Minimum Time For K Virus Variants to Spread

This problem asks to find the minimum number of days for any point to contain at least k unique virus variants. The virus spreads in four directions daily. We can solve this using a binary search approach combined with a check function.

Approach:

1. Binary Search: The core idea is to use binary search on the number of days. The search space is from 0 to the maximum possible distance between any two virus starting points (as this represents the maximum time it could take for them to meet).

2. Check Function: The binary search relies on a check function that determines if at least k variants can be found at any point within the given number of days. This is done by simulating the virus spread for the specified number of days. For each virus variant's starting point, we calculate the reachable area, which is a square centered at the point. The size of this square is twice the number of days. Then we count the number of variants within each cell's reachable area.

3. Optimization: We can optimize the check function by using a 2D array (or a hash map) to track the number of times each point is reached by a virus. This avoids redundant calculations.

Time Complexity Analysis:

• Binary Search: O(log(max_distance)), where max_distance is the maximum distance between any two points. This is because we're performing a binary search on the number of days.

• Check Function: O(n * d^2), where 'n' is the number of virus variants and 'd' is the number of days (the size of the area to check). This is because we iterate through each variant and check the cells in its reachable area. The area of that square is proportional to d^2.

• Overall: The overall time complexity is dominated by the check function, making it O(n * d^2 * log(max_distance)). In the worst case, d can be proportional to the maximum distance between points.

Space Complexity Analysis:

The space complexity is primarily determined by the 2D array used to track the number of times each point is reached by a virus, which is O(max_x * max_y), where max_x and max_y are the maximum x and y coordinates among all virus starting points.

Code Implementation (Python):

import itertools
 
def min_days(points, k):
    max_dist = 0
    for i in range(len(points)):
        for j in range(i + 1, len(points)):
            max_dist = max(max_dist, abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]))
 
    def check(days):
        grid = {}
        for x, y in points:
            for i in range(x - days, x + days + 1):
                for j in range(y - days, y + days + 1):
                    grid[(i, j)] = grid.get((i, j), 0) + 1
        for count in grid.values():
            if count >= k:
                return True
        return False
 
    left, right = 0, max_dist
    ans = max_dist
    while left <= right:
        mid = (left + right) // 2
        if check(mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1
    return ans
 

Code Implementation (Java):

import java.util.*;
 
class Solution {
    public int minDays(int[][] points, int k) {
        int maxDist = 0;
        for (int i = 0; i < points.length; i++) {
            for (int j = i + 1; j < points.length; j++) {
                maxDist = Math.max(maxDist, Math.abs(points[i][0] - points[j][0]) + Math.abs(points[i][1] - points[j][1]));
            }
        }
 
        int left = 0, right = maxDist;
        int ans = maxDist;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (check(points, k, mid)) {
                ans = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
 
    private boolean check(int[][] points, int k, int days) {
        Map<String, Integer> grid = new HashMap<>();
        for (int[] point : points) {
            for (int i = point[0] - days; i <= point[0] + days; i++) {
                for (int j = point[1] - days; j <= point[1] + days; j++) {
                    grid.put(i + "," + j, grid.getOrDefault(i + "," + j, 0) + 1);
                }
            }
        }
        for (int count : grid.values()) {
            if (count >= k) return true;
        }
        return false;
    }
}

This detailed explanation, including the optimized approach, time and space complexity analysis, and code in both Python and Java, provides a comprehensive solution to the problem. Remember to handle edge cases and potential integer overflows appropriately in your implementation.