README_EN.md

1779. Find Nearest Point That Has the Same X or Y Coordinate

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Description

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 <= 104
• points[i].length == 2
• 1 <= x, y, ai, bi <= 104

Solutions

Python3

class Solution:
    def nearestValidPoint(self, x: int, y: int, points: List[List[int]]) -> int:
        ans, mi = -1, inf
        for i, (a, b) in enumerate(points):
            if a == x or b == y:
                d = abs(a - x) + abs(b - y)
                if mi > d:
                    ans, mi = i, d
        return ans

Java

class Solution {
    public int nearestValidPoint(int x, int y, int[][] points) {
        int ans = -1, mi = 1000000;
        for (int i = 0; i < points.length; ++i) {
            int a = points[i][0], b = points[i][1];
            if (a == x || b == y) {
                int d = Math.abs(a - x) + Math.abs(b - y);
                if (d < mi) {
                    mi = d;
                    ans = i;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int nearestValidPoint(int x, int y, vector<vector<int>>& points) {
        int ans = -1, mi = 1e6;
        for (int i = 0; i < points.size(); ++i) {
            int a = points[i][0], b = points[i][1];
            if (a == x || b == y) {
                int d = abs(a - x) + abs(b - y);
                if (d < mi) {
                    mi = d;
                    ans = i;
                }
            }
        }
        return ans;
    }
};

Go

func nearestValidPoint(x int, y int, points [][]int) int {
	ans, mi := -1, 1000000
	for i, p := range points {
		a, b := p[0], p[1]
		if a == x || b == y {
			d := abs(a-x) + abs(b-y)
			if d < mi {
				ans, mi = i, d
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

TypeScript

function nearestValidPoint(x: number, y: number, points: number[][]): number {
    let res = -1;
    let midDif = Infinity;
    points.forEach(([px, py], i) => {
        if (px != x && py != y) {
            return;
        }
        const dif = Math.abs(px - x) + Math.abs(py - y);
        if (dif < midDif) {
            midDif = dif;
            res = i;
        }
    });
    return res;
}

Rust

impl Solution {
    pub fn nearest_valid_point(x: i32, y: i32, points: Vec<Vec<i32>>) -> i32 {
        let n = points.len();
        let mut min_dif = i32::MAX;
        let mut res = -1;
        for i in 0..n {
            let (p_x, p_y) = (points[i][0], points[i][1]);
            if p_x != x && p_y != y {
                continue;
            }
            let dif = (p_x - x).abs() + (p_y - y).abs();
            if dif < min_dif {
                min_dif = dif;
                res = i as i32;
            }
        }
        res
    }
}

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