1779 - Find Nearest Point That Has the Same X or Y Coordinate (Easy)

Problem Link

https://leetcode.com/problems/find-nearest-point-that-has-the-same-x-or-y-coordinate/

Problem Statement

You are given two integers, x and y, which represent your current location on a Cartesian grid: (x, y). You are also given an array points where each points[i] = [ai, bi] represents that a point exists at (ai, bi). A point is valid if it shares the same x-coordinate or the same y-coordinate as your location.

Return the index (0-indexed) of the valid point with the smallest Manhattan distance from your current location. If there are multiple, return the valid point with the smallest index. If there are no valid points, return -1.

The Manhattan distance between two points (x1, y1) and (x2, y2) is abs(x1 - x2) + abs(y1 - y2).

Example 1:

Input: x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]]
Output: 2
Explanation: Of all the points, only [3,1], [2,4] and [4,4] are valid. Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. [2,4] has the smallest index, so return 2.

Example 2:

Input: x = 3, y = 4, points = [[3,4]]
Output: 0
Explanation: The answer is allowed to be on the same location as your current location.

Example 3:

Input: x = 3, y = 4, points = [[2,3]]
Output: -1
Explanation: There are no valid points.

Constraints:

• 1 <= points.length <= 10^4
• points[i].length == 2
• 1 <= x, y, ai, bi <= 10^4

Approach 1: Brute Force

For each point, we simply check if the current point has the same X or Y coordinate and calculate the distance between that and the target point $(x, y)$. If we find the one with smaller distance, update the distance and answer.

C++

Written by @wingkwong

class Solution {
public:
    int nearestValidPoint(int x, int y, vector<vector<int>>& points) {
        int ans = -1, mi = INT_MAX;
        for(int i = 0; i < points.size(); i++) {
            auto p = points[i];
            if(x == p[0] || y == p[1]) {
                if(abs(x - p[0]) + abs(y - p[1]) < mi) {
                    mi = abs(x - p[0]) + abs(y - p[1]);
                    ans = i;
                }
            }
        }
        return ans;
    }
};

Rust

Written by @wingkwong

impl Solution {
    pub fn nearest_valid_point(x: i32, y: i32, points: Vec<Vec<i32>>) -> i32 {
        let mut mi = std::i32::MAX;
        let mut ans = -1;
        for (i, p) in points.into_iter().enumerate() {
            if p[0] == x || p[1] == y {
                let d = (p[0] - x).abs() + (p[1] - y).abs();
                if (d < mi) {
                    mi = d;
                    ans = i as i32;
                }
            }
        }
        ans
    }
}