[Codeforces] Round 2239 D1 — XOR Sorting (Easy)

Codeforces Round 2239 — Problem D1: XOR Sorting (Easy).

This was one of those problems where the algorithm is hidden behind an unusual graph defined on indices rather than on the values themselves.

At first glance, the condition

We may swap indices i and j only if i XOR j <= k

looks difficult to reason about. The trick is to completely ignore the array for a while and study the graph induced by this condition.

Step 1: Think of indices as a graph

For a fixed k, create an undirected graph:

• Vertices = indices 0 ... n-1
• Edge between i and j iff i XOR j <= k

Since we may perform any number of swaps, what matters is not the individual edges but the connected components.

Any value can move anywhere inside its connected component, but it can never leave it.

Therefore:

An array is sortable iff every element’s current index and its final sorted index belong to the same connected component.

Step 2: Understanding the connected components

Let’s examine small values of k.

k = 1

Components:

{0,1}
{2,3}
{4,5}
...

k = 2

New edges merge adjacent pairs:

{0,1,2,3}
{4,5,6,7}
...

k = 3

Nothing new happens.

Components remain

{0,1,2,3}
{4,5,6,7}
...

k = 4

The previous groups merge again:

{0,1,2,3,4,5,6,7}
{8,9,10,11,12,13,14,15}
...

A clear pattern emerges.

If the highest set bit of k is b, then the connected components are contiguous blocks of size

2^(b+1)

This means two indices belong to the same component iff they lie inside the same block.

Step 3: Where should every element go?

Let

a = current array
b = sorted(a)

For every element, determine the index where it should finally appear.

For duplicate values, we must be careful to match occurrences consistently.

A convenient implementation is:

pos = defaultdict(list)

for i in range(n - 1, -1, -1):
    pos[b[i]].append(i)

Then

j = pos[a[i]].pop()

returns the correct destination index for each occurrence.

Step 4: What constraint does one element impose?

Suppose an element must move from index

i -> j

These two indices must lie inside the same connected component.

Instead of thinking about blocks directly, consider

i XOR j

The important observation is:

Only the highest differing bit matters.

If

i XOR j = 13 = 1101₂

then the highest differing bit is

8 = 1000₂

Therefore this move requires the answer to be at least 8.

Notice that we do not require

i XOR j <= k

directly.

The element may travel through intermediate indices.

Step 5: Combining all constraints

Every element contributes one required move.

We compute

answer_mask |= (current_index ^ destination_index)

At the end,

highest_power_of_two(answer_mask)

is exactly the minimum valid answer.

In Python:

if answer_mask == 0:
    print(0)
else:
    print(1 << (answer_mask.bit_length() - 1))

Complete Python Solution

from collections import defaultdict

t = int(input())

for _ in range(t):
    n, q = map(int, input().split())
    a = list(map(int, input().split()))

    b = sorted(a)

    if a == b:
        print(0)
        continue

    pos = defaultdict(list)

    for i in range(n - 1, -1, -1):
        pos[b[i]].append(i)

    mask = 0

    for i in range(n):
        j = pos[a[i]].pop()
        mask |= i ^ j

    if mask == 0:
        print(0)
    else:
        print(1 << (mask.bit_length() - 1))

Complexity

Sorting:

O(n log n)

Matching indices:

O(n)

Overall:

O(n log n)

with linear extra memory.

TODO: Hard Version (D2)

The hard version introduces up to 10⁶ point updates.

Recomputing the sorted order and all destination indices after every update is far too slow.

The key observation from the easy version still holds:

• every element contributes a destination index,
• every move contributes its highest differing bit,
• the answer is determined by the maximum active bit.

The remaining challenge is to maintain these constraints dynamically after each update.

TODO: Derive an efficient data structure that supports updates while maintaining the answer in sublinear time.