2453. Destroy Sequential Targets

Explanation:

To solve this problem, we can iterate through each number in the nums array and calculate the number of targets that can be destroyed if we seed the machine with that number. We can do this by considering all possible values that can be destroyed based on the given space.

Create a map to store the count of each possible target value.
Iterate through each number in nums and calculate the min seed value that can destroy the maximum number of targets.
For each number in nums, calculate the minimum seed value by iterating through all possible targets that can be destroyed.
Update the count of destroyed targets in the map.
Return the number with the maximum count of destroyed targets.

Time complexity: O(n^2) where n is the length of the nums array.
Space complexity: O(n) for the map storing the count of destroyed targets.

View Leetcode Problem

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minSeed(int[] nums, int space) {
        Map<Integer, Integer> targetCount = new HashMap<>();
        int maxDestroyed = 0;
        int minSeed = Integer.MAX_VALUE;

        for (int i = 0; i < nums.length; i++) {
            int count = 0;
            for (int j = 0; j < nums.length; j++) {
                int target = nums[j];
                int seed = (int) Math.ceil((double)(target - nums[i]) / space);
                targetCount.put(seed, targetCount.getOrDefault(seed, 0) + 1);
                count = Math.max(count, targetCount.get(seed));
            }

            if (count > maxDestroyed || (count == maxDestroyed && nums[i] < minSeed)) {
                maxDestroyed = count;
                minSeed = nums[i];
            }

            targetCount.clear();
        }

        return minSeed;
    }
}

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Explanation:

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