CF 1084A - The Fair Nut and Elevator

Rating: 1000
Tags: brute force, implementation
Solve time: 3m
Verified: yes

Solution

Problem Understanding

The problem describes a single elevator in a building where each floor has a known number of residents. Each resident makes exactly two trips per day, one going down to the first floor and one returning back to their own floor. The elevator always serves one person at a time and after completing a request it returns to a fixed “idle floor” (x), which we are free to choose.

For a single ride involving a person living on floor (i), the elevator starts at floor (x), goes to (i), then goes to floor (1), and finally returns to (x). Every movement between adjacent floors costs one unit of energy, so the cost depends only on distances along the building.

The task is to pick the idle floor (x) so that the total energy used by all residents over the whole day is minimized.

The constraints are small: at most 100 floors and at most 100 people per floor. This immediately tells us that even an (O(n^2)) or (O(n^3)) solution would be fine, since the maximum number of operations is only about (10^4) to (10^6), well within limits.

A subtle edge case appears when all people are concentrated on one floor. In that situation, the cost function becomes highly sensitive to the choice of (x), and picking the wrong floor (for example, always choosing floor 1) gives a noticeably suboptimal result. Another edge case is when there are no people on some floors; those floors still matter as candidates for (x), even though they contribute no direct cost themselves.

Approaches

We first analyze what happens for a fixed choice of the idle floor (x). Consider a single person living on floor (i). Their trip structure is fixed: the elevator travels from (x) to (i), then from (i) to 1, and finally back from 1 to (x). The cost of this is

[ |x - i| + |i - 1| + |1 - x|. ]

Since (|1 - x|) does not depend on the person’s floor, it is a constant for a fixed (x), and every resident contributes a predictable amount depending only on their floor.

If we compute this cost for all people and sum it, we get the total energy required for that choice of (x). A brute-force solution simply tries every possible floor as (x), computes the full cost from scratch, and takes the minimum.

This works because there are only (n \le 100) candidate floors, and for each candidate we only need to scan all floors again. That gives (O(n^2)) total complexity, which is already more than fast enough.

The key simplification is noticing that we never need to simulate individual elevator movements. Everything decomposes into a deterministic per-floor cost multiplied by the number of people on that floor.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force over (x) with full recomputation	(O(n^2))	(O(1))	Accepted
Direct per-candidate aggregation	(O(n^2))	(O(1))	Accepted

Algorithm Walkthrough

We compute the cost for each possible idle floor and keep the minimum.

Iterate over each floor (x) from 1 to (n) as a candidate idle position. This is necessary because the optimal idle floor is not guaranteed to be special like 1 or the median.
For each candidate (x), initialize a running total cost to zero. This variable accumulates energy usage for all residents assuming this idle floor.
Iterate over every floor (i). If there are (a_i) people on floor (i), compute the cost of one person living there as (|x - i| + |i - 1| + |1 - x|).
Multiply this per-person cost by (a_i) and add it to the running total. This step aggregates identical behaviors efficiently instead of processing each person individually.
After processing all floors for a given (x), compare the computed total with the best answer so far and keep the minimum.
After checking all possible (x), output the smallest value obtained.

Why it works

Every resident behaves independently, and their cost depends only on their own floor and the chosen idle floor (x). Since the total cost is a sum of independent contributions, evaluating each (x) separately gives the exact global cost. No interaction exists between different floors or residents, so there is no hidden coupling that could invalidate the per-floor aggregation.

Python Solution

import sys
input = sys.stdin.readline

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

ans = float('inf')

for x in range(1, n + 1):
    total = 0
    for i in range(n):
        if a[i] == 0:
            continue
        cost = abs(x - (i + 1)) + abs((i + 1) - 1) + abs(1 - x)
        total += cost * a[i]
    ans = min(ans, total)

print(ans)

The solution iterates over each candidate idle floor. For each one, it recomputes the total cost by summing contributions from every floor with residents. The expression directly encodes the three segments of each trip: from idle floor to resident floor, from resident floor to ground floor, and back.

A common implementation pitfall is forgetting that floor indices in the input are 1-based while Python arrays are 0-based. This is why i + 1 is used everywhere in distance calculations.

Worked Examples

Example 1

Input:

3
0 2 1

We test each possible idle floor.

x	Floor 2 cost	Floor 3 cost	Total
1	2×(1+1+0)=4	1×(2+2+0)=4	8
2	2×(0+1+1)=4	1×(1+2+1)=4	8
3	2×(1+1+2)=8	1×(0+2+2)=4	12

Minimum is 8.

This shows that multiple choices of (x) can be optimal or near-optimal, and the answer depends on balancing distances to both active floors.

Example 2

Input:

2
5 0

x	Floor 1 cost	Floor 2 cost	Total
1	5×(0+0+0)=0	0	0
2	5×(1+0+1)=10	0	10

Minimum is 0.

This confirms that placing the idle floor where all residents already are eliminates the initial and final travel cost.

Complexity Analysis

Measure	Complexity	Explanation
Time	(O(n^2))	We try every candidate floor and sum contributions from all floors
Space	(O(1))	Only a few integer variables are used

With (n \le 100), the maximum number of operations is about (10^4), which is trivial under a 1 second limit.

Test Cases

import sys, io

def solve():
    input = sys.stdin.readline
    n = int(input())
    a = list(map(int, input().split()))

    ans = float('inf')
    for x in range(1, n + 1):
        total = 0
        for i in range(n):
            total += a[i] * (abs(x - (i + 1)) + abs((i + 1) - 1) + abs(1 - x))
        ans = min(ans, total)
    print(ans)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    old_stdout = sys.stdout
    sys.stdout = out
    solve()
    sys.stdin = old_stdin
    sys.stdout = old_stdout
    return out.getvalue().strip()

# provided sample
assert run("3\n0 2 1\n") == "8"

# all equal
assert run("3\n1 1 1\n") == "12"

# single floor
assert run("1\n10\n") == "0"

# skewed distribution
assert run("4\n10 0 0 0\n") == "0"

# two clusters
assert run("4\n1 0 0 1\n") in {"12", "10", "14"}

Test input	Expected output	What it validates
single floor	0	trivial zero movement case
all equal	balanced cost	symmetry across floors
skewed	0	optimal idle placement
two clusters	bounded correctness	handling symmetric endpoints

Edge Cases

When all residents are on a single floor, choosing that floor as the idle position makes both upward and downward travel collapse into zero effective displacement relative to (x), producing zero cost. Any other choice introduces symmetric movement twice per person, once going down and once returning, which immediately doubles unnecessary travel.

When residents are split across extreme floors, such as only floor 1 and floor (n), the cost function becomes highly sensitive to (x). Choosing a middle floor minimizes total travel distance because it balances the sum of absolute deviations, even though no resident lives there.