2593

2593. Find Score of an Array After Marking All Elements (Medium)

You are given an array nums consisting of positive integers.

Starting with score = 0, apply the following algorithm:

• Choose the smallest integer of the array that is not marked. If there is a tie, choose the one with the smallest index.
• Add the value of the chosen integer to score.
• Mark the chosen element and its two adjacent elements if they exist.
• Repeat until all the array elements are marked.

Return the score you get after applying the above algorithm.

 

Example 1:

Input: nums = [2,1,3,4,5,2]
Output: 7
Explanation: We mark the elements as follows:
- 1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,1,3,4,5,2].
- 2 is the smallest unmarked element, so we mark it and its left adjacent element: [2,1,3,4,5,2].
- 4 is the only remaining unmarked element, so we mark it: [2,1,3,4,5,2].
Our score is 1 + 2 + 4 = 7.

Example 2:

Input: nums = [2,3,5,1,3,2]
Output: 5
Explanation: We mark the elements as follows:
- 1 is the smallest unmarked element, so we mark it and its two adjacent elements: [2,3,5,1,3,2].
- 2 is the smallest unmarked element, since there are two of them, we choose the left-most one, so we mark the one at index 0 and its right adjacent element: [2,3,5,1,3,2].
- 2 is the only remaining unmarked element, so we mark it: [2,3,5,1,3,2].
Our score is 1 + 2 + 2 = 5.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 106

Related Topics:
Array, Sorting, Heap (Priority Queue), Simulation

Similar Questions:

• Sort Integers by The Power Value (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/find-score-of-an-array-after-marking-all-elements
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    long long findScore(vector<int>& A) {
        long long N = A.size(), ans = 0;
        vector<int> id(N);
        iota(begin(id), end(id), 0);
        sort(begin(id), end(id), [&](int a, int b) { return A[a] != A[b] ? A[a] < A[b] : a < b; });
        for (int i = 0; i < N; ++i) {
            int j = id[i];
            if (A[j] == 0) continue;
            ans += A[j];
            if (j - 1 >= 0) A[j - 1] = 0;
            if (j + 1 < N) A[j + 1] = 0;
        }
        return ans;
    }
};