What is the time complexity of Find Score of an Array After Marking All Elements?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Find Score of an Array After Marking All Elements?

This problem uses Array, Hash Table, Sorting, Heap (Priority Queue), Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find Score of an Array After Marking All Elements?

Insight 1: We need to keep track of which elements are marked. A boolean array is a good way to do this. Insight 2: To efficiently find the smallest unmarked element, sorting the array along with indices is crucial. This allows us to iterate through the sorted array and quickly identify the smallest unmarked element.

Which companies ask LeetCode 2593. Find Score of an Array After Marking All Elements in interviews?

This problem is frequently asked at Visa.

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

Find Score of an Array After Marking All Elements - Complete Solution Guide

Find Score of an Array After Marking All Elements is LeetCode problem 2593, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to calculate a 'score' from an array of positive integers. We start with a score of 0. We repeatedly find the smallest *unmarked* integer in the array, add its value to the score, and then mark that element, along with its immediate neighbors (if they exist) as 'marked'. If there's a tie for the smallest unmarked integer, we choose the one with the smallest index. We continue this process until all elements in the array are marked. The final score is the value we return.

Solution Approach

The solution involves sorting the input array 'nums' along with their original indices. This sorted list allows us to iterate through the numbers in ascending order while still knowing their original positions. A boolean array, 'marked', is used to track which elements have been marked. We iterate through the sorted (value, index) pairs. If an element at a particular index is not marked, we add its value to the 'score', mark the element itself and its adjacent elements (if they exist) as marked. This continues until all elements are processed. The final 'score' is then returned.

Step-by-Step Algorithm

Step 1: Create a list of pairs/tuples where each pair consists of (value, index) from the original input array 'nums'.
Step 2: Sort this list of pairs based on the value in ascending order. If there is a tie, the pair with the smaller index should come first. Most sorting algorithms maintain the original order if elements are the same.
Step 3: Initialize a boolean array 'marked' of the same size as 'nums', initially filled with 'false' (indicating that no elements are initially marked).
Step 4: Initialize 'score' to 0.
Step 5: Iterate through the sorted list of (value, index) pairs.
Step 6: In each iteration, check if the element at the current 'index' is marked (i.e., marked[index] is true).
Step 7: If marked[index] is false (meaning the element is not marked), perform the following:
Step 8: Add the 'value' of the element to the 'score'.
Step 9: Mark the current element as marked (marked[index] = true).
Step 10: Mark the element to the left of the current element as marked if it exists (i.e., if index > 0, then marked[index - 1] = true).
Step 11: Mark the element to the right of the current element as marked if it exists (i.e., if index < n - 1, then marked[index + 1] = true).
Step 12: After iterating through all the pairs, return the final 'score'.

Key Insights

Insight 1: We need to keep track of which elements are marked. A boolean array is a good way to do this.
Insight 2: To efficiently find the smallest unmarked element, sorting the array along with indices is crucial. This allows us to iterate through the sorted array and quickly identify the smallest unmarked element.
Insight 3: The index information needs to be preserved during sorting, as ties are broken by the index. Using a pair (or similar construct) to store both the value and the index is vital.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sorting, Heap (Priority Queue), Simulation.

Companies

Asked at: Visa.

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

Medium

Array Hash Table Sorting Heap (Priority Queue)Simulation

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public long findScore(int[] nums) {
        int n = nums.length;
        
        Integer[] indices = new Integer[n];
        for (int i = 0; i < n; i++) {
            indices[i] = i;
        }

        Arrays.sort(indices, Comparator.comparingInt(i -> nums[i]));
        
        boolean[] marked = new boolean[n];
        long score = 0;
        
        for (int i = 0; i < n; i++) {
            int index = indices[i];
            if (!marked[index]) {
                score += nums[index];
                marked[index] = true;
                
                if (index > 0) {
                    marked[index - 1] = true;
                }
                
                if (index < n - 1) {
                    marked[index + 1] = true;
                }
            }
        }
        
        return score;
    }
}

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Problem Info

Description

Problem 2593: Find Score of an Array After Marking All Elements - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Heap (Priority Queue)Simulation

Companies

Visa

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2593

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Create a list of pairs/tuples where each pair consists of (value, index) from the original input array 'nums'.
2
Step 2: Sort this list of pairs based on the value in ascending order. If there is a tie, the pair with the smaller index should come first. Most sorting algorithms maintain the original order if elements are the same.
3
Step 3: Initialize a boolean array 'marked' of the same size as 'nums', initially filled with 'false' (indicating that no elements are initially marked).
4
Step 4: Initialize 'score' to 0.
5
Step 5: Iterate through the sorted list of (value, index) pairs.
6
Step 6: In each iteration, check if the element at the current 'index' is marked (i.e., marked[index] is true).
7
Step 7: If marked[index] is false (meaning the element is not marked), perform the following:
8
Step 8: Add the 'value' of the element to the 'score'.
9
Step 9: Mark the current element as marked (marked[index] = true).
10
Step 10: Mark the element to the left of the current element as marked if it exists (i.e., if index > 0, then marked[index - 1] = true).
11
Step 11: Mark the element to the right of the current element as marked if it exists (i.e., if index < n - 1, then marked[index + 1] = true).
12
Step 12: After iterating through all the pairs, return the final 'score'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sorting step. The space complexity is determined by the additional data structures used to store the indexed numbers and the marked flags.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Heap (Priority Queue)

Instead of sorting, we can use a min-heap (priority queue) to store the (value, index) pairs. This allows us to retrieve the smallest unmarked element in O(log n) time. We iterate through the heap, marking elements and adding to the score as needed.

Time: O(n log n)Space: O(n)

Trade-offs: The heap-based approach has the same time complexity as the sorting-based approach (O(n log n)). The main advantage of using a heap is when the input array is very large and only a small fraction of elements need to be processed. However, in this problem, we need to process almost all elements, so the sorting-based approach is generally preferred due to its simplicity and lower constant factors.

✨

Key Insights

1Insight 1: We need to keep track of which elements are marked. A boolean array is a good way to do this.
2Insight 2: To efficiently find the smallest unmarked element, sorting the array along with indices is crucial. This allows us to iterate through the sorted array and quickly identify the smallest unmarked element.
3Insight 3: The index information needs to be preserved during sorting, as ties are broken by the index. Using a pair (or similar construct) to store both the value and the index is vital.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty array: The code should handle an empty array gracefully. In this implementation, an empty array would result in an empty indexed_nums, and the loop would not execute, correctly returning a score of 0.
Edge case 2: Array with a single element: The code correctly handles this case because after processing the single element, there are no adjacent elements to mark, and the loop finishes.
Edge case 3: Array with duplicate values: The sorting algorithm ensures that if there are ties for the smallest value, the element with the smaller index will be chosen first, as per the problem requirements.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not preserving the original indices during sorting: If you only sort the numbers themselves, you'll lose the information about their original positions, making it difficult to break ties correctly.

✗Common mistake 2: Incorrectly handling the marking of adjacent elements: It's important to check the boundaries (i.e., whether the current element is at the beginning or end of the array) before attempting to mark adjacent elements to avoid out-of-bounds errors.

✗Common mistake 3: Not initializing the 'marked' array correctly: If the 'marked' array isn't properly initialized to 'false' for all elements, you might incorrectly skip elements or add them to the score multiple times.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation with dependencies: Imagine allocating resources to tasks, where selecting a task also requires you to allocate resources to dependent tasks nearby in a queue. The problem is finding the most efficient set of tasks to maximize a score based on resource usage, while accounting for these dependencies.

🏦Real-world scenario 2: Selecting locations for services along a street: Suppose you're placing services (e.g., charging stations for electric cars) along a road. Placing a service at a location may also require placing them nearby, due to infrastructure limitations. The problem becomes selecting the optimal locations to maximize coverage (score), considering these spatial constraints.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find Score of an Array After Marking All Elements"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?