2683. Neighboring Bitwise XOR - Detailed Explanation

Problem Statement

Description:
You are given a 0-indexed integer array arr. A triplet of indices (i, j, k) is considered valid if it satisfies:

• ( 0 \le i < j \le k < \text{arr.length} )
• The bitwise XOR of the subarray arr[i] to arr[j-1] is equal to the bitwise XOR of the subarray arr[j] to arr[k].

In other words, let
[ \text{xor1} = \text{arr}[i] \oplus \text{arr}[i+1] \oplus \dots \oplus \text{arr}[j-1] ]
and
[ \text{xor2} = \text{arr}[j] \oplus \text{arr}[j+1] \oplus \dots \oplus \text{arr}[k] ]
the triplet ((i, j, k)) is valid if: [ \text{xor1} = \text{xor2}. ]

Return the number of such valid triplets.

Example 1:

• Input: arr = [2, 3, 1, 6, 7]
• Output: 4
• Explanation:
  The valid triplets are:
  • ((0, 1, 2)): (2 = 3 \oplus 1) (because (2 = 2))
  • ((0, 2, 3)): (2 \oplus 3 = 1 \oplus 6) (both equal (1))
  • ((2, 3, 4)): (1 = 6 \oplus 7) (because (1 = 1))
  • ((0, 1, 4)): (2 = 3 \oplus 1 \oplus 6 \oplus 7) (both equal (2))
  (Note: The above explanation is illustrative; the main idea is that the XOR on the left of index j equals the XOR on the right for a valid triplet.)

Example 2:

• Input: arr = [1, 1, 1, 1, 1]
• Output: 10
• Explanation:
  Every possible triplet works because every subarray XOR is 1 or 0 in a way that the condition is satisfied. (There are 10 valid triplets in total.)

Constraints:

• ( 1 \le \text{arr.length} \le 300 )
• ( 1 \le \text{arr}[i] \le 10^8 )

Hints

• Hint 1:
  Notice that the XOR operation is associative and that if you define a prefix XOR array ( p ) where ( p[0] = 0 ) and
  [ p[i+1] = p[i] \oplus \text{arr}[i] ] then the XOR of a subarray from index ( i ) to ( j-1 ) is given by: [ p[j] \oplus p[i] ]

• Hint 2:
  For two subarrays to have the same XOR, you need: [ p[j] \oplus p[i] = p[k+1] \oplus p[j] ] Rearranging (using the property that ( a \oplus a = 0 ) and ( a \oplus 0 = a )), you can show that: [ p[i] = p[k+1] ]

• Hint 3:
  Once you observe that for every pair of indices ( (i, k+1) ) where ( p[i] = p[k+1] ), any index ( j ) such that ( i < j \le k ) will satisfy the condition, you can count the number of valid triplets contributed by this pair.

3. Approach Explanation

Using Prefix XOR and Counting

1. Prefix XOR Array:
   Build an array ( p ) of length ( n+1 ) (where ( n = \text{arr.length} )) such that:

   • ( p[0] = 0 )
   • For each ( i ) from 0 to ( n-1 ), ( p[i+1] = p[i] \oplus \text{arr}[i] )

2. Observation:
   For any valid triplet ( (i, j, k) ) (with ( 0 \le i < j \le k < n )), the condition that: [ \text{arr}[i] \oplus \dots \oplus \text{arr}[j-1] = \text{arr}[j] \oplus \dots \oplus \text{arr}[k] ] is equivalent to: [ p[i] = p[k+1] ]

3. Counting Triplets:
   For every pair ( (i, k+1) ) with ( p[i] = p[k+1] ) and ( i < k+1 ), every index ( j ) with ( i < j \le k ) forms a valid triplet. The number of valid ( j ) values is ( k - i ).

4. Efficient Counting Using Hash Map:
   Instead of checking all pairs explicitly (which would be ( O(n^2) )), we can use a hash map (or dictionary) to keep track of:

   • The number of times a particular prefix XOR value has occurred (its frequency).
   • The cumulative sum of indices at which it occurred.

   Then, for each index ( k ) (corresponding to ( p[k+1] )), if the prefix XOR value has occurred ( \text{count} ) times at earlier indices, with the sum of those indices being ( \text{sum} ), then the number of valid triplets contributed by the current index is: [ \text{count} \times (k) - \text{sum} ] (Here, ( k ) represents the current index in the original array where ( p[k+1] ) is computed.)

5. Total Count:
   Sum the contributions for every index to get the final answer.

Code Implementations

Python Code Implementation