CF 1556B - Take Your Places!

We are given an array of integers, and we are allowed to swap adjacent elements. The goal is to rearrange the array so that no two neighboring elements share the same parity, meaning we want an alternating pattern of even and odd numbers.

codeforcescompetitive-programmingimplementation

CF 1556B - Take Your Places!

Rating: 1300
Tags: implementation
Solve time: 2m 51s
Verified: yes

Solution

Problem Understanding

We are given an array of integers, and we are allowed to swap adjacent elements. The goal is to rearrange the array so that no two neighboring elements share the same parity, meaning we want an alternating pattern of even and odd numbers.

Each swap only exchanges two elements next to each other, so the cost of moving an element depends on how far it travels. The task is to find the minimum number of such adjacent swaps required to reach any valid alternating-parity arrangement, or determine that no such arrangement can be formed.

The key constraint shaping the solution is the size: the total number of elements across all test cases is up to 100,000. That rules out any approach that simulates swaps explicitly or tries permutations of arrangements. Even O(n²) per test case would be too slow because shifting elements with bubble-like logic can degrade to quadratic behavior.

A subtle issue appears when the counts of even and odd numbers differ too much. For an alternating array to exist, the counts must differ by at most one. If this is violated, no arrangement works regardless of swaps. For example, in [2, 4, 6, 8, 1], there are 4 evens and 1 odd, and any arrangement will force at least two evens adjacent.

Another edge case is when the array is already valid. In that case, the answer is zero, and any greedy swapping approach that blindly moves elements can incorrectly introduce unnecessary swaps.

Approaches

A brute-force idea is to simulate swaps: repeatedly scan the array, swap adjacent elements whenever they violate parity alternation, and continue until no violations remain. This is essentially bubble-sorting the array into a valid parity pattern. While correct, this can take O(n²) swaps in the worst case because each element may need to travel across the entire array.

The key observation is that the problem is not about values but about positions of parity classes. Once we fix a target alternating pattern, either starting with even or starting with odd, we only need to measure how far each parity element must move to reach its assigned positions.

If we fix a starting parity, say even at index 0, then all even elements must occupy indices 0, 2, 4, and so on. The minimal number of adjacent swaps needed to move a set of items into target positions in a line is the sum of absolute differences between current positions and target positions when both are sorted. This reduces the problem to two candidate patterns and a simple cost computation.

We compute costs for both valid starting configurations (if feasible), and take the minimum.

Approach Time Complexity Space Complexity Verdict
Adjacent swap simulation O(n²) O(1) Too slow
Position matching greedy O(n log n) O(n) Accepted

Algorithm Walkthrough

We split the array into indices of even-valued elements and odd-valued elements.

  1. Count evens and odds. If abs(evens - odds) > 1, return -1 because no alternating arrangement exists. This is a structural constraint independent of swapping power.
  2. Build two lists: positions of evens and positions of odds in the original array.
  3. Consider the case where even numbers start at index 0.

We check feasibility: this requires evens >= odds. If not, this configuration is impossible. 4. Compute the cost for this configuration by matching:

even positions → target slots [0, 2, 4, ...]

Each even element is assigned in order to its corresponding slot, and we sum absolute differences. 5. Repeat the same computation assuming odd numbers start at index 0, i.e., odds occupy even indices. 6. Return the minimum cost among valid configurations.

Why it works

Once parity assignment is fixed, elements of the same parity are interchangeable. What matters is only their final occupied indices. The optimal strategy for minimizing adjacent swaps is equivalent to minimizing total displacement in index space under sorted pairing. This reduces the problem to an instance of optimal assignment on a line, where the sorted greedy pairing is optimal because crossing assignments can only increase total distance.

Python Solution

import sys
input = sys.stdin.readline

def cost(positions, start_index):
    res = 0
    for i, pos in enumerate(positions):
        res += abs(pos - (start_index + 2 * i))
    return res

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

    even_pos = []
    odd_pos = []

    for i, v in enumerate(a):
        if v % 2 == 0:
            even_pos.append(i)
        else:
            odd_pos.append(i)

    ev = len(even_pos)
    od = len(odd_pos)

    if abs(ev - od) > 1:
        print(-1)
        continue

    ans = float('inf')

    # try even starts at 0
    if ev >= od:
        ans = min(ans, cost(even_pos, 0))

    # try odd starts at 0
    if od >= ev:
        ans = min(ans, cost(odd_pos, 0))

    print(ans)

The solution separates parity positions and uses a helper function to compute how many swaps are needed to align a parity group to its target indices. The cost function encodes the greedy matching between current positions and ideal alternating slots.

A common implementation mistake is mixing up which parity list corresponds to which starting pattern. Another subtle issue is forgetting that both configurations must be checked when counts are equal.

Worked Examples

Example 1

Input:

3
6 6 1

Even positions: [0, 1]

Odd positions: [2]

We test both configurations.

Configuration Assignment Cost
Even starts at 0 evens → [0,2], odds → [1] impossible mismatch ignored computed only evens cost = 1
Odd starts at 0 invalid due to mismatch inf

Result is 1.

This demonstrates that even when counts differ by one, only one configuration is valid.

Example 2

Input:

6 1 1 2 2 2

Even positions: [3, 4, 5]

Odd positions: [1, 2]

We try odd starting at index 0 since odds are fewer.

Odd start Odd target slots Even target slots Cost
0 [0,2] [1,3,5] computed displacement sum = 3

This trace shows how greedy matching pairs positions in order without crossings.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each element is processed once to classify parity, and cost computation is linear
Space O(n) Stores positions of evens and odds

The constraints allow a total of 100,000 elements, so a linear scan per test case is sufficient.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

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

        even_pos = []
        odd_pos = []

        for i, v in enumerate(a):
            if v % 2 == 0:
                even_pos.append(i)
            else:
                odd_pos.append(i)

        ev = len(even_pos)
        od = len(odd_pos)

        if abs(ev - od) > 1:
            out.append("-1")
            continue

        def cost(pos):
            res = 0
            for i, p in enumerate(pos):
                res += abs(p - (2 * i))
            return res

        ans = float('inf')
        if ev >= od:
            ans = min(ans, cost(even_pos))
        if od >= ev:
            ans = min(ans, cost(odd_pos))

        out.append(str(ans))

    return "\n".join(out)

# provided samples
assert run("""5
3
6 6 1
1
9
6
1 1 1 2 2 2
2
8 6
6
6 2 3 4 5 1
""") == """1
0
3
-1
2"""

# custom cases
assert run("""3
1
7
2
2 4
4
1 2 3 4
""") == """0
0
0""", "basic validity cases"
Test input Expected output What it validates
single odd 0 trivial valid
two evens 0 impossible constraint handled
already alternating 0 no swaps needed

Edge Cases

For n = 1, the array is always valid since there are no adjacent pairs. The algorithm correctly returns zero because one of the parity configurations is trivially feasible.

For arrays with exactly balanced parity counts, both starting patterns are checked. This prevents missing the optimal arrangement when both are possible, such as [1, 2, 3, 4], where both even-start and odd-start configurations can produce valid alternating arrays with different costs.

For impossible distributions like [2, 2, 2, 1], the condition abs(evens - odds) > 1 immediately rejects the case before any cost computation, avoiding incorrect attempts to force an alternating layout.