1442. number of triples forming two exclusive or equal arrays

Give you an array of integers arr .

Now we need to take three subscripts from the array i、j and k , among (0 <= i < j <= k < arr.length) .

a and b The definition is as follows ：

a = arr[i] ^ arr[i + 1] ^ … ^ arr[j - 1]
b = arr[j] ^ arr[j + 1] ^ … ^ arr[k]
Be careful ：^ Express Bitwise XOR operation .

Please return to can make a == b Set up a triple (i, j , k) Number of .

Example 1：

 Input ：arr = [2,3,1,6,7]
 Output ：4
 explain ： The triples that satisfy the meaning of the question are  (0,1,2), (0,2,2), (2,3,4)  as well as  (2,4,4)

Example 2：

 Input ：arr = [1,1,1,1,1]
 Output ：10

Example 3：

 Input ：arr = [2,3]
 Output ：0

Example 5：

 Input ：arr = [7,11,12,9,5,2,7,17,22]
 Output ：8

Tips ：

1 <= arr.length <= 300
1 <= arr[i] <= 10^8

1. Triple cycle

class Solution {
    
public:
    int countTriplets(vector<int>& arr) {
    
        int n=arr.size();
        vector<int>s(n+1);
        for(int i=0;i<n;i++){
    
            s[i+1]=s[i]^arr[i];
        }
        int cnt=0;

        for(int i=0;i<n;i++){
    
            for(int j=i+1;j<n;j++){
    
                for(int k=j;k<n;k++){
    
                    if(s[i]==s[k+1]){
    
                        cnt++;
                    }
                }
            }
        }

        return cnt;

    }
};

2. Double cycle

class Solution {
    
public:
    int countTriplets(vector<int>& arr) {
    
        int n=arr.size();
        vector<int>s(n+1);
        for(int i=0;i<n;i++){
    
            s[i+1]=s[i]^arr[i];
        }
        int cnt=0;

        for(int i=0;i<n;i++){
    
            for(int k=i+1;k<n;k++){
    
                if(s[i]==s[k+1]){
    
                    cnt+=(k-i);
                }
            }
        }

        return cnt;

    }
};

3. Hashtable （ A heavy cycle ）

class Solution {
    
public:
    int countTriplets(vector<int>& arr) {
    
        int n=arr.size();
        vector<int>s(n+1);
        for(int i=0;i<n;i++){
    
            s[i+1]=s[i]^arr[i];
        }
        int cnt=0;
        map<int,int>m1;
        map<int,int>m2;

        for(int k=0;k<n;k++){
    
            if(m1.count(s[k+1])){
    
                cnt+=m1[s[k+1]]*k-m2[s[k+1]];
            }
            m1[s[k]]++;
            m2[s[k]]+=k;
        }
        return cnt;

    }
};

4. Prefixes and optimizations

class Solution {
    
public:
    int countTriplets(vector<int>& arr) {
    
        int n=arr.size();
        int cnt=0;
        map<int,int>m1;
        map<int,int>m2;
        int s1=0;
        int s2=0;

        for(int k=0;k<n;k++){
    
            s2=s1^arr[k];

            if(m1.count(s2)){
    
                cnt+=m1[s2]*k-m2[s2];
            }
            m1[s1]++;
            m2[s1]+=k;
            s1=s2;
        }
        return cnt;

    }
};