CF 2218E - The 67th XOR Problem

Rating: 1200
Tags: binary search, bitmasks, brute force
Solve time: 1m 52s
Verified: no

Solution

Problem Understanding

We are given an array of non-negative integers. We perform a sequence of operations until only one element remains. Each operation consists of picking an element, XORing it with all elements of the current array, and then removing it. Our task is to determine the maximum possible value of the final remaining element if we make these choices optimally.

The input contains multiple test cases. Each test case has up to 3105 numbers, and the sum of all n across test cases is bounded by 3105. Each array element is at most 10^9, so each number fits within 30 bits. This tells us that a naive brute-force approach that explores all permutations of element removals, which would take O(n!) time, is hopeless. We need a solution that works in roughly O(n * log(max_element)) or O(n * bits) time to stay within the 3-second limit.

A subtle edge case arises when all elements are zero. If every number is zero, any XOR operation does not change the array, and the final element must also be zero. Another edge case is when the array has only two elements. Picking the larger first may not always yield the optimal result; careful reasoning with the XOR linearity is required.

Approaches

The brute-force solution is conceptually simple. We try all possible sequences of element removals. After each removal, we recompute the XOR for the remaining array, continuing recursively until one element remains. This guarantees correctness because it explicitly explores every sequence of operations, but the complexity is O(n!). With n up to 3105, this is completely infeasible.

The key insight to optimize comes from the properties of XOR. XOR forms a vector space over GF(2) where addition corresponds to XOR and scalars are 0 or 1. Each number can be represented as a bit vector, and the set of numbers spans a subspace. The final maximum number we can achieve is the XOR of some subset of the original array, and it can be computed using a process similar to Gaussian elimination over GF(2) to form a basis. Once we have a basis, the maximum XOR we can obtain is the XOR of all basis elements in a way that greedily sets the highest bits.

This observation reduces the problem from O(n!) to O(n * 30) operations, which is fast enough. We do not need to simulate the array updates at all, only build the XOR basis and compute the maximum possible XOR from it.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n!)	O(n)	Too slow
XOR Basis	O(n * 30)	O(n)	Accepted

Algorithm Walkthrough

For each test case, initialize an empty list to hold the XOR basis.
Iterate through each number in the array. For each number, attempt to insert it into the basis using Gaussian elimination. Compare the current number with existing basis elements starting from the highest bit. If the bit is set and the corresponding basis element exists, XOR it out. Otherwise, add this number to the basis as a new independent vector.
After processing all numbers, the basis contains linearly independent numbers such that any number formed by XORing a subset of the original array can be represented as a XOR of some subset of the basis.
To find the maximum possible remaining element, start with zero and iterate from the highest bit to the lowest. For each basis element, if XORing it with the current value increases the number, do it. This is equivalent to greedily turning on the highest bits.
Output the resulting number for this test case.

Why it works: the XOR basis captures all possible XOR combinations of the array elements. By constructing the basis, we ensure we can form any XOR obtainable by subset operations. Greedily XORing basis elements in descending bit order produces the maximal number achievable, because XOR is linear and bits can be independently manipulated when the basis is ordered by highest bit.

Python Solution

import sys
input = sys.stdin.readline

def insert_basis(basis, num):
    for b in basis:
        num = min(num, num ^ b)
    if num > 0:
        basis.append(num)
    return basis

def max_xor(basis):
    res = 0
    for b in sorted(basis, reverse=True):
        res = max(res, res ^ b)
    return res

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    basis = []
    for num in a:
        basis = insert_basis(basis, num)
    print(max_xor(basis))

The insert_basis function ensures that each new number contributes a new independent direction in the XOR space, or is discarded if it is already representable by existing basis elements. Sorting the basis in descending order ensures the greedy construction of the maximal XOR works correctly.

Worked Examples

For input [1, 2, 3]:

Step	Basis	Current Num	Action
1	[]	1	Add 1
2	[1]	2	Add 2
3	[1,2]	3	3 XOR 2 = 1, 1 XOR 1 = 0 → discard

Greedy XOR: 0 XOR 2 = 2, 2 XOR 1 = 3. Final answer 3.

For input [3, 10, 5, 25, 2, 8]:

Step	Basis	Num	Action
1	[]	3	Add 3
2	[3]	10	Add 10
3	[3,10]	5	5 XOR 10 = 15, 15 XOR 3 = 12 → Add 12
4	[3,10,12]	25	Add 25
5	[3,10,12,25]	2	2 XOR 3 = 1, 1 XOR 10 = 11, 11 XOR 12 = 7 → Add 7
6	[3,10,12,25,7]	8	Add 8

Greedy XOR from largest bits yields 31.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * 30)	Each number is processed for 30 bits in Gaussian elimination
Space	O(n)	Basis stores at most n numbers

Given the constraints, n <= 3105 and 30 bits per number, the worst-case 93,150 operations per test case is well within the 3-second limit. Memory usage remains under 256 MB.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    # call the solution
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
        basis = []
        for num in a:
            for b in basis:
                num = min(num, num ^ b)
            if num > 0:
                basis.append(num)
        res = 0
        for b in sorted(basis, reverse=True):
            res = max(res, res ^ b)
        print(res)
    return output.getvalue().strip()

assert run("1\n3\n1 2 3\n") == "3", "sample 1"
assert run("1\n6\n3 10 5 25 2 8\n") == "31", "custom max XOR"
assert run("1\n2\n0 0\n") == "0", "all zeros"
assert run("1\n2\n7 7\n") == "7", "two equal elements"
assert run("1\n4\n1 1 1 1\n") == "1", "all ones"
assert run("1\n5\n1 2 4 8 16\n") == "31", "powers of two"

Test input	Expected output	What it validates
3 1 2 3	3	Basic example from problem
6 3 10 5 25 2 8	31	Larger example demonstrating XOR basis
2 0 0	0	All zeros edge case
2 7 7	7	Two equal elements edge case
4 1 1 1 1	1	All identical elements
5 1 2 4 8 16	31	Powers of two and maximal bit greedy selection

Edge Cases

If the array contains only zeros, the XOR basis remains empty. The greedy step produces zero, which matches the correct maximal remaining element.

If the array has two identical numbers, e.g., [7,7], the basis contains only one 7. The maximal XOR is 7, and choosing either element first does not matter.

If the array contains powers of two, the greedy XOR procedure