LeetCode #1442 — MEDIUM

Count Triplets That Can Form Two Arrays of Equal XOR

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

Given an array of integers arr.

We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).

Let's define a and b as follows:

a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]

Note that ^ denotes the bitwise-xor operation.

Return the number of triplets (i, j and k) Where a == b.

Example 1:

Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)

Example 2:

Input: arr = [1,1,1,1,1]
Output: 10

Constraints:

1 <= arr.length <= 300
1 <= arr[i] <= 10⁸

Step 01

Brute Force Baseline

Problem summary: Given an array of integers arr. We want to select three indices i, j and k where (0 <= i < j <= k < arr.length). Let's define a and b as follows: a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1] b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k] Note that ^ denotes the bitwise-xor operation. Return the number of triplets (i, j and k) Where a == b.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math · Bit Manipulation

Example 1

[2,3,1,6,7]

Example 2

[1,1,1,1,1]

Core Insight

What unlocks the optimal approach

We are searching for sub-array of length ≥ 2 and we need to split it to 2 non-empty arrays so that the xor of the first array is equal to the xor of the second array. This is equivalent to searching for sub-array with xor = 0.
Keep the prefix xor of arr in another array, check the xor of all sub-arrays in O(n^2), if the xor of sub-array of length x is 0 add x-1 to the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1442: Count Triplets That Can Form Two Arrays of Equal XOR
class Solution {
    public int countTriplets(int[] arr) {
        int ans = 0, n = arr.length;
        for (int i = 0; i < n; ++i) {
            int s = arr[i];
            for (int k = i + 1; k < n; ++k) {
                s ^= arr[k];
                if (s == 0) {
                    ans += k - i;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1442: Count Triplets That Can Form Two Arrays of Equal XOR
func countTriplets(arr []int) (ans int) {
	for i, x := range arr {
		s := x
		for k := i + 1; k < len(arr); k++ {
			s ^= arr[k]
			if s == 0 {
				ans += k - i
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1442: Count Triplets That Can Form Two Arrays of Equal XOR
class Solution:
    def countTriplets(self, arr: List[int]) -> int:
        ans, n = 0, len(arr)
        for i, x in enumerate(arr):
            s = x
            for k in range(i + 1, n):
                s ^= arr[k]
                if s == 0:
                    ans += k - i
        return ans

// Accepted solution for LeetCode #1442: Count Triplets That Can Form Two Arrays of Equal XOR
impl Solution {
    pub fn count_triplets(arr: Vec<i32>) -> i32 {
        let mut ans = 0;
        let n = arr.len();

        for i in 0..n {
            let mut s = arr[i];
            for k in (i + 1)..n {
                s ^= arr[k];
                if s == 0 {
                    ans += (k - i) as i32;
                }
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #1442: Count Triplets That Can Form Two Arrays of Equal XOR
function countTriplets(arr: number[]): number {
    const n = arr.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        let s = arr[i];
        for (let k = i + 1; k < n; ++k) {
            s ^= arr[k];
            if (s === 0) {
                ans += k - i;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.