2453

2453. Destroy Sequential Targets (Medium)

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 <= 109

Companies: Intuit

Related Topics:
Array, Hash Table, Counting

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Solution 1.

// OJ: https://leetcode.com/problems/destroy-sequential-targets
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int destroyTargets(vector<int>& A, int space) {
        unordered_map<int, int> cnt, mn;
        int maxCnt = 0, ans = 0;
        for (int n : A) {
            int r = n % space;
            ++cnt[r];
            if (mn.count(r)) mn[r] = min(mn[r], n);
            else mn[r] = n;
        }
        for (auto &[r, c] : cnt) {
            if (c > maxCnt || (c == maxCnt && mn[r] < ans)) {
                maxCnt = c;
                ans = mn[r];
            }
        }
        return ans;
    }
};