Overview

This problem asks for the minimum movement cost to start from 0 degrees, visit all N temperatures, and return to 0 degrees. By thinking of temperatures as points on a number line, this reduces to finding the optimal path.

Analysis

Problem Reformulation

This problem can be thought of as a shortest path problem on a number line: - Starting point: 0 - Points to visit: \(H_1, H_2, \ldots, H_N\) - Ending point: 0

Key Insight

When moving on a number line, repeatedly traversing the same interval back and forth is wasteful.

Let’s consider a concrete example. For \(H = \{-3, 2, 5\}\): - Points to visit are -3, 2, 5 - Minimum is -3, maximum is 5

The optimal strategy is to go from 0 degrees to one end (e.g., 5), then to the opposite end (-3), and finally return to 0 degrees.

0 → 5 → -3 → 0
Cost = 5 + 8 + 3 = 16

With a different order \(0 → -3 → 5 → 0\):

Cost = 3 + 8 + 5 = 16

Both yield the same result!

Why This Is Optimal

• Making one round trip each through the intervals \([\min, 0]\) and \([0, \max]\) is minimal
• Since 0 degrees is both the starting and ending point, we need to go and return in both the positive and negative directions
• Intermediate points are naturally visited along this path

Formula for Minimum Cost

Let \(\max\_positive\) be the maximum positive temperature and \(\min\_negative\) be the minimum negative temperature:

\[\text{Minimum Cost} = 2 \times \max\_positive + 2 \times (-\min\_negative)\]

This is equivalent to \(2 \times (\max\_positive - \min\_negative)\).

Example: For \(H = \{-3, 2, 5\}\) - \(\max\_positive = 5\) - \(\min\_negative = -3\) - Cost \(= 2 \times 5 + 2 \times 3 = 16\)

Algorithm

1. Iterate through all \(H_i\)
2. Update the maximum positive temperature \(\max\_positive\) (initial value 0)
3. Update the minimum negative temperature \(\min\_negative\) (initial value 0)
4. The answer is \(2 \times \max\_positive - 2 \times \min\_negative\)

Key Point: By initializing to 0, we correctly handle cases where all temperatures are positive or all are negative.

• If all positive: \(\min\_negative = 0\), resulting in a round trip \(0 → \max → 0\)
• If all negative: \(\max\_positive = 0\), resulting in a round trip \(0 → \min → 0\)

Complexity

• Time Complexity: \(O(N)\) - just one pass through the array
• Space Complexity: \(O(N)\) - storing the input array (can be \(O(1)\) if only computing max/min)

Implementation Notes

• Since the range of \(H_i\) values is \(-10^9 \leq H_i \leq 10^9\), the answer can be up to approximately \(4 \times 10^9\). In Python there’s no concern about integer overflow, but in other languages you need to use long long type
• By initializing \(\max\_positive\) and \(\min\_negative\) to 0, we naturally handle the starting/ending point of 0 degrees
• Sorting is unnecessary; we only need to find the maximum and minimum values

Source Code

def solve():
    N = int(input())
    H = list(map(int, input().split()))
    
    # Find minimum energy consumption
    # Start from 0 degrees, visit all temperatures, return to 0 degrees
    # 
    # Consider sorting the temperatures
    # The optimal path is: go from 0 degrees in one direction, come back and go in the other direction
    # Or: go from 0 degrees in only one direction and return
    
    # Find the maximum and minimum of all temperatures
    # Need to consider 0 degrees as well (starting and ending point)
    
    if N == 0:
        print(0)
        return
    
    # Separate into positive and negative temperatures
    max_positive = 0  # Maximum positive temperature (0 if none)
    min_negative = 0  # Minimum negative temperature (0 if none)
    
    for h in H:
        if h > max_positive:
            max_positive = h
        if h < min_negative:
            min_negative = h
    
    # Minimum cost starting from 0 degrees and returning to 0 degrees
    # Pattern 1: 0 -> positive max -> negative min -> 0
    #   Cost = max_positive + (max_positive - min_negative) + (-min_negative) = 2*max_positive - 2*min_negative
    # Pattern 2: 0 -> negative min -> positive max -> 0
    #   Cost = (-min_negative) + (max_positive - min_negative) + max_positive = 2*max_positive - 2*min_negative
    # 
    # Both are the same: 2 * (max_positive - min_negative)
    # However, since max_positive >= 0, min_negative <= 0
    # Cost = 2 * max_positive + 2 * (-min_negative) = 2 * max_positive - 2 * min_negative
    
    # In practice, making one round trip through the range [min_negative, max_positive] is optimal
    # Since 0 degrees is included, starting from 0, go to one end, go to the other end, return to 0
    # This allows visiting all temperatures
    
    result = 2 * max_positive - 2 * min_negative
    print(result)

solve()

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