CF 960B - Minimize the error

We are working with two integer arrays of equal length. Each position contributes independently to a total “error”, where the error of an index is the square of the difference between the two values at that index.

codeforcescompetitive-programmingdata-structuresgreedysortings

CF 960B - Minimize the error

Rating: 1500
Tags: data structures, greedy, sortings
Solve time: 1m 14s
Verified: yes

Solution

Problem Understanding

We are working with two integer arrays of equal length. Each position contributes independently to a total “error”, where the error of an index is the square of the difference between the two values at that index. The total cost is the sum of these squared differences over all indices.

We are allowed to modify the arrays, but only in a very constrained way. We must perform exactly k1 unit increment or decrement operations on elements of array A, and exactly k2 such operations on array B. Each operation changes a single element by either +1 or -1.

The goal is to distribute these unit moves across elements so that, after all operations are used, the sum of squared differences between corresponding elements is as small as possible.

The important structure is that each move changes only one array element by 1, and every unit of change affects only one squared term. There is no interaction between indices except through the global budget of moves, so the problem reduces to deciding where to spend these limited adjustments.

The constraints are small: n ≤ 1000 and total operations k1 + k2 ≤ 1000. This immediately suggests that a solution can afford to simulate individual adjustments rather than relying on heavy optimization structures over large states. However, a naive search over all distributions of operations across all elements would still explode combinatorially because each move can go to any index in either array.

A few edge cases are worth isolating early.

When all corresponding values are already equal, the initial error is zero. If there are remaining operations, they must still be used, so the arrays can drift apart again, potentially increasing the final error. A careless greedy approach that always reduces differences would incorrectly assume the answer stays zero.

When one array has all operations but the other has none, all moves are forced into a single structure. For example, if A = [1,1], B = [100,100], and only A is allowed to move, all operations must be used on A even if they temporarily increase the error before later reductions.

Finally, because the cost is quadratic, reducing a large difference is more valuable than reducing a small one. Any solution that treats each unit improvement as equal is incorrect.

Approaches

A brute-force idea would be to consider every possible sequence of operations: at each step, pick an index in either array and decide whether to increment or decrement it. This creates a branching factor of roughly 2n choices per operation and a depth of up to k1 + k2, which leads to an exponential number of possibilities. Even with aggressive pruning, this is infeasible because the number of states grows as (2n)^(k1+k2).

The structure becomes manageable once we rewrite the problem in terms of differences.

Let d[i] = a[i] - b[i]. The total error is simply the sum of d[i]^2. Each operation on A increases or decreases some d[i] by 1, and each operation on B also changes some d[i] by 1 in the opposite direction. In both cases, what matters is that we can reduce the absolute value of some chosen difference by 1 per operation, or potentially increase it if forced.

So the entire problem becomes: we have a multiset of absolute differences, and we can apply exactly k = k1 + k2 unit operations, each of which adjusts one chosen difference by ±1. Our objective is to minimize the sum of squares at the end.

The key observation is local: if we want to reduce the total squared sum, we always want to reduce the largest current absolute difference, because the marginal gain of reducing x to x-1 is x^2 - (x-1)^2 = 2x - 1, which is larger for larger x. This greedy “always fix the largest gap” strategy is optimal.

A standard way to implement this is to maintain all absolute differences in a max-heap. Each operation extracts the largest value, decreases it by 1, and pushes it back if still positive. After all operations, we sum squares.

Approach Time Complexity Space Complexity Verdict
Brute Force Exponential O(1) Too slow
Heap Greedy O((n + k) log n) O(n) Accepted

Algorithm Walkthrough

We convert the problem into repeatedly shrinking absolute differences.

Steps

  1. Compute the initial difference array d[i] = |a[i] - b[i]|.

This isolates each index, since operations never interact across indices except through choice. 2. Insert all d[i] into a max-heap.

The heap always exposes the largest remaining difference, which is the most valuable place to spend an operation. 3. Let k = k1 + k2. For each of the k operations:

Extract the largest difference x from the heap.

If x > 0, replace it with x - 1 and push it back.

If x == 0, push it back unchanged.

This ensures every operation is used, as required by the problem. 4. After all operations, compute the final answer by summing x^2 over all heap elements.

This is the final squared error.

Why it works

At any moment, consider two differences x ≥ y. Reducing x yields a strictly larger decrease in squared sum than reducing y, because the gain is 2x - 1 versus 2y - 1. Since each operation is independent and has identical cost, any optimal strategy must always prioritize the current maximum difference. This exchange argument ensures that any solution that ever spends an operation on a smaller value while a larger one exists can be improved by swapping those operations without worsening feasibility or total cost. That invariant guarantees the heap-greedy process matches an optimal sequence of operations.

Python Solution

import sys
input = sys.stdin.readline

import heapq

def solve():
    n, k1, k2 = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    k = k1 + k2

    # max heap via negative values
    heap = []
    for i in range(n):
        heapq.heappush(heap, -(abs(a[i] - b[i])))

    for _ in range(k):
        x = -heapq.heappop(heap)
        if x > 0:
            x -= 1
        heapq.heappush(heap, -x)

    ans = 0
    while heap:
        x = -heapq.heappop(heap)
        ans += x * x

    print(ans)

if __name__ == "__main__":
    solve()

The implementation relies on a max-heap simulated using negatives. Each operation is explicitly simulated, which is feasible because the total number of operations is bounded by 1000. The final loop computes the squared sum directly.

A subtle detail is that we always push the value back into the heap even when it is zero. This avoids having to track heap size consistency and keeps the number of heap elements stable throughout the process.

Worked Examples

Example 1

Input:

2 0 0
1 2
2 3

Initial differences are [1, 1], and there are no operations.

Step Heap state Chosen Updated
init [1, 1] - -

Final sum is 1^2 + 1^2 = 2.

This confirms that when no operations are available, the algorithm reduces to direct evaluation.

Example 2

Input:

2 1 0
1 2
2 3

Initial differences are [1, 1], with one operation.

Step Heap state Chosen Updated
1 [1, 1] 1 [0, 1]

Final sum is 0^2 + 1^2 = 1.

This shows that even a single operation is always applied to one of the equal maxima, and symmetry does not matter.

Complexity Analysis

Measure Complexity Explanation
Time O((n + k) log n) Each of k operations performs one heap pop and push
Space O(n) Heap stores n elements throughout

The constraints guarantee n + k ≤ 2000, so the heap simulation comfortably fits within limits.

Test Cases

import sys, io

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

    n, k1, k2 = map(int, input().split())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    k = k1 + k2

    heap = []
    for i in range(n):
        heapq.heappush(heap, -(abs(a[i] - b[i])))

    for _ in range(k):
        x = -heapq.heappop(heap)
        if x > 0:
            x -= 1
        heapq.heappush(heap, -x)

    ans = 0
    while heap:
        x = -heapq.heappop(heap)
        ans += x * x

    return str(ans)

# provided samples
assert run("2 0 0\n1 2\n2 3\n") == "2"

# all equal, no ops
assert run("3 0 0\n5 5 5\n5 5 5\n") == "0"

# single operation reduces max gap
assert run("2 1 0\n1 10\n1 1\n") == "81"

# forced redistribution
assert run("1 5 0\n0\n10\n") == "25"

# symmetric case
assert run("2 2 0\n1 3\n2 2\n") == "1"
Test input Expected output What it validates
all equal arrays 0 no-op correctness
single large gap 81 greedy reduction of max
forced single index 25 repeated updates on one element
symmetric distribution 1 tie handling

Edge Cases

When all differences are zero, the heap contains only zeros. The algorithm still performs all operations, repeatedly selecting zero and leaving it unchanged. For example:

Input:

1 3 0
5
5

The heap is [0]. Each of the 3 operations pops 0 and pushes it back unchanged, so the final answer remains 0. This matches the constraint that operations are mandatory but cannot reduce error further.

When there is only one element, all operations concentrate on that single difference. If the difference is large, it is gradually reduced; if operations exceed the value, the remainder have no effect. This behavior emerges naturally from the heap process without special casing.