FindNearestPointThatHasTheSameXOrYCoordinate.cpp

// Source : https://leetcode.com/problems/find-nearest-point-that-has-the-same-x-or-y-coordinate/
// Author : Hao Chen
// Date   : 2021-03-13

/***************************************************************************************************** 
 *
 * 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
 ******************************************************************************************************/

class Solution {
public:
    int nearestValidPoint(int x, int y, vector<vector<int>>& points) {
        int result = -1;
        int distance = INT_MAX;
        for(int i=0; i<points.size(); i++) {
            if ( x  != points[i][0]  && y != points[i][1] ) continue;
            int dist = abs(x - points[i][0]) + abs( y - points[i][1]);
            if ( distance > dist ){
                distance = dist;
                result = i;
            }
        }
        return result;
    }
};