LeetCode 2593 - Find Score of an Array After Marking All Elements

Here is a comprehensive, detailed technical solution guide for LeetCode 2593 - Find Score of an Array After Marking All Elements, following your requested format precisely. The problem provides an array of positive integers, nums.

leetcodemediumarrayhash-tablesortingheap-(priority-queue)simulation

LeetCode Problem 2593

Difficulty: 🟡 Medium
Topics: Array, Hash Table, Sorting, Heap (Priority Queue), Simulation

Solution

Here is a comprehensive, detailed technical solution guide for LeetCode 2593 - Find Score of an Array After Marking All Elements, following your requested format precisely.

Problem Understanding

The problem provides an array of positive integers, nums. We are asked to simulate a scoring process where at each step we select the smallest unmarked element in the array. If multiple elements tie for the smallest value, we select the left-most index. Once selected, the value of that element is added to a running score, and the element itself, as well as its immediate neighbors (if they exist), are marked. This process repeats until all elements are marked.

The input is a standard integer array, and the output is a single integer representing the total score accumulated according to this marking rule.

Constraints indicate that nums can be up to 10^5 elements long, and each element can be up to 10^6. This rules out naive approaches that repeatedly scan the array to find the minimum, since repeated full scans would result in O(n^2) time.

Key edge cases include arrays of length 1, arrays where multiple minimum elements exist, and arrays with alternating values where marking neighbors could skip over other small elements.

Approaches

Brute Force Approach: The naive method is to repeatedly scan the array to find the smallest unmarked element. After selecting it, mark it and its neighbors, then repeat. While this approach correctly implements the rules, it requires scanning the array O(n) times for n elements, resulting in O(n^2) time complexity. For n = 10^5, this is far too slow.

Optimal Approach: The key observation is that we can pre-sort the array while keeping track of indices and then use a priority queue (min-heap) or sorted list to efficiently select the next smallest unmarked element. By maintaining a boolean array for marking and iterating through elements in ascending order of value, we can achieve O(n log n) time complexity.

Approach Time Complexity Space Complexity Notes
Brute Force O(n^2) O(n) Repeatedly scan array for smallest unmarked element
Optimal (Sorting / Heap) O(n log n) O(n) Sort elements by value and process greedily with marking array

Algorithm Walkthrough

  1. Create a list of tuples (value, index) for all elements in nums. This allows us to sort by value while remembering original positions.
  2. Sort the list by value. Python and Go sorts are stable, so ties are broken by index automatically.
  3. Initialize a boolean array marked of the same length as nums to track which elements are marked.
  4. Initialize score = 0.
  5. Iterate through the sorted list. For each element (value, index):
  • If the element at index is already marked, skip it.
  • Otherwise, add value to score.
  • Mark the current index and its neighbors if they exist (indices index - 1 and index + 1), ensuring we stay within array bounds.
  1. Continue until all elements have been processed. Return score.

Why it works: The algorithm maintains the invariant that at each step, the smallest unmarked element is chosen, exactly as the problem requires. Marking neighbors ensures that future iterations respect the marking rule. Sorting guarantees we process elements in the correct order without rescanning the array.

Python Solution

from typing import List

class Solution:
    def findScore(self, nums: List[int]) -> int:
        n = len(nums)
        indexed_nums = [(num, i) for i, num in enumerate(nums)]
        indexed_nums.sort()  # sort by value, ties broken by index due to stable sort

        marked = [False] * n
        score = 0

        for value, i in indexed_nums:
            if not marked[i]:
                score += value
                marked[i] = True
                if i > 0:
                    marked[i - 1] = True
                if i < n - 1:
                    marked[i + 1] = True

        return score

Explanation: We first pair each value with its index and sort the pairs. marked keeps track of which elements have been scored or invalidated by adjacency marking. By iterating through sorted elements and marking neighbors, we avoid repeated scans, achieving O(n log n) efficiency.

Go Solution

import (
    "sort"
)

func findScore(nums []int) int64 {
    n := len(nums)
    type pair struct {
        value int
        index int
    }

    pairs := make([]pair, n)
    for i, v := range nums {
        pairs[i] = pair{v, i}
    }

    sort.SliceStable(pairs, func(i, j int) bool {
        return pairs[i].value < pairs[j].value
    })

    marked := make([]bool, n)
    var score int64 = 0

    for _, p := range pairs {
        i := p.index
        if !marked[i] {
            score += int64(p.value)
            marked[i] = true
            if i > 0 {
                marked[i-1] = true
            }
            if i < n-1 {
                marked[i+1] = true
            }
        }
    }

    return score
}

Go Notes: int64 is used for score to avoid potential overflow for large sums. sort.SliceStable ensures stable sort to respect tie-breaking by index. Marking is done in the same way as Python.

Worked Examples

Example 1: nums = [2,1,3,4,5,2]

Step Selected Score Marked Array
1 1 at index 1 1 [True, True, True, False, False, False]
2 2 at index 0 3 [True, True, True, False, False, False] (already marked by neighbor)
3 4 at index 3 7 [True, True, True, True, True, False]

Final score: 7

Example 2: nums = [2,3,5,1,3,2]

Step Selected Score Marked Array
1 1 at index 3 1 [False, False, True, True, True, False]
2 2 at index 0 3 [True, True, True, True, True, False]
3 2 at index 5 5 [True, True, True, True, True, True]

Final score: 5

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting n elements dominates; marking is O(n)
Space O(n) Extra arrays for indexed_nums and marked

Sorting ensures efficient selection of the next minimum element. Space is linear to maintain marking status and index-value pairs.

Test Cases

# Provided examples
assert Solution().findScore([2,1,3,4,5,2]) == 7  # example 1
assert Solution().findScore([2,3,5,1,3,2]) == 5  # example 2

# Edge cases
assert Solution().findScore([1]) == 1  # single element
assert Solution().findScore([1,1,1,1]) == 1  # all elements same
assert Solution().findScore([5,4,3,2,1]) == 9  # descending order
assert Solution().findScore([1,2,3,4,5]) == 9  # ascending order
assert Solution().findScore([1,1000000,1,1000000,1]) == 3  # large numbers with small numbers in between
Test Why
[2,1,3,4,5,2] Standard example with neighbors
[2,3,5,1,3,2] Tie-breaking by index
[1] Single element edge case
[1,1,1,1] All equal elements, testing marking correctness
[5,4,3,2,1] Descending order to test greedy selection
[1,2,3,4,5] Ascending order, ensure marking skips neighbors properly
[1,1000000,1,1000000,1] Large numbers with interspersed small numbers to check overflow and correctness

Edge Cases

Single Element: When the array has only one element, the algorithm correctly adds that element to the score and marks it, with no neighbors to consider.

All Elements Equal: When all elements are the same, only the left-most unmarked element contributes to the score in each iteration because neighbors are immediately marked, ensuring no double-counting occurs.

Adjacent Small Elements: If multiple small values are adjacent, marking neighbors may cause some of them to be skipped. The algorithm handles this because it checks marked[i] before adding the value to the score,