CF 1698A - XOR Mixup

CF 1698A - XOR Mixup

Rating: 800
Tags: bitmasks, brute force
Solve time: 2m 49s
Verified: no

Solution

Problem Understanding

We are given an array of length n that was generated by taking an original array of length n-1, computing the XOR of all its elements, appending that XOR to the array, and then shuffling the result. The task is to recover the XOR value that was appended, given only the shuffled array. Each test case gives one such array, and we must produce one integer per test case corresponding to the XOR that was added.

The constraints are modest: n can go up to 100 and there can be up to 1000 test cases. Each element in the array is a small integer between 0 and 127. This implies that any algorithm with complexity O(n) per test case will run efficiently, because even in the worst case we only need about 100,000 operations, which is trivial for modern CPUs. This also rules out the need for complex data structures or optimizations.

Edge cases include arrays where all elements are identical, where x could coincide with one of the original elements, or where x is zero. For instance, if the array is [6, 6, 6, 6, 6, 6], the correct answer is 6. Another tricky case is [100, 100, 0], where the XOR of the original two elements is 0. A naive approach that, for example, tries to remove duplicates or assumes x is the maximum element could fail on these inputs.

Approaches

The brute-force approach would be to try removing each element in turn, computing the XOR of the rest, and checking if it matches the removed element. For each of the n elements, we would compute an XOR over n-1 elements, leading to O(n²) operations per test case. This is technically feasible for the given constraints, but it is unnecessary because the XOR operation has a property that makes a simpler solution possible.

The key insight is that XOR is both associative and commutative. This means that XORing all elements of the final array gives us the same result as XORing the original array twice and the appended value once. If we let the XOR of the original array be x, the final array contains all original elements plus x. XORing everything together gives x ⊕ a1 ⊕ a2 ⊕ ... ⊕ a(n-1). But a1 ⊕ a2 ⊕ ... ⊕ a(n-1) is just x, so the total XOR is x ⊕ x = 0 plus the appended x, leaving x. Therefore, the XOR of all elements of the final array directly gives the value of x.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Works but unnecessary
XOR All Elements	O(n)	O(1)	Accepted and simple

Algorithm Walkthrough

1. Read the number of test cases.
2. For each test case, read n and the array of n integers.
3. Initialize a variable x to 0. This will hold the XOR of all elements.
4. Iterate through the array, XORing each element into x. After this loop, x contains the XOR of all n elements.
5. Output x for the current test case.

Why it works: XOR has the property that a ⊕ a = 0 and a ⊕ 0 = a. In the final array, every original element appears once and x appears once. XORing all of them together cancels out all original elements because a1 ⊕ a2 ⊕ ... ⊕ a(n-1) ⊕ x = x. This guarantees that we will always recover the correct appended XOR, regardless of the shuffle.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    x = 0
    for num in a:
        x ^= num
    print(x)

The solution reads the number of test cases and then loops over them. For each array, it computes the XOR of all elements using a simple for loop. The XOR accumulator starts at 0, ensuring that it correctly handles arrays where x might be 0. No special handling of the shuffle or repeated elements is necessary because XOR automatically cancels duplicates.

Worked Examples

Sample Input 1

4
4
4 3 2 5
5
6 1 10 7 10
6
6 6 6 6 6 6
3
100 100 0

Test Case	Array	XOR of All Elements	Output x
1	4 3 2 5	4 ⊕ 3 ⊕ 2 ⊕ 5 = 3	3
2	6 1 10 7 10	6 ⊕ 1 ⊕ 10 ⊕ 7 ⊕ 10 = 7	7
3	6 6 6 6 6 6	6 ⊕ 6 ⊕ 6 ⊕ 6 ⊕ 6 ⊕ 6 = 6	6
4	100 100 0	100 ⊕ 100 ⊕ 0 = 0	0

These traces show that XOR automatically cancels the original array elements, leaving the appended XOR value intact, even if there are repeated numbers or zeros.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(t * n)	Each test case requires one pass through the array to XOR elements
Space	O(n)	Only the array needs to be stored; the XOR accumulator uses constant space

Given n <= 100 and t <= 1000, the solution runs comfortably in under 1 second and uses minimal memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
        x = 0
        for num in a:
            x ^= num
        print(x)
    return output.getvalue().strip()

# provided samples
assert run("4\n4\n4 3 2 5\n5\n6 1 10 7 10\n6\n6 6 6 6 6 6\n3\n100 100 0\n") == "3\n7\n6\n0", "sample 1"

# custom cases
assert run("1\n2\n0 0\n") == "0", "minimum size, zero"
assert run("1\n2\n1 1\n") == "1", "minimum size, non-zero"
assert run("1\n3\n127 127 0\n") == "0", "max element boundary"
assert run("1\n5\n1 2 3 0 0\n") == "0", "zero among small numbers"

Test input	Expected output	What it validates
2 0 0	0	Minimum array size, zero XOR
2 1 1	1	Minimum array size, non-zero XOR
3 127 127 0	0	Maximum element, handling XOR with boundary
5 1 2 3 0 0	0	Zero mixed with non-zero numbers

Edge Cases

For an array [6, 6, 6, 6, 6, 6], the algorithm computes the XOR as 6 ⊕ 6 ⊕ 6 ⊕ 6 ⊕ 6 ⊕ 6 = 6. The repeated elements cancel in pairs, leaving the appended XOR intact. For [100, 100, 0], the XOR 100 ⊕ 100 ⊕ 0 = 0 correctly recovers the XOR appended to the original two-element array [100, 100]. These examples confirm that the solution handles duplicates and zeros correctly without additional logic.