Count Triplets That Can Form Two Arrays of Equal XOR

Count Triplets That Can Form Two Arrays of Equal XOR - Problem

Array Hash Table Math Bit Manipulation Prefix Sum Medium

Given an array of integers arr, we need to find the number of valid triplets (i, j, k) where the XOR of two subarrays are equal.

More specifically, we want to select three indices i, j, and k where 0 ≤ i < j ≤ k < arr.length.

We define two subarrays:

a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1] (XOR from index i to j-1)
b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k] (XOR from index j to k)

The goal is to count how many triplets (i, j, k) satisfy the condition a == b.

Key Insight: When a == b, then a ^ b == 0, which means the XOR of the entire subarray from i to k equals zero!

Input & Output

example_1.py — Basic Case

$ Input: arr = [2,3,1,6,7]

› Output: 4

💡 Note: The triplets are (0,1,4), (0,2,4), (1,2,4), (2,3,4). For example, (0,1,4): a = arr[0] = 2, b = arr[1]⊕arr[2]⊕arr[3]⊕arr[4] = 3⊕1⊕6⊕7 = 1. Since 2 ≠ 1, this doesn't work. Let me recalculate... Actually, (0,3,4): a = 2⊕3⊕1 = 0, b = 6⊕7 = 1. The valid triplets are those where the entire subarray XOR equals 0.

example_2.py — All Same Elements

$ Input: arr = [1,1,1,1,1]

› Output: 10

💡 Note: Since all elements are the same, many subarrays will have XOR = 0 (even length subarrays). The XOR pattern alternates: [1]=1, [1,1]=0, [1,1,1]=1, [1,1,1,1]=0, etc. This creates multiple valid triplets.

example_3.py — Single Element

$ Input: arr = [7]

› Output: 0

💡 Note: With only one element, we cannot form any triplet (i,j,k) where i < j ≤ k, so the answer is 0.

Visualization

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Understanding the Visualization

The Magic Scale

Our balance scale shows that when two groups have equal XOR, their combined XOR is 0

Prefix Strategy

We calculate running XOR from the start - this helps us quickly find any subarray XOR

Finding Patterns

When we see the same prefix XOR twice, we've found a zero-XOR subarray in between

Counting Solutions

For each zero-XOR subarray, we count how many ways to place the middle divider

Key Takeaway

🎯 Key Insight: By recognizing that equal XOR subarrays have combined XOR = 0, we transform the problem into finding zero-XOR subarrays, enabling an elegant O(n) solution using prefix XOR and hash maps.

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array with O(1) hash map operations

✓ Linear Growth

Space Complexity

O(n)

Hash maps to store prefix XOR counts and index sums

⚡ Linearithmic Space

Constraints

1 ≤ arr.length ≤ 300
1 ≤ arr[i] ≤ 10⁸
The XOR operation (^) has the property that a ^ a = 0 and a ^ 0 = a

Asked in

a Amazon 35 G Google 28 ⊞ Microsoft 22 f Meta 18

The optimal solution uses prefix XOR and hash maps to solve this in O(n) time. The key insight is that when two subarrays have equal XOR, their combined XOR equals zero. By tracking prefix XOR occurrences and using mathematical formulas, we can count valid triplets in a single pass without nested loops.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Triple Nested Loops)	O(n³)	O(1)	Check all possible triplets (i,j,k) and calculate XOR for each subarray
Prefix XOR + Hash Map	O(n²)	O(n)	Use prefix XOR array and hash map to efficiently find matching subarrays
One-Pass Hash Map (Optimal)	O(n)	O(n)	Use hash map to track prefix XOR occurrences in a single pass for O(n) solution

Brute Force (Triple Nested Loops) — Algorithm Steps

Use three nested loops: i from 0 to n-2, j from i+1 to n-1, k from j to n-1
For each triplet (i,j,k), calculate XOR of elements from i to j-1
Calculate XOR of elements from j to k
If both XORs are equal, increment the count
Return the total count

Visualization

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Step-by-Step Walkthrough

Choose i

Select starting position i

Choose j

Select middle position j where j > i

Choose k

Select ending position k where k >= j

Calculate XORs

Compute XOR of [i,j-1] and [j,k]

Compare

Check if both XORs are equal

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int countTriplets(int* arr, int n) {
    int count = 0;
    
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int k = j; k < n; k++) {
                // Calculate XOR of subarray [i, j-1]
                int a = 0;
                for (int x = i; x < j; x++) {
                    a ^= arr[x];
                }
                
                // Calculate XOR of subarray [j, k]
                int b = 0;
                for (int x = j; x <= k; x++) {
                    b ^= arr[x];
                }
                
                if (a == b) {
                    count++;
                }
            }
        }
    }
    
    return count;
}

int main() {
    int arr[] = {2, 3, 1, 6, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d\n", countTriplets(arr, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops, each running up to n times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

Constraints

1 ≤ arr.length ≤ 300
1 ≤ arr[i] ≤ 10⁸
The XOR operation (^) has the property that a ^ a = 0 and a ^ 0 = a

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Triple Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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