LeetCode #2593 — MEDIUM

Find Score of an Array After Marking All Elements

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[2,1,3,4,5,2]

Example 2

[2,3,5,1,3,2]

Core Insight

What unlocks the optimal approach

Try simulating the process of marking the elements and their adjacent.
If there is an element that was already marked, then you skip it.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2593: Find Score of an Array After Marking All Elements
class Solution {
    public long findScore(int[] nums) {
        int n = nums.length;
        boolean[] vis = new boolean[n];
        PriorityQueue<int[]> q
            = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        for (int i = 0; i < n; ++i) {
            q.offer(new int[] {nums[i], i});
        }
        long ans = 0;
        while (!q.isEmpty()) {
            var p = q.poll();
            ans += p[0];
            vis[p[1]] = true;
            for (int j : List.of(p[1] - 1, p[1] + 1)) {
                if (j >= 0 && j < n) {
                    vis[j] = true;
                }
            }
            while (!q.isEmpty() && vis[q.peek()[1]]) {
                q.poll();
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2593: Find Score of an Array After Marking All Elements
func findScore(nums []int) (ans int64) {
	h := hp{}
	for i, x := range nums {
		heap.Push(&h, pair{x, i})
	}
	n := len(nums)
	vis := make([]bool, n)
	for len(h) > 0 {
		p := heap.Pop(&h).(pair)
		x, i := p.x, p.i
		ans += int64(x)
		vis[i] = true
		for _, j := range []int{i - 1, i + 1} {
			if j >= 0 && j < n {
				vis[j] = true
			}
		}
		for len(h) > 0 && vis[h[0].i] {
			heap.Pop(&h)
		}
	}
	return
}

type pair struct{ x, i int }
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].x < h[j].x || (h[i].x == h[j].x && h[i].i < h[j].i) }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)        { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any          { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

# Accepted solution for LeetCode #2593: Find Score of an Array After Marking All Elements
class Solution:
    def findScore(self, nums: List[int]) -> int:
        n = len(nums)
        vis = [False] * n
        q = [(x, i) for i, x in enumerate(nums)]
        heapify(q)
        ans = 0
        while q:
            x, i = heappop(q)
            ans += x
            vis[i] = True
            for j in (i - 1, i + 1):
                if 0 <= j < n:
                    vis[j] = True
            while q and vis[q[0][1]]:
                heappop(q)
        return ans

// Accepted solution for LeetCode #2593: Find Score of an Array After Marking All Elements
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2593: Find Score of an Array After Marking All Elements
// class Solution {
//     public long findScore(int[] nums) {
//         int n = nums.length;
//         boolean[] vis = new boolean[n];
//         PriorityQueue<int[]> q
//             = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
//         for (int i = 0; i < n; ++i) {
//             q.offer(new int[] {nums[i], i});
//         }
//         long ans = 0;
//         while (!q.isEmpty()) {
//             var p = q.poll();
//             ans += p[0];
//             vis[p[1]] = true;
//             for (int j : List.of(p[1] - 1, p[1] + 1)) {
//                 if (j >= 0 && j < n) {
//                     vis[j] = true;
//                 }
//             }
//             while (!q.isEmpty() && vis[q.peek()[1]]) {
//                 q.poll();
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2593: Find Score of an Array After Marking All Elements
interface pair {
    x: number;
    i: number;
}

function findScore(nums: number[]): number {
    const q = new PriorityQueue({
        compare: (a: pair, b: pair) => (a.x === b.x ? a.i - b.i : a.x - b.x),
    });
    const n = nums.length;
    const vis: boolean[] = new Array(n).fill(false);
    for (let i = 0; i < n; ++i) {
        q.enqueue({ x: nums[i], i });
    }
    let ans = 0;
    while (!q.isEmpty()) {
        const { x, i } = q.dequeue()!;
        if (vis[i]) {
            continue;
        }
        ans += x;
        vis[i] = true;
        if (i - 1 >= 0) {
            vis[i - 1] = true;
        }
        if (i + 1 < n) {
            vis[i + 1] = true;
        }
        while (!q.isEmpty() && vis[q.front()!.i]) {
            q.dequeue();
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.