给定一个包含N个整数的数组arr [] ，并且存在Q个查询，其中每个查询包含一个范围[L，R] 。任务是查找给定索引范围内的所有元素是否具有偶数频率。
例子：
Input: arr[] = {1, 1, 2, 2, 1}, Q[][] = {{1, 5}, {1, 4}, {3, 4}}
Output:
No
Yes
Yes
Input: arr[] = {100, 100, 200, 100}, Q[][] = {{1, 4}, {1, 2}}
Output:
No
Yes
天真的方法：一种简单的方法是从每个查询的索引范围[L，R]进行迭代，并计算该范围内每个元素的出现频率，并检查每个元素是否发生偶数次。这种方法最坏的情况是时间复杂度为O(Q * N) ，其中Q是查询数， N是数组中元素的数。
高效的方法：由于两个相等数字的XOR为0，即，如果数组的所有元素出现偶数次，则整个数组的XOR将为0 。对于给定范围[L，R]中的元素可以说相同。现在，要检查给定范围内所有元素的XOR是否为0 ，可以创建一个前缀XOR数组，其中preXor [i]将存储元素arr [0…i]的XOR。元素arr [i…j]的XOR可以很容易地计算为preXor [j] ^ preXor [i – 1] 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to perform the given queries
void performQueries(vector& A,
vector >& q)
{
int n = (int)A.size();
// Making array 1-indexed
A.insert(A.begin(), 0);
// To store the cumulative xor
vector pref_xor(n + 1, 0);
// Taking cumulative Xor
for (int i = 1; i <= n; ++i) {
pref_xor[i]
= pref_xor[i - 1] ^ A[i];
}
// Iterating over the queries
for (auto i : q) {
int L = i.first, R = i.second;
if (L > R)
swap(L, R);
// If both indices are different and xor
// in the range [L, R] is 0
if (L != R and pref_xor[R] == pref_xor[L - 1])
cout << "Yes\n";
else
cout << "No\n";
}
}
// Driver code
int main()
{
vector Arr = { 1, 1, 2, 2, 1 };
vector > q = {
{ 1, 5 },
{ 1, 4 },
{ 3, 4 }
};
performQueries(Arr, q);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to perform the given queries
static void performQueries(int []A,
pair[] q)
{
int n = A.length;
// To store the cumulative xor
int []pref_xor = new int[n + 1];
// Taking cumulative Xor
for (int i = 1; i <=n; ++i)
{
pref_xor[i] = pref_xor[i - 1] ^
A[i - 1];
}
// Iterating over the queries
for (pair i : q)
{
int L = i.first, R = i.second;
if (L > R)
{
int temp = L;
L = R;
R = temp;
}
// If both indices are different and xor
// in the range [L, R] is 0
if (L != R && pref_xor[R] == pref_xor[L - 1])
System.out.println("Yes");
else
System.out.println("No");
}
}
// Driver code
static public void main (String []arg)
{
int []Arr = { 1, 1, 2, 2, 1 };
pair[] q = { new pair(1, 5 ),
new pair(1, 4 ),
new pair(3, 4 ) };
performQueries(Arr, q);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementation of the approach
# Function to perform the given queries
def performQueries(A, q):
n = len(A)
# To store the cumulative xor
pref_xor = [0 for i in range(n + 1)]
# Taking cumulative Xor
for i in range(1, n + 1):
pref_xor[i] = pref_xor[i - 1] ^ A[i - 1]
# Iterating over the queries
for i in q:
L = i[0]
R = i[1]
if (L > R):
L, R = R, L
# If both indices are different and
# xor in the range [L, R] is 0
if (L != R and
pref_xor[R] == pref_xor[L - 1]):
print("Yes")
else:
print("No")
# Driver code
Arr = [1, 1, 2, 2, 1]
q = [[ 1, 5 ],
[ 1, 4 ],
[ 3, 4 ]]
performQueries(Arr, q);
# This code is contributed by Mohit KumarC#
// C# implementation of the approach
using System;
class GFG
{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to perform the given queries
static void performQueries(int []A,
pair[] q)
{
int n = A.Length;
// To store the cumulative xor
int []pref_xor = new int[n + 1];
// Taking cumulative Xor
for (int i = 1; i <= n; ++i)
{
pref_xor[i] = pref_xor[i - 1] ^
A[i - 1];
}
// Iterating over the queries
foreach (pair k in q)
{
int L = k.first, R = k.second;
if (L > R)
{
int temp = L;
L = R;
R = temp;
}
// If both indices are different and xor
// in the range [L, R] is 0
if (L != R && pref_xor[R] == pref_xor[L - 1])
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// Driver code
static public void Main (String []arg)
{
int []Arr = { 1, 1, 2, 2, 1 };
pair[] q = {new pair(1, 5 ),
new pair(1, 4 ),
new pair(3, 4 )};
performQueries(Arr, q);
}
}
// This code is contributed by Rajput-Ji输出：
No
Yes
Yes
时间复杂度： O(N)