1779. Find Nearest Point That Has the Same X or Y Coordinate
Explanation
To solve this problem, we iterate through each point in the points array and calculate the Manhattan distance for each point. We keep track of the minimum distance and the index of the point with the minimum distance that shares the same x-coordinate or y-coordinate with the given location (x, y).
We initialize the minimum distance as infinity and the result index as -1. We then iterate through each point, calculate the Manhattan distance, and update the result index if the current point has a smaller Manhattan distance and shares the same x-coordinate or y-coordinate.
After iterating through all points, we return the result index as the answer.
- Time complexity: O(n), where n is the number of points
- Space complexity: O(1)
class Solution {
public int nearestValidPoint(int x, int y, int[][] points) {
int minDistance = Integer.MAX_VALUE;
int resultIndex = -1;
for (int i = 0; i < points.length; i++) {
int px = points[i][0];
int py = points[i][1];
if (px == x || py == y) {
int distance = Math.abs(px - x) + Math.abs(py - y);
if (distance < minDistance) {
minDistance = distance;
resultIndex = i;
}
}
}
return resultIndex;
}
}Code Editor (Testing phase)
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