Compute optimal locker placement with obstacles

Given house coordinates H = {(xi, yi)} and tree coordinates T = {(xj, yj)} on a 2D integer grid, choose a locker location L = (x, y) such that L ∉ T and the sum of Manhattan distances from L to all houses is minimized. Return the minimal total distance and one optimal location. Discuss an efficient algorithm that leverages medians and how to handle cases where the unconstrained median lies on a tree cell.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Optimal Locker Placement with Obstacles

You are given a list of houses, where houses[i] = [xi, yi] are the integer grid coordinates of the i-th house, and a list of trees, where trees[j] = [xj, yj] are the coordinates of tree cells. Tree cells are blocked: a locker cannot be placed on them.

Choose a locker location L = (x, y) on the integer grid such that L is NOT a tree cell, and the sum of Manhattan distances from L to every house is minimized. The Manhattan distance between (x, y) and (xi, yi) is |x - xi| + |y - yi|.

Return the minimal achievable total Manhattan distance. If houses is empty, return 0.

Key idea: because Manhattan cost separates into independent x- and y-terms, the unconstrained optimum is (median(xs), median(ys)). The cost surface is convex and separable, so when that median cell (or any candidate) is blocked by a tree, an optimal valid cell lies in the small band around the median/house/tree coordinates — checking that finite candidate set yields the true minimum.

Constraints

• 0 <= len(houses) <= 10^4
• 0 <= len(trees) <= 10^4
• -10^6 <= xi, yi, xj, yj <= 10^6
• A house and a tree may share the same cell; the locker still may not be placed on a tree cell.
• If houses is empty, the answer is 0.

Examples

Example 1

Input: ([[0,0],[0,4],[2,2]], [])
Output: 6

Explanation: Median is (0, 2). No trees block it. Cost at (0,2) = (0+2) + (0+2) + (2+0) = 6.

Example 2

Input: ([[0,0],[0,4],[2,2]], [[0,2]])
Output: 7

Explanation: The unconstrained optimum (0,2) is now a tree. The cheapest non-tree cell (e.g. (1,2) or (0,1) or (0,3)) costs 7.

Hints

1. Manhattan distance is separable: the total cost = (sum over houses of |x - xi|) + (sum over houses of |y - yi|). Optimize the x-coordinate and y-coordinate independently.
2. Each one-dimensional sum |x - xi| is minimized when x is a median of the house x-coordinates (likewise for y). That gives the unconstrained optimum (median(xs), median(ys)).
3. The constraint is that the chosen cell can't be a tree. Because the cost grows monotonically as you move away from the median along each axis, an optimal valid cell is close to the median — check the median, the house coordinates, and the cells one step around them (and around trees) and take the cheapest non-tree cell.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Given house coordinates H = {(xi, yi)} and tree coordinates T = {(xj, yj)} on a 2D integer grid, choose a locker location L = (x, y) such that L ∉ T and the sum of Manhattan distances from L to all houses is minimized. Return the minimal total distance and one optimal location. Discuss an efficient algorithm that leverages medians and how to handle cases where the unconstrained median lies on a tree cell.

Recommended Approach

Choose traversal based on the required output. DFS is natural for subtree computations, reconstruction, and range pruning; BFS is natural for level order or side views. Keep per-depth or per-position state when the output depends on columns, rows, or depths.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

Most tree traversals are O(n) time and O(h) recursion stack for DFS or O(w) queue space for BFS, where h is height and w is maximum width.

Edge Cases and Tests

Empty tree, one node, skewed tree, duplicate values when reconstruction assumes uniqueness, deep recursion, and tie-breaking for same row/column nodes.