给定一个整数数组,arr []的大小为N。arr []的任何子数组的XOR值定义为该子数组中所有整数的异或。任务是找到XOR值为2的幂的子数组的数量。(1、2、4、8、16,…。)
例子:
Input : arr[] = {2, 6, 7, 5, 8}
Output : 6
Subarrays : {2, 6}, {2}, {6, 7}, {6, 7, 5}, {7, 5}, {8}
Input : arr[] = {2, 4, 8, 16}
Output : 4
方法:
- 维护一个哈希图,该哈希图存储所有前缀XOR值及其计数。
- 在任何索引i处,我们都需要找到在i处结束并具有XOR值2的幂的子数组的数量。对于所有2的幂,我们需要找到可能的子数组的数量。例如。 :假设当前XOR值直到索引i为X,我们需要找到结果为16的子数组的数量(例如S),因此我们需要对前缀XOR Y进行计数,以使(X ^ Y)= S或Y = S ^ X。可以使用哈希图找到Y。
- 对所有索引执行步骤2,添加输出。
下面是上述方法的实现:
C++
// C++ Program to count number of subarrays
// with Bitwise-XOR as power of 2
#include
#define ll long long int
#define MAX 10
using namespace std;
// Function to find number of subarrays
int findSubarray(int array[], int n)
{
// Hash Map to store prefix XOR values
unordered_map mp;
// When no element is selected
mp.insert({ 0, 1 });
int answer = 0;
int preXor = 0;
for (int i = 0; i < n; i++) {
int value = 1;
preXor ^= array[i];
// Check for all the powers of 2,
// till a MAX value
for (int j = 1; j <= MAX; j++) {
int Y = value ^ preXor;
if (mp.find(Y) != mp.end()) {
answer += mp[Y];
}
value *= 2;
}
// Insert Current prefixxor in Hash Map
if (mp.find(preXor) != mp.end()) {
mp[preXor]++;
}
else {
mp.insert({ preXor, 1 });
}
}
return answer;
}
// Driver Code
int main()
{
int array[] = { 2, 6, 7, 5, 8 };
int n = sizeof(array) / sizeof(array[0]);
cout << findSubarray(array, n) << endl;
return 0;
}
Java
// Java Program to count number of subarrays
// with Bitwise-XOR as power of 2
import java.util.*;
class GFG
{
static int MAX = 10;
// Function to find number of subarrays
static int findSubarray(int array[], int n)
{
// Hash Map to store prefix XOR values
HashMap mp = new HashMap();
// When no element is selected
mp.put(0, 1);
int answer = 0;
int preXor = 0;
for (int i = 0; i < n; i++)
{
int value = 1;
preXor ^= array[i];
// Check for all the powers of 2,
// till a MAX value
for (int j = 1; j <= MAX; j++)
{
int Y = value ^ preXor;
if (mp.containsKey(Y))
{
answer += mp.get(Y);
}
value *= 2;
}
// Insert Current prefixxor in Hash Map
if (mp.containsKey(preXor))
{
mp.put(preXor,mp.get(preXor) + 1);
}
else
{
mp.put(preXor, 1);
}
}
return answer;
}
// Driver Code
public static void main (String[] args)
{
int array[] = { 2, 6, 7, 5, 8 };
int n = array.length;
System.out.println(findSubarray(array, n));
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 Program to count number of subarrays
# with Bitwise-XOR as power of 2
MAX = 10
# Function to find number of subarrays
def findSubarray(array, n):
# Hash Map to store prefix XOR values
mp = dict()
# When no element is selected
mp[0] = 1
answer = 0
preXor = 0
for i in range(n):
value = 1
preXor ^= array[i]
# Check for all the powers of 2,
# till a MAX value
for j in range(1, MAX + 1):
Y = value ^ preXor
if ( Y in mp.keys()):
answer += mp[Y]
value *= 2
# Insert Current prefixxor in Hash Map
if (preXor in mp.keys()):
mp[preXor] += 1
else:
mp[preXor] = 1
return answer
# Driver Code
array = [2, 6, 7, 5, 8]
n = len(array)
print(findSubarray(array, n))
# This code is contributed by Mohit Kumar
C#
// C# Program to count number of subarrays
// with Bitwise-XOR as power of 2
using System;
using System.Collections.Generic;
class GFG
{
static int MAX = 10;
// Function to find number of subarrays
static int findSubarray(int []array, int n)
{
// Hash Map to store prefix XOR values
Dictionary mp = new Dictionary();
// When no element is selected
mp.Add(0, 1);
int answer = 0;
int preXor = 0;
for (int i = 0; i < n; i++)
{
int value = 1;
preXor ^= array[i];
// Check for all the powers of 2,
// till a MAX value
for (int j = 1; j <= MAX; j++)
{
int Y = value ^ preXor;
if (mp.ContainsKey(Y))
{
answer += mp[Y];
}
value *= 2;
}
// Insert Current prefixxor in Hash Map
if (mp.ContainsKey(preXor))
{
mp[preXor] = mp[preXor] + 1;
}
else
{
mp.Add(preXor, 1);
}
}
return answer;
}
// Driver Code
public static void Main (String[] args)
{
int []array = { 2, 6, 7, 5, 8 };
int n = array.Length;
Console.WriteLine(findSubarray(array, n));
}
}
// This code is contributed by 29AjayKumar
输出:
6