Leetcode #2593: Find Score of an Array After Marking All Elements
In this guide, we solve Leetcode #2593 Find Score of an Array After Marking All Elements in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Sorting, Simulation, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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
Complexity
The time complexity is and the space complexity is , where is the length of the array. The space complexity is , where is the length of the array.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.