Dada una array de enteros. encuentre el valor máximo del subarreglo XOR en el arreglo dado. Complejidad del tiempo esperado O(n).
Ejemplos:
Entrada: arr[] = {1, 2, 3, 4}
Salida: 7
Explicación : el subarreglo {3, 4} tiene un valor XOR máximoEntrada: arr[] = {8, 1, 2, 12, 7, 6}
Salida: 15
Explicación : el subarreglo {1, 2, 12} tiene un valor XOR máximoEntrada: arr[] = {4, 6}
Salida: 6
Explicación: El subarreglo {6} tiene un valor XOR máximo
Una solución simple es usar dos bucles para encontrar XOR de todos los subarreglos y devolver el máximo.
C++
// A simple C++ program to find max subarray XOR
#include<bits/stdc++.h>
using namespace std;
int maxSubarrayXOR(int arr[], int n)
{
int ans = INT_MIN; // Initialize result
// Pick starting points of subarrays
for (int i=0; i<n; i++)
{
int curr_xor = 0; // to store xor of current subarray
// Pick ending points of subarrays starting with i
for (int j=i; j<n; j++)
{
curr_xor = curr_xor ^ arr[j];
ans = max(ans, curr_xor);
}
}
return ans;
}
// Driver program to test above functions
int main()
{
int arr[] = {8, 1, 2, 12};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Max subarray XOR is " << maxSubarrayXOR(arr, n);
return 0;
}
Java
// A simple Java program to find max subarray XOR
class GFG {
static int maxSubarrayXOR(int arr[], int n)
{
int ans = Integer.MIN_VALUE; // Initialize result
// Pick starting points of subarrays
for (int i=0; i<n; i++)
{
// to store xor of current subarray
int curr_xor = 0;
// Pick ending points of subarrays starting with i
for (int j=i; j<n; j++)
{
curr_xor = curr_xor ^ arr[j];
ans = Math.max(ans, curr_xor);
}
}
return ans;
}
// Driver program to test above functions
public static void main(String args[])
{
int arr[] = {8, 1, 2, 12};
int n = arr.length;
System.out.println("Max subarray XOR is " +
maxSubarrayXOR(arr, n));
}
}
//This code is contributed by Sumit Ghosh
Python3
# A simple Python program
# to find max subarray XOR
def maxSubarrayXOR(arr,n):
ans = -2147483648 #Initialize result
# Pick starting points of subarrays
for i in range(n):
# to store xor of current subarray
curr_xor = 0
# Pick ending points of
# subarrays starting with i
for j in range(i,n):
curr_xor = curr_xor ^ arr[j]
ans = max(ans, curr_xor)
return ans
# Driver code
arr = [8, 1, 2, 12]
n = len(arr)
print("Max subarray XOR is ",
maxSubarrayXOR(arr, n))
# This code is contributed
# by Anant Agarwal.
C#
// A simple C# program to find
// max subarray XOR
using System;
class GFG
{
// Function to find max subarray
static int maxSubarrayXOR(int []arr, int n)
{
int ans = int.MinValue;
// Initialize result
// Pick starting points of subarrays
for (int i = 0; i < n; i++)
{
// to store xor of current subarray
int curr_xor = 0;
// Pick ending points of
// subarrays starting with i
for (int j = i; j < n; j++)
{
curr_xor = curr_xor ^ arr[j];
ans = Math.Max(ans, curr_xor);
}
}
return ans;
}
// Driver code
public static void Main()
{
int []arr = {8, 1, 2, 12};
int n = arr.Length;
Console.WriteLine("Max subarray XOR is " +
maxSubarrayXOR(arr, n));
}
}
// This code is contributed by Sam007.
PHP
<?php
// A simple PHP program to
// find max subarray XOR
function maxSubarrayXOR($arr, $n)
{
// Initialize result
$ans = PHP_INT_MIN;
// Pick starting points
// of subarrays
for ($i = 0; $i < $n; $i++)
{
// to store xor of
// current subarray
$curr_xor = 0;
// Pick ending points of
// subarrays starting with i
for ($j = $i; $j < $n; $j++)
{
$curr_xor = $curr_xor ^ $arr[$j];
$ans = max($ans, $curr_xor);
}
}
return $ans;
}
// Driver Code
$arr = array(8, 1, 2, 12);
$n = count($arr);
echo "Max subarray XOR is "
, maxSubarrayXOR($arr, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// A simple Javascript program to find
// max subarray XOR
function maxSubarrayXOR(arr, n)
{
// Initialize result
let ans = Number.MIN_VALUE;
// Pick starting points of subarrays
for(let i = 0; i < n; i++)
{
// To store xor of current subarray
let curr_xor = 0;
// Pick ending points of subarrays
// starting with i
for(let j = i; j < n; j++)
{
curr_xor = curr_xor ^ arr[j];
ans = Math.max(ans, curr_xor);
}
}
return ans;
}
// Driver code
let arr = [ 8, 1, 2, 12 ];
let n = arr.length;
document.write("Max subarray XOR is " +
maxSubarrayXOR(arr, n));
// This code is contributed by divyesh072019
</script>
Producción
Max subarray XOR is 15
Complejidad Temporal: O(n 2 ).
Espacio Auxiliar: O(1)
Una solución eficiente puede resolver el problema anterior en tiempo O(n) bajo el supuesto de que los números enteros toman una cantidad fija de bits para almacenar. La idea es utilizar Trie Data Structure. A continuación se muestra el algoritmo.
1) Create an empty Trie. Every node of Trie is going to
contain two children, for 0 and 1 value of bit.
2) Initialize pre_xor = 0 and insert into the Trie.
3) Initialize result = minus infinite
4) Traverse the given array and do following for every
array element arr[i].
a) pre_xor = pre_xor ^ arr[i]
pre_xor now contains xor of elements from
arr[0] to arr[i].
b) Query the maximum xor value ending with arr[i]
from Trie.
c) Update result if the value obtained in step
4.b is more than current value of result.
¿Cómo funciona 4.b?
Podemos observar desde el algoritmo anterior que construimos un Trie que contiene XOR de todos los prefijos de la array dada. Para encontrar el subarreglo XOR máximo que termina en arr[i], puede haber dos casos.
i) El prefijo en sí tiene el valor XOR máximo que termina con arr[i]. Por ejemplo, si i=2 en {8, 2, 1, 12}, entonces el subarreglo máximo xor que termina en arr[2] es el prefijo completo.
ii) Necesitamos eliminar algunos prefijos (que terminan en el índice de 0 a i-1). Por ejemplo, si i=3 en {8, 2, 1, 12}, entonces el subarreglo máximo xor que termina en arr[3] comienza con arr[1] y debemos eliminar arr[0].
Para encontrar el prefijo a eliminar, buscamos la entrada en Trie que tiene el valor XOR máximo con el prefijo actual. Si hacemos XOR de dicho prefijo anterior con el prefijo actual, obtenemos el valor máximo de XOR que termina con arr[i].
Si no hay ningún prefijo que eliminar (caso i), devolvemos 0 (es por eso que insertamos 0 en Trie).
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program for a Trie based O(n) solution to find max
// subarray XOR
#include<bits/stdc++.h>
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function to create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<<i);
// Create a new node if needed
if (temp->arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR 'pre_xor' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<<i);
// Traverse Trie, first look for a
// prefix that has opposite bit
if (temp->arr[1-val]!=NULL)
temp = temp->arr[1-val];
// If there is no prefix with opposite
// bit, then look for same bit.
else if (temp->arr[val] != NULL)
temp = temp->arr[val];
}
return pre_xor^(temp->value);
}
// Returns maximum XOR value of a subarray in arr[0..n-1]
int maxSubarrayXOR(int arr[], int n)
{
// Create a Trie and insert 0 into it
TrieNode *root = newNode();
insert(root, 0);
// Initialize answer and xor of current prefix
int result = INT_MIN, pre_xor =0;
// Traverse all input array element
for (int i=0; i<n; i++)
{
// update current prefix xor and insert it into Trie
pre_xor = pre_xor^arr[i];
insert(root, pre_xor);
// Query for current prefix xor in Trie and update
// result if required
result = max(result, query(root, pre_xor));
}
return result;
}
// Driver program to test above functions
int main()
{
int arr[] = {8, 1, 2, 12};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Max subarray XOR is " << maxSubarrayXOR(arr, n);
return 0;
}
Java
// Java program for a Trie based O(n) solution to
// find max subarray XOR
class GFG
{
// Assumed int size
static final int INT_SIZE = 32;
// A Trie Node
static class TrieNode
{
int value; // Only used in leaf nodes
TrieNode[] arr = new TrieNode[2];
public TrieNode() {
value = 0;
arr[0] = null;
arr[1] = null;
}
}
static TrieNode root;
// Inserts pre_xor to trie with given root
static void insert(int pre_xor)
{
TrieNode temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
int val = (pre_xor & (1<<i)) >=1 ? 1 : 0;
// Create a new node if needed
if (temp.arr[val] == null)
temp.arr[val] = new TrieNode();
temp = temp.arr[val];
}
// Store value at leaf node
temp.value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR 'pre_xor' and returns the XOR of this
// maximum with pre_xor which is maximum XOR ending
// with last element of pre_xor.
static int query(int pre_xor)
{
TrieNode temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
int val = (pre_xor & (1<<i)) >= 1 ? 1 : 0;
// Traverse Trie, first look for a
// prefix that has opposite bit
if (temp.arr[1-val] != null)
temp = temp.arr[1-val];
// If there is no prefix with opposite
// bit, then look for same bit.
else if (temp.arr[val] != null)
temp = temp.arr[val];
}
return pre_xor^(temp.value);
}
// Returns maximum XOR value of a subarray in
// arr[0..n-1]
static int maxSubarrayXOR(int arr[], int n)
{
// Create a Trie and insert 0 into it
root = new TrieNode();
insert(0);
// Initialize answer and xor of current prefix
int result = Integer.MIN_VALUE;
int pre_xor =0;
// Traverse all input array element
for (int i=0; i<n; i++)
{
// update current prefix xor and insert it
// into Trie
pre_xor = pre_xor^arr[i];
insert(pre_xor);
// Query for current prefix xor in Trie and
// update result if required
result = Math.max(result, query(pre_xor));
}
return result;
}
// Driver program to test above functions
public static void main(String args[])
{
int arr[] = {8, 1, 2, 12};
int n = arr.length;
System.out.println("Max subarray XOR is " +
maxSubarrayXOR(arr, n));
}
}
// This code is contributed by Sumit Ghosh
Python3
"""Python implementation for a Trie based solution
to find max subArray XOR"""
"""structure of Trie Node"""
class Node:
def __init__(self, data):
self.data = data
self.left = None # left node for 0
self.right = None # right node for 1
""" class for implementing Trie """
class Trie:
def __init__(self):
self.root = Node(0)
"""insert pre_xor to trie with given root"""
def insert(self, pre_xor):
self.temp = self.root
"""start from msb, insert all bits of pre_xor
into the Trie"""
for i in range(31, -1, -1):
"""Find current bit in prefix sum"""
val = pre_xor & (1<<i)
if val :
"""create new node if needed"""
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
if not val:
"""create new node if needed"""
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
"""store value at leaf node"""
self.temp.data = pre_xor
"""find the maximum xor ending with last number
in prefix XOR and return the XOR of this"""
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
"""find the current bit in prefix xor"""
val = xor & (1<<i)
"""Traverse the trie, first look for opposite bit
and then look for same bit"""
if val:
if self.temp.left:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
return xor ^ self.temp.data
"""returns maximum XOR value of subarray"""
def maxSubArrayXOR(self, n, Arr):
"""insert 0 in the trie"""
self.insert(0)
"""initialize result and pre_xor"""
result = -float('inf')
pre_xor = 0
"""traverse all input array element"""
for i in range(n):
"""update current prefix xor and insert it into Trie"""
pre_xor = pre_xor ^ Arr[i]
self.insert(pre_xor)
"""Query for current prefix xor in Trie and update result"""
result = max(result, self.query(pre_xor))
return result
"""Driver program to test above functions"""
if __name__ == "__main__":
Arr = [8, 1, 2, 12]
n = len(Arr)
trie = Trie()
print(trie.maxSubArrayXOR(n, Arr))
# This code is contributed by chaudhary_19
C#
using System;
// C# program for a Trie based O(n) solution to
// find max subarray XOR
public class GFG
{
// Assumed int size
public const int INT_SIZE = 32;
// A Trie Node
public class TrieNode
{
public int value; // Only used in leaf nodes
public TrieNode[] arr = new TrieNode[2];
public TrieNode()
{
value = 0;
arr[0] = null;
arr[1] = null;
}
}
public static TrieNode root;
// Inserts pre_xor to trie with given root
public static void insert(int pre_xor)
{
TrieNode temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i = INT_SIZE-1; i >= 0; i--)
{
// Find current bit in given prefix
int val = (pre_xor & (1 << i)) >= 1 ? 1 : 0;
// Create a new node if needed
if (temp.arr[val] == null)
{
temp.arr[val] = new TrieNode();
}
temp = temp.arr[val];
}
// Store value at leaf node
temp.value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR 'pre_xor' and returns the XOR of this
// maximum with pre_xor which is maximum XOR ending
// with last element of pre_xor.
public static int query(int pre_xor)
{
TrieNode temp = root;
for (int i = INT_SIZE-1; i >= 0; i--)
{
// Find current bit in given prefix
int val = (pre_xor & (1 << i)) >= 1 ? 1 : 0;
// Traverse Trie, first look for a
// prefix that has opposite bit
if (temp.arr[1 - val] != null)
{
temp = temp.arr[1 - val];
}
// If there is no prefix with opposite
// bit, then look for same bit.
else if (temp.arr[val] != null)
{
temp = temp.arr[val];
}
}
return pre_xor ^ (temp.value);
}
// Returns maximum XOR value of a subarray in
// arr[0..n-1]
public static int maxSubarrayXOR(int[] arr, int n)
{
// Create a Trie and insert 0 into it
root = new TrieNode();
insert(0);
// Initialize answer and xor of current prefix
int result = int.MinValue;
int pre_xor = 0;
// Traverse all input array element
for (int i = 0; i < n; i++)
{
// update current prefix xor and insert it
// into Trie
pre_xor = pre_xor ^ arr[i];
insert(pre_xor);
// Query for current prefix xor in Trie and
// update result if required
result = Math.Max(result, query(pre_xor));
}
return result;
}
// Driver program to test above functions
public static void Main(string[] args)
{
int[] arr = new int[] {8, 1, 2, 12};
int n = arr.Length;
Console.WriteLine("Max subarray XOR is " + maxSubarrayXOR(arr, n));
}
}
// This code is contributed by Shrikant13
Producción
Max subarray XOR is 15
Complejidad temporal: O(n).
Espacio Auxiliar: O(n)
Ejercicio: Extienda la solución anterior para que también imprima los índices inicial y final del subarreglo con el valor máximo (Sugerencia: podemos agregar un campo más al Node Trie para lograr esto
Este artículo es una contribución de Aarti_Rathi y Romil Punetha. Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA