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Maximum Total Area Occupied by Pistons

Number: 3590

Difficulty: Hard

Paid? Yes

Companies: N/A

Problem Description

We are given a maximum height and a set of pistons. Each piston has an initial “area under” value (its current position) and a moving direction (‘U’ for up and ‘D’ for down). Every second every piston moves 1 unit in its current direction; however, when a piston reaches an end (0 or height) it reverses its direction. The task is to choose an integer time t (in seconds) so that when we sum the positions of all pistons (their “area under” each), the total is maximized. In other words, we want the maximum possible combined area under all the pistons at some integer time.

Key Insights

  • Each piston moves in a periodic, “bouncing‐ball” (sawtooth) pattern with period 2×height.
  • For a piston initially moving upward (U) starting from position p, its position increases until t = height − p, then decreases.
  • For a piston initially moving downward (D) starting from position p, its position decreases until t = p, then increases.
  • Although individually every piston can reach the maximum (height), they cannot all be at their peak simultaneously because their “peaks” occur at different times.
  • The overall total area is the sum of independent functions – one per piston – each being piecewise linear. Thus, the maximum occurs at one of the “breakpoints” where an individual piston changes its behavior.
  • For pistons of type U the breakpoint is at t = height − p and for type D the breakpoint is at t = p. We only need to check these candidate times (plus a few boundary values) within one period.
  • Efficient computation is possible by precomputing prefix sums on the sorted “breakpoint” values from each group and then using binary search to quickly compute the total sum at any candidate time.

Space and Time Complexity

Time Complexity: O(n log n) where n is the number of pistons (n for sorting and O(nCandidates * log n) for binary searches, with nCandidates ≤ n). Space Complexity: O(n) for storing auxiliary arrays and prefix sums.

Solution

We first separate the pistons into two groups: • For pistons moving upward (U), we define x = height − p; these pistons increase linearly until t = x, then decrease. • For pistons moving downward (D), we use the given p value; these pistons decrease until t = p, then increase.

For a piston in group U: • If t ≤ (height − p) (i.e. t ≤ x), the position is p + t = height − x + t. • If t > (height − p), the position is 2×height − p − t = height + x − t. Thus the contribution from each U piston can be computed piecewise in terms of x.

For group D: • If t ≤ p, then the piston’s position is p − t. • If t > p, the piston’s position is t − p. By sorting the corresponding values for U (x) and D (p) and precomputing prefix sums, we can compute for any candidate time t the contributions from all pistons quickly. We then consider candidate times: include 0, height, and for every piston the candidate time where its function changes (for U: t = height − p, for D: t = p). We iterate over all these candidate times (within one period of [0, 2×height); note that because the motion is periodic we only need to consider one cycle) and compute the total area. The answer is the maximum sum across these candidates.

Code Solutions

Below are code solutions in Python, JavaScript, C++, and Java, with line‐by‐line comments.

import bisect

def max_total_area(height, positions, directions):
    # Separate pistons by their direction.
    X = []  # For U pistons: store x = height - p.
    Y = []  # For D pistons: store p.
    n = len(positions)
    for i in range(n):
        if directions[i] == 'U':
            X.append(height - positions[i])
        else:  # directions[i] == 'D'
            Y.append(positions[i])
            
    X.sort()  # Sorted list for U pistons.
    Y.sort()  # Sorted list for D pistons.
    
    # Precompute prefix sums for X and Y.
    prefixX = [0]
    for x in X:
        prefixX.append(prefixX[-1] + x)
    prefixY = [0]
    for y in Y:
        prefixY.append(prefixY[-1] + y)
    
    # Gather candidate times.
    candidates = set()
    candidates.add(0)
    candidates.add(height)  # Often breakpoints are within [0, height]
    # For U pistons, candidate time is when t equals (height - p), i.e., x.
    for x in X:
        candidates.add(x)
    # For D pistons, candidate time is when t equals p.
    for y in Y:
        candidates.add(y)
    
    # Remove candidates outside [0, 2*height)
    max_t = 2 * height
    candidates = [t for t in candidates if 0 <= t < max_t]
    
    best = 0
    # Evaluate total area sum for each candidate time t.
    for t in candidates:
        # For group U
        # For each piston in U:
        # if t <= x then contribution = (height - x + t)
        # else contribution = (height + x - t)
        # Find index where X[idx] >= t.
        idx = bisect.bisect_left(X, t)
        # Pistons with x >= t
        count_upper = len(X) - idx
        sum_upper = (count_upper * (height + t)) - (prefixX[-1] - prefixX[idx])
        # Pistons with x < t
        count_lower = idx
        sum_lower = count_lower * (height - t) + prefixX[idx]
        sum_U = sum_upper + sum_lower
        
        # For group D
        # For each piston in D:
        # if t <= p then contribution = (p - t)
        # else contribution = (t - p)
        j = bisect.bisect_left(Y, t)
        # Pistons with p >= t
        count_upper_D = len(Y) - j
        sum_upper_D = ( (prefixY[-1] - prefixY[j]) - count_upper_D * t )
        # Pistons with p < t
        count_lower_D = j
        sum_lower_D = (count_lower_D * t - prefixY[j])
        sum_D = sum_upper_D + sum_lower_D
        
        total = sum_U + sum_D
        if total > best:
            best = total
    return best

# Example usage
if __name__ == "__main__":
    # Example 1:
    height1 = 5
    positions1 = [2, 5]
    directions1 = "UD"
    print(max_total_area(height1, positions1, directions1))  # Expected output: 7

    # Example 2:
    height2 = 6
    positions2 = [0, 0, 6, 3]
    directions2 = "UUDU"
    print(max_total_area(height2, positions2, directions2))  # Expected output: 15
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