr[j] # update maxCount when reach window size i.e. total count of ones in array if ((j - i + 1) == totalCount): maxCount = max(maxCount, currCount) if (arr[i] == 1): currCount -= 1 # decrease current count if first element of window is 1 i += 1 # slide window j += 1 return totalCount - maxCount # return total no of ones in array - maximum count of ones out of all windows # Driver Code a = [1, 0, 1, 0, 1, 1] n = len(a) print(minSwaps(a, n)) # this code is contributed by shivanisighss2110
C#
using System;
class GFG{
static int minSwaps(int[] arr, int n)
{
// To store total number of ones
int totalCount = 0;
// Count total no of ones
int i;
for(i = 0; i < n; i++)
totalCount += arr[i];
int currCount = 0; // To store count of ones in current window
int maxCount = 0; // To store maximum count ones out
// of all windows
// start of window
i = 0;
// end of window
int j = 0;
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e.
// total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = Math.Max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count
// if first element of
// window is 1
// slide window
i++;
}
j++;
}
return totalCount - maxCount; // return total no of ones in array
// - maximum count of ones out of
// all windows
}
// Driver Code
public static void Main()
{
int[] a = { 1, 0, 1, 0, 1, 1 };
int n = a.Length;
Console.WriteLine(minSwaps(a, n));
}
}
// This code is contributed by ukasp
Javascript
<script>
function minSwaps(arr, n)
{
// To store total number of ones
let totalCount = 0;
// Count total no of ones
let i;
for(i = 0; i < n; i++)
totalCount += arr[i];
let currCount = 0; // To store count of ones in current window
let maxCount = 0; // To store maximum count ones out
// of all windows
// start of window
i = 0;
// end of window
let j = 0;
while (j < n)
{
currCount += arr[j];
// update maxCount when reach window size i.e.
// total count of ones in array
if ((j - i + 1) == totalCount)
{
maxCount = Math.max(maxCount, currCount);
if (arr[i] == 1)
currCount--; // decrease current count
// if first element of
// window is 1
// slide window
i++;
}
j++;
}
return totalCount - maxCount; // return total no of ones in array
// - maximum count of ones out of
// all windows
}
// Driver Code
let a = [ 1, 0, 1, 0, 1, 1 ];
let n = a.length;
document.write(minSwaps(a, n));
// This code is contributed by shivanisinghss2110
</script>
Producción
1
Complejidades:
Tiempo en)
Espacio: O(1)
Publicación traducida automáticamente
Artículo escrito por harsh.agarwal0 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA