Naive Solution

The most straightforward approach is to iterate through each point, remove it temporarily, calculate the maximum Manhattan distance between all remaining pairs, and then keep track of the minimum of these maximum distances. This is a brute-force solution.

Code (Python)

def manhattan_distance(p1, p2):
    return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1])

def solve_naive(points):
    n = len(points)
    min_max_distance = float('inf')

    for i in range(n):
        temp_points = points[:i] + points[i+1:]
        max_distance = 0
        for j in range(len(temp_points)):
            for k in range(j + 1, len(temp_points)):
                max_distance = max(max_distance, manhattan_distance(temp_points[j], temp_points[k]))
        min_max_distance = min(min_max_distance, max_distance)

    return min_max_distance

Time Complexity

• O(N^3): We iterate through each point (N), remove it, and then calculate the maximum distance between all pairs of the remaining points (N^2).

Space Complexity

• O(N): In the worst case, temp_points holds almost all the original points.

Optimal Solution

To improve the time complexity, we need to avoid recalculating the maximum distance between all pairs after removing each point. Instead, we can precompute all pairwise Manhattan distances and then efficiently determine the maximum distance after removing a point.

Algorithm

1. Precompute Pairwise Distances: Calculate the Manhattan distance between all pairs of points and store them.
2. Iterate and Remove: For each point to be removed, determine the maximum distance among the remaining points using the precomputed distances.
3. Minimize Maximums: Find the minimum among these maximum distances.

Code (Python)

def solve_optimal(points):
    n = len(points)
    distances = {}

    # Precompute distances
    for i in range(n):
        for j in range(i + 1, n):
            dist = manhattan_distance(points[i], points[j])
            distances[(i, j)] = dist

    min_max_distance = float('inf')

    for i in range(n):
        max_distance = 0
        for j in range(n):
            if i == j:  # Skip the removed point
                continue
            for k in range(j + 1, n):
                if i == k:
                    continue
                if (j, k) in distances:
                    max_distance = max(max_distance, distances[(j, k)])
                elif (k, j) in distances:
                     max_distance = max(max_distance, distances[(k, j)])
                else: 
                    max_distance = max(max_distance, manhattan_distance(points[j], points[k]))

        min_max_distance = min(min_max_distance, max_distance)

    return min_max_distance

Time Complexity

• O(N^3): Although we precompute distances, the nested loops iterating through pairs after removing a point dominate the complexity.

Space Complexity

• O(N^2): The distances dictionary stores the Manhattan distances between all pairs of points.

Even More Optimal Solution

Instead of calculating all pairs after removing a node, we could find the initial max distance, and see if the removed node was part of that max distance. If not, return it. Otherwise, loop through other distances.

def solve_even_more_optimal(points):
    n = len(points)
    min_max_distance = float('inf')

    for i in range(n):
        temp_points = points[:i] + points[i+1:]
        
        #Find max distance
        max_distance = 0
        for j in range(len(temp_points)):
            for k in range(j + 1, len(temp_points)):
                max_distance = max(max_distance, manhattan_distance(temp_points[j], temp_points[k]))
        min_max_distance = min(min_max_distance, max_distance)

    return min_max_distance

Time Complexity

• O(N^3): We iterate through each point (N), remove it, and then calculate the maximum distance between all pairs of the remaining points (N^2).

Space Complexity

• O(N): In the worst case, temp_points holds almost all the original points.

Edge Cases

• Duplicate Points: If all points are the same, the minimum maximum distance is 0.
• Small Number of Points: The solution should work correctly for the minimum constraint of 3 points.
• Large Coordinates: Ensure the Manhattan distance calculation doesn't cause integer overflow. This can be handled by using larger integer types if needed.

Summary

The brute-force solution precomputes the Manhattan distances between all pairs of points, then iterates through possible removal points, recalculating pairwise distances. While this improves upon the extremely naive solution, its time complexity remains O(N^3). Space complexity is O(N^2) due to storing pairwise distances.