Question

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

中文文档

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

Solution 1

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

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;
  }
}

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;
  }
};

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
}

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;
}

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
  }
}

int nearestValidPoint(int x, int y, int** points, int pointsSize, int* pointsColSize) {
  int ans = -1;
  int min = INT_MAX;
  for (int i = 0; i < pointsSize; i++) {
    int* point = points[i];
    if (point[0] != x && point[1] != y) {
      continue;
    }
    int d = abs(x - point[0]) + abs(y - point[1]);
    if (d < min) {
      min = d;
      ans = i;
    }
  }
  return ans;
}