2593. Find Score of an Array After Marking All Elements

Description

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

 

Solutions

Solution: Simulation

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

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const findScore = function (nums) {
  const n = nums.length;
  const elements = nums.map((num, index) => ({ num, index }));
  const visited = Array.from({ length: n }, () => false);
  let result = 0;

  elements.sort((a, b) => a.num - b.num);

  for (const { num, index } of elements) {
    if (visited[index]) continue;

    result += num;
    visited[index] = true;

    if (index - 1 > -1) {
      visited[index - 1] = true;
    }
    if (index + 1 < n) {
      visited[index + 1] = true;
    }
  }
  return result;
};