How should I practice Count Valid Bitwise Pairs?

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Problem · Array

Count Valid Bitwise Pairs

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Problem statement

Stripe's analytics team is studying relationships between values stored in an array.

Given an array A of N non-negative integers and a non-negative integer K, determine the number of index pairs (i, j) such that 1 <= i < j <= N and:

3(A_i | A_j) - (A_i & A_j) - A_i - A_j + (A_i XOR A_j)^2 - ((A_i | A_j) - (A_i & A_j))^2 + (A_i XOR A_j) + (A_i & A_j) + (A_i | A_j) = K + A_j + 2(A_i XOR A_j)

Here |, &, and XOR denote bitwise OR, AND, and XOR. Return the total number of valid pairs.

Equivalent condition

Using the standard bitwise identities A_i | A_j = (A_i & A_j) + (A_i XOR A_j) and A_i + A_j = (A_i | A_j) + (A_i & A_j), the source equation simplifies to:

A_i + (A_i XOR A_j) = K

The order remains important: A_i is the earlier value because the source requires i < j.

Original input and output format

The source uses T test cases. Each case contains N, K, and the array A, and outputs one count. The FastPrep function below represents one source test case.

Function

countValidBitwisePairs(A: long[], K: long) → long

Examples

Example 1

A = [1,2,3]K = 4return = 1

The pair at 1-based indices (1, 2) is valid because 1 + (1 XOR 2) = 4. The other two pairs produce 3.

Example 2

A = [0,15]K = 15return = 1

The only pair is valid: 0 + (0 XOR 15) = 15.

Example 3

A = [15,0]K = 15return = 0

The condition is asymmetric. Here 15 + (15 XOR 0) = 30, so reversing the values from Example 2 changes the answer to zero.

Constraints

For the FastPrep one-case interface, 1 <= A.length <= 10^5.
0 <= K <= 10^18.
0 <= A[i] <= 10^18.
In the source batch format, 1 <= T <= 200 and the sum of N over all test cases does not exceed 3 * 10^5.