3102. Minimize Manhattan Distances

Description

You are given an array points representing integer coordinates of some points on a 2D plane, where points[i] = [x_i, y_i].

The distance between two points is defined as their Manhattan distance.

Return the minimum possible value for maximum distance between any two points by removing exactly one point.

Example 1:

Input: points = [[3,10],[5,15],[10,2],[4,4]]

Output: 12

Explanation:

The maximum distance after removing each point is the following:

After removing the 0^th point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18.
After removing the 1^st point the maximum distance is between points (3, 10) and (10, 2), which is |3 - 10| + |10 - 2| = 15.
After removing the 2^nd point the maximum distance is between points (5, 15) and (4, 4), which is |5 - 4| + |15 - 4| = 12.
After removing the 3^rd point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18.

12 is the minimum possible maximum distance between any two points after removing exactly one point.

Example 2:

Input: points = [[1,1],[1,1],[1,1]]

Output: 0

Explanation:

Removing any of the points results in the maximum distance between any two points of 0.

Constraints:

3 <= points.length <= 10⁵
points[i].length == 2
1 <= points[i][0], points[i][1] <= 10⁸

Solutions

Solution 1: Ordered Set

For two points $(x_1, y_1)$ and $(x_2, y_2)$, their Manhattan distance is $|x_1 - x_2| + |y_1 - y_2|$. We can transform it into $\max(x_1 - x_2, x_2 - x_1) + \max(y_1 - y_2, y_2 - y_1)$, which is:

$$ |x_1 - x_2| + |y_1 - y_2| = \max \begin{cases} x_1 - x_2 + y_1 - y_2 \ x_2 - x_1 + y_2 - y_1 \ x_1 - x_2 + y_2 - y_1 \ x_2 - x_1 + y_1 - y_2 \end{cases} $$

This can be simplified to:

$$ |x_1 - x_2| + |y_1 - y_2| = \max \begin{cases} (x_1 + y_1) - (x_2 + y_2) \ (x_2 + y_2) - (x_1 + y_1) \ (x_1 - y_1) - (x_2 - y_2) \ (x_2 - y_2) - (x_1 - y_1) \end{cases} $$

Here, the first two cases can be represented as $\max(\max(x_1 + y_1, x_2 + y_2) - \min(x_1 + y_1, x_2 + y_2))$, and the last two cases can be represented as $\max(\max(x_1 - y_1, x_2 - y_2) - \min(x_1 - y_1, x_2 - y_2))$.

Therefore, we can store all points according to the values of $x + y$ and $x - y$ in two ordered sets, respectively. Then, for each point, we remove it, update the values in the ordered sets, calculate the difference between the maximum and minimum values, and take the minimum value.

The time complexity is $O(n \log n)$, and the space complexity is $O(n)$. Here, $n$ is the number of points.

Python3JavaC++GoTypeScript

from sortedcontainers import SortedList


class Solution:
    def minimumDistance(self, points: List[List[int]]) -> int:
        sl1 = SortedList()
        sl2 = SortedList()
        for x, y in points:
            sl1.add(x + y)
            sl2.add(x - y)
        ans = inf
        for x, y in points:
            sl1.remove(x + y)
            sl2.remove(x - y)
            ans = min(ans, max(sl1[-1] - sl1[0], sl2[-1] - sl2[0]))
            sl1.add(x + y)
            sl2.add(x - y)
        return ans

class Solution {
    public int minimumDistance(int[][] points) {
        TreeMap<Integer, Integer> tm1 = new TreeMap<>();
        TreeMap<Integer, Integer> tm2 = new TreeMap<>();
        for (int[] p : points) {
            int x = p[0], y = p[1];
            tm1.merge(x + y, 1, Integer::sum);
            tm2.merge(x - y, 1, Integer::sum);
        }
        int ans = Integer.MAX_VALUE;
        for (int[] p : points) {
            int x = p[0], y = p[1];
            if (tm1.merge(x + y, -1, Integer::sum) == 0) {
                tm1.remove(x + y);
            }
            if (tm2.merge(x - y, -1, Integer::sum) == 0) {
                tm2.remove(x - y);
            }
            ans = Math.min(
                ans, Math.max(tm1.lastKey() - tm1.firstKey(), tm2.lastKey() - tm2.firstKey()));
            tm1.merge(x + y, 1, Integer::sum);
            tm2.merge(x - y, 1, Integer::sum);
        }
        return ans;
    }
}

class Solution {
public:
    int minimumDistance(vector<vector<int>>& points) {
        multiset<int> st1;
        multiset<int> st2;
        for (auto& p : points) {
            int x = p[0], y = p[1];
            st1.insert(x + y);
            st2.insert(x - y);
        }
        int ans = INT_MAX;
        for (auto& p : points) {
            int x = p[0], y = p[1];
            st1.erase(st1.find(x + y));
            st2.erase(st2.find(x - y));
            ans = min(ans, max(*st1.rbegin() - *st1.begin(), *st2.rbegin() - *st2.begin()));
            st1.insert(x + y);
            st2.insert(x - y);
        }
        return ans;
    }
};