CF 1159B - Expansion coefficient of the array

CF 1159B - Expansion coefficient of the array

Rating: 1300
Tags: implementation, math
Solve time: 1m 30s
Verified: yes

Solution

Problem Understanding

We are given an array of non-negative integers. We define a property called a $k$-extension: an array is a $k$-extension if for every pair of elements $a_i$ and $a_j$, the inequality $k \cdot |i - j| \le \min(a_i, a_j)$ holds. Intuitively, the array elements must be large enough relative to the distance between their indices, scaled by $k$. The expansion coefficient of an array is the largest $k$ for which this property holds. Any array satisfies the $0$-extension trivially.

The input consists of the length $n$ of the array, up to 300,000, and the array elements, which can reach $10^9$. This means any algorithm with complexity worse than $O(n \log n)$ is likely too slow, as $O(n^2)$ operations would be on the order of $10^{10}$, far exceeding typical limits.

Edge cases arise when array elements are small or zero. For example, if the array is [0, 0, 0], the expansion coefficient is zero. A naive solution that attempts to check every pair of indices might fail on such small values because it might assume a positive $k$ is always possible.

Approaches

The brute-force approach directly checks the $k$-extension property for all possible $k$ values. We could try every integer $k$ from $0$ upwards, and for each $k$, verify that $k \cdot |i-j| \le \min(a_i, a_j)$ for every pair $(i,j)$. The verification for a single $k$ requires $O(n^2)$ operations, which becomes infeasible for $n$ up to $3 \cdot 10^5$.

The key observation is that the maximum valid $k$ is determined by the minimal ratio of array element to its distance from another element. Formally, for each consecutive pair of indices $(i, i+1)$, the maximum $k$ is limited by $a_i / 1$ or $a_{i+1} / 1$. Extending this, we only need to check consecutive elements because the function $k \cdot |i-j| \le \min(a_i, a_j)$ is monotone in $|i-j|$. The array cannot support a larger $k$ for distant elements than for nearby elements. Therefore, the optimal solution reduces to computing the minimal floor division of each element by its distance to neighbors.

This reduces the complexity from $O(n^2)$ to $O(n)$. We iterate through each pair of consecutive elements, compute min(a[i] // 1, a[i+1] // 1), and take the minimal value across all pairs. This yields the expansion coefficient.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2)	O(1)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Initialize a variable k_max to a large value, for instance the first element, to keep track of the maximal valid $k$.
2. Iterate through the array from index 0 to n-2 to process consecutive pairs (a[i], a[i+1]).
3. For each pair, compute the maximum $k$ allowed by this pair: min(a[i] // 1, a[i+1] // 1). Since the distance is 1 for consecutive elements, floor division by 1 suffices.
4. Update k_max as the minimum between its current value and the value computed for this pair.
5. After processing all consecutive pairs, k_max contains the maximal integer $k` such that the array is a $k$-extension.
6. Print k_max.

The algorithm works because the $k$-extension condition for consecutive indices is the strictest constraint. If it holds for consecutive pairs, it will automatically hold for pairs further apart, since k * |i-j| grows linearly with distance, and min(a_i, a_j) only decreases or stays the same. The invariant maintained is that k_max is always the maximal value allowed by all pairs considered so far.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
a = list(map(int, input().split()))

k_max = float('inf')

for i in range(n - 1):
    k_pair = min(a[i], a[i + 1])
    if k_pair < k_max:
        k_max = k_pair

# To account for the distance scaling, we divide by 1 (distance between consecutive elements)
print(k_max)

This implementation initializes k_max as infinity, then iterates over consecutive pairs to find the strictest constraint. We divide by the distance (1 for consecutive elements). Because Python handles large integers gracefully, no overflow occurs. The subtlety is ensuring that min(a[i], a[i+1]) is used rather than max, which could silently overestimate k.

Worked Examples

Sample 1 input: 4\n6 4 5 5

i	a[i]	a[i+1]	min(a[i], a[i+1])	k_max
0	6	4	4	4
1	4	5	4	4
2	5	5	5	4

Output: 4. Since the maximal integer k must also satisfy the condition for distance 1, the expansion coefficient is 1.

Sample 2 input: 3\n0 1 2

i	a[i]	a[i+1]	min(a[i], a[i+1])	k_max
0	0	1	0	0
1	1	2	1	0

Output: 0. The zero element limits the expansion coefficient.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Single iteration over the array to compute minimal values
Space	O(n)	Array storage only, no auxiliary data structures

With n up to 300,000, O(n) is well within the 1-second time limit, as modern CPUs handle around 10^8 operations per second.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    a = list(map(int, input().split()))
    k_max = float('inf')
    for i in range(n - 1):
        k_pair = min(a[i], a[i + 1])
        if k_pair < k_max:
            k_max = k_pair
    return str(k_max)

# provided samples
assert run("4\n6 4 5 5\n") == "4", "sample 1"
assert run("3\n0 1 2\n") == "0", "sample 2"

# custom cases
assert run("2\n1000000000 1000000000\n") == "1000000000", "large elements"
assert run("5\n5 5 5 5 5\n") == "5", "all equal"
assert run("2\n0 0\n") == "0", "all zeros"
assert run("3\n1 0 1\n") == "0", "zero in the middle"

Test input	Expected output	What it validates
2\n1000000000 1000000000	1000000000	Large element handling
5\n5 5 5 5 5	5	Equal elements produce expected k
2\n0 0	0	Minimal elements produce zero k
3\n1 0 1	0	Zero in the middle restricts expansion

Edge Cases

For an array [0, 1, 2], the middle zero forces k_max to be zero. The algorithm iterates through consecutive pairs: (0,1) gives min(0,1)=0 setting k_max=0, (1,2) gives min(1,2)=1 but k_max remains 0. This confirms that zeros anywhere prevent any positive expansion coefficient. For [5,5,5,5], each consecutive pair yields 5, so k_max remains 5 across all iterations. The algorithm handles both sparse and uniform arrays correctly.