2453. Destroy Sequential Targets

Problem

You are given a 0-indexed array nums consisting of positive integers, representing targets on a number line. You are also given an integer space.

You have a machine which can destroy targets. Seeding the machine with some nums[i] allows it to destroy all targets with values that can be represented as nums[i] + c * space, where c is any non-negative integer. You want to destroy the maximum number of targets in nums.

Return** the minimum value of nums[i] you can seed the machine with to destroy the maximum number of targets.**

  Example 1:

Input: nums = [3,7,8,1,1,5], space = 2
Output: 1
Explanation: If we seed the machine with nums[3], then we destroy all targets equal to 1,3,5,7,9,... 
In this case, we would destroy 5 total targets (all except for nums[2]). 
It is impossible to destroy more than 5 targets, so we return nums[3].

Example 2:

Input: nums = [1,3,5,2,4,6], space = 2
Output: 1
Explanation: Seeding the machine with nums[0], or nums[3] destroys 3 targets. 
It is not possible to destroy more than 3 targets.
Since nums[0] is the minimal integer that can destroy 3 targets, we return 1.

Example 3:

Input: nums = [6,2,5], space = 100
Output: 2
Explanation: Whatever initial seed we select, we can only destroy 1 target. The minimal seed is nums[1].

  Constraints:

• 1 <= nums.length <= 105

• 1 <= nums[i] <= 109

• 1 <= space <=&nbsp;109

Solution (Java)

class Solution {
    public int destroyTargets(int[] nums, int space) {
        Map<Integer, Integer> freqs = new HashMap<>();
        int maxFreq = 0;
        for (int n : nums) {
            n %= space;
            maxFreq = Math.max(maxFreq, freqs.merge(n, 1, Integer::sum));
        }
        int min = Integer.MAX_VALUE;
        for (int n : nums) {
            if (freqs.get(n % space) == maxFreq && min > n) {
                min = n;
            }
        }
        return min;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).