Problem Description

You are given an array of integers arr. The task is to count how many triplets of indices (i, j, k) exist such that a specific condition is met.

For each triplet, you need to select three indices where 0 <= i < j <= k < arr.length. These indices divide the array into two segments:

• Segment a: XOR of all elements from index i to j-1 (i.e., arr[i] ^ arr[i+1] ^ ... ^ arr[j-1])
• Segment b: XOR of all elements from index j to k (i.e., arr[j] ^ arr[j+1] ^ ... ^ arr[k])

The goal is to find all triplets where a == b.

The key insight is that if a == b, then a ^ b = 0. This means the XOR of all elements from index i to k equals 0. When this happens, we can place j at any position between i+1 and k (inclusive), and the condition a == b will be satisfied.

The solution iterates through each possible starting position i, then for each ending position k, it calculates the XOR sum of elements from i to k. When this XOR sum equals 0, it means we've found k - i valid triplets (since j can be placed at any of the k - i positions between i+1 and k).

For example, if arr = [2, 3, 1, 6, 7] and we find that XOR from index 0 to 2 equals 0 (i.e., 2 ^ 3 ^ 1 = 0), then we have 2 valid triplets: (0, 1, 2) and (0, 2, 2).

Intuition

The key observation is understanding what it means for a == b in terms of XOR operations. Since a represents the XOR of elements from index i to j-1, and b represents the XOR from index j to k, when a == b, something interesting happens mathematically.

If we XOR equal values together, we get 0. So if a == b, then a ^ b = 0. But what is a ^ b? It's actually the XOR of all elements from index i to k. This is because:

• a ^ b = (arr[i] ^ ... ^ arr[j-1]) ^ (arr[j] ^ ... ^ arr[k])
• This simplifies to arr[i] ^ arr[i+1] ^ ... ^ arr[k]

So the problem reduces to finding all subarrays from index i to k where the XOR of all elements equals 0.

Once we find such a subarray where XOR equals 0, we need to count how many valid triplets it contains. The beauty here is that when the total XOR from i to k is 0, we can split this subarray at any position j (where i < j <= k), and the two parts will always have equal XOR values. This is because if the total XOR is 0, then a ^ b = 0, which means a = b.

There are exactly k - i positions where we can place j (from i+1 to k inclusive). Each position gives us a valid triplet (i, j, k).

This insight allows us to simplify our approach: instead of checking all possible combinations of i, j, and k separately, we only need to find pairs (i, k) where the XOR from i to k equals 0, and then we automatically know there are k - i valid triplets for that pair.

Learn more about Math and Prefix Sum patterns.

Solution Approach

The implementation uses a straightforward enumeration approach based on our insight that we need to find subarrays with XOR sum equal to 0.

The algorithm works as follows:

1. Initialize counters: We start with ans = 0 to track the total count of valid triplets, and get the array length n.

2. Enumerate starting position i: We iterate through each possible starting index i from 0 to n-1. For each starting position, we extract the element x = arr[i] and initialize our XOR sum s with this value.

3. Enumerate ending position k: For each starting position i, we iterate through all possible ending positions k from i+1 to n-1. This ensures we have at least two elements in our subarray (required since i < j <= k).

4. Calculate running XOR: As we extend k, we continuously update our XOR sum by computing s ^= arr[k]. This gives us the XOR of all elements from index i to k.

5. Check for valid subarrays: Whenever s == 0, we've found a subarray from i to k where the total XOR is 0. This means we can place j at any of the k - i positions (from i+1 to k), and each placement gives us a valid triplet where a == b.

6. Count the triplets: When we find s == 0, we add k - i to our answer, representing all valid triplets for this (i, k) pair.

The time complexity is O(n²) since we have two nested loops iterating through the array. The space complexity is O(1) as we only use a constant amount of extra space for variables.

This approach is efficient because instead of checking all O(n³) possible triplets directly, we reduce the problem to checking O(n²) pairs of (i, k) and mathematically determining the count of valid j values for each pair.

Example Walkthrough

Let's walk through the solution with arr = [2, 3, 1, 6, 7].

Step 1: Start with i = 0

• Initialize x = arr[0] = 2, s = 2
• Check k = 1: s = 2 ^ 3 = 1 (not 0, skip)
• Check k = 2: s = 1 ^ 1 = 0 ✓
  • Found XOR from i=0 to k=2 equals 0
  • Can place j at positions 1 or 2
  • Add k - i = 2 - 0 = 2 triplets: (0,1,2) and (0,2,2)
• Check k = 3: s = 0 ^ 6 = 6 (not 0, skip)
• Check k = 4: s = 6 ^ 7 = 1 (not 0, skip)

Step 2: Start with i = 1

• Initialize x = arr[1] = 3, s = 3
• Check k = 2: s = 3 ^ 1 = 2 (not 0, skip)
• Check k = 3: s = 2 ^ 6 = 4 (not 0, skip)
• Check k = 4: s = 4 ^ 7 = 3 (not 0, skip)

Step 3: Start with i = 2

• Initialize x = arr[2] = 1, s = 1
• Check k = 3: s = 1 ^ 6 = 7 (not 0, skip)
• Check k = 4: s = 7 ^ 7 = 0 ✓
  • Found XOR from i=2 to k=4 equals 0
  • Can place j at positions 3 or 4
  • Add k - i = 4 - 2 = 2 triplets: (2,3,4) and (2,4,4)

Step 4: Start with i = 3

• Initialize x = arr[3] = 6, s = 6
• Check k = 4: s = 6 ^ 7 = 1 (not 0, skip)

Total count: 2 + 2 = 4 triplets

The valid triplets are:

• (0,1,2): a = 2, b = 3^1 = 2 ✓
• (0,2,2): a = 2^3 = 1, b = 1 ✓
• (2,3,4): a = 1, b = 6^7 = 1 ✓
• (2,4,4): a = 1^6 = 7, b = 7 ✓

Solution Implementation

Time and Space Complexity

The time complexity is O(n²), where n is the length of the array arr. This is because we have two nested loops: the outer loop iterates through each element in the array (running n times), and for each element i, the inner loop iterates from i+1 to n-1. In the worst case, this gives us approximately n * (n-1) / 2 iterations, which simplifies to O(n²).

The space complexity is O(1). The algorithm only uses a constant amount of extra space for variables ans, n, i, x, s, and k, regardless of the input array size. No additional data structures that scale with input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect XOR Initialization

A common mistake is initializing the XOR sum as 0 instead of arr[start_index]. This leads to incorrect XOR calculations.

Incorrect:

for start_index in range(array_length):
    xor_sum = 0  # Wrong! Missing arr[start_index]
    for end_index in range(start_index + 1, array_length):
        xor_sum ^= arr[end_index]

Correct:

for start_index in range(array_length):
    xor_sum = arr[start_index]  # Start with first element
    for end_index in range(start_index + 1, array_length):
        xor_sum ^= arr[end_index]

2. Misunderstanding Index Relationships

Many incorrectly assume j must be strictly between i and k (i.e., i < j < k), but the problem states i < j <= k, meaning j can equal k.

Wrong interpretation leads to:

if xor_sum == 0:
    total_count += end_index - start_index - 1  # Wrong count!

Correct understanding:

if xor_sum == 0:
    total_count += end_index - start_index  # j can be from i+1 to k inclusive

3. Attempting O(n³) Brute Force

Some try to directly enumerate all triplets, which is inefficient and unnecessary:

Inefficient approach:

for i in range(n):
    for j in range(i + 1, n):
        for k in range(j, n):
            a = calculate_xor(i, j-1)
            b = calculate_xor(j, k)
            if a == b:
                count += 1

Better approach: Use the XOR property that a == b implies a ^ b == 0, reducing to O(n²).

4. Off-by-One Errors in Loop Bounds

Forgetting that end_index must start from start_index + 1 to ensure i < j:

Wrong:

for end_index in range(start_index, array_length):  # Allows i == j

Correct:

for end_index in range(start_index + 1, array_length):  # Ensures i < j

5. Not Handling Edge Cases

Failing to consider arrays with length less than 2:

Solution:

if len(arr) < 2:
    return 0  # No valid triplets possible