What is the time complexity of Mark Elements on Array by Performing Queries?

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

What data structures are used to solve Mark Elements on Array by Performing Queries?

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 Mark Elements on Array by Performing Queries?

Insight 1: Sorting the array is crucial for efficiently finding the smallest unmarked elements. Insight 2: Using a boolean array (or similar) to track marked elements avoids repeated calculations and ensures elements are only marked once.

Which companies ask LeetCode 3080. Mark Elements on Array by Performing Queries in interviews?

This problem is frequently asked at Barclays, Samsung.

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Mark Elements on Array by Performing Queries - LeetCode 3080 Solution

Mark Elements on Array by Performing Queries - Complete Solution Guide

Mark Elements on Array by Performing Queries is LeetCode problem 3080, 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 a 0-indexed array nums of size n consisting of positive integers. You are also given a 2D array queries of size m where queries[i] = [index i , k i ] . Initially all elements of the array are unmarked . You need to apply m queries on the array in order, where on the i th query you do the following: Mark the element at index index i if it is not already marked. Then mark k i unmarked elements in the array with the smallest values. If multiple such elements exist, mark the ones with

Detailed Explanation

The problem requires us to simulate marking elements in an array based on a series of queries. Each query consists of an index and a count 'k'. First, if the element at the given index is unmarked, we mark it. Then, we need to find 'k' unmarked elements with the smallest values and mark them as well. If there are fewer than 'k' unmarked elements, we mark all remaining unmarked elements. The goal is to return an array where each element represents the sum of all unmarked elements in the array after processing each query.

Solution Approach

The provided solution uses a straightforward simulation approach. First, it calculates the initial sum of all elements in the array. Then, it creates a boolean array to keep track of which elements are marked. The core idea is to sort the array elements along with their indices to quickly find the smallest unmarked values. For each query, it first marks the element at the given index (if it's not already marked) and updates the total sum. Then, it iterates through the sorted elements and marks the 'k' smallest unmarked elements, updating the total sum accordingly. Finally, the current total sum is appended to the answer array.

Step-by-Step Algorithm

Step 1: Calculate the initial sum of all elements in the `nums` array.
Step 2: Create a boolean array `marked` of the same size as `nums`, initialized to `false`, to track marked elements.
Step 3: Create a sorted list (or array) of pairs, where each pair contains an element from `nums` and its original index.
Step 4: Iterate through the `queries` array.
Step 5: For each query `(index, k)`, check if `marked[index]` is `false`. If it is, mark the element at `index` by setting `marked[index] = true` and subtract `nums[index]` from the `total_sum`.
Step 6: Iterate through the sorted list of pairs, marking the `k` smallest unmarked elements. Use a `ptr` to keep track of the current position in the sorted list. For each unmarked element encountered, mark it, subtract its value from `total_sum`, and increment a `marked_count`.
Step 7: Append the current `total_sum` to the answer array.
Step 8: Return the answer array.

Key Insights

Insight 1: Sorting the array is crucial for efficiently finding the smallest unmarked elements.
Insight 2: Using a boolean array (or similar) to track marked elements avoids repeated calculations and ensures elements are only marked once.
Insight 3: The total sum needs to be maintained and updated as elements are marked to efficiently calculate the sum of unmarked elements after each query.

Complexity Analysis

Time Complexity: O(q*n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Barclays, Samsung.

Mark Elements on Array by Performing Queries | LeetCode 3080

Medium

Array Hash Table Sorting Heap (Priority Queue)Simulation

Time: O(q*n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public long[] unmarkedSumArray(int[] nums, int[][] queries) {
        int n = nums.length;
        long total_sum = 0;
        for (int num : nums) {
            total_sum += num;
        }
        boolean[] marked = new boolean[n];

        List<Pair> sorted_elements = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            sorted_elements.add(new Pair(nums[i], i));
        }
        sorted_elements.sort((a, b) -> Integer.compare(a.value, b.value));

        int ptr = 0;
        List<Long> answerList = new ArrayList<>();

        for (int[] query : queries) {
            int index = query[0];
            int k = query[1];

            if (!marked[index]) {
                marked[index] = true;
                total_sum -= nums[index];
            }

            int marked_count = 0;
            while (marked_count < k && ptr < n) {
                Pair pair = sorted_elements.get(ptr);
                int value = pair.value;
                int original_index = pair.index;

                if (!marked[original_index]) {
                    marked[original_index] = true;
                    total_sum -= value;
                    marked_count += 1;
                }

                ptr += 1;
            }

            answerList.add(total_sum);
        }

        long[] answer = new long[answerList.size()];
        for (int i = 0; i < answerList.size(); i++) {
            answer[i] = answerList.get(i);
        }

        return answer;
    }

    static class Pair {
        int value;
        int index;

        public Pair(int value, int index) {
            this.value = value;
            this.index = index;
        }
    }
}

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Problem Info

Description

Problem 3080: Mark Elements on Array by Performing Queries - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Heap (Priority Queue)Simulation

Companies

BarclaysSamsung

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3080

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

8 steps

1
Step 1: Calculate the initial sum of all elements in the `nums` array.
2
Step 2: Create a boolean array `marked` of the same size as `nums`, initialized to `false`, to track marked elements.
3
Step 3: Create a sorted list (or array) of pairs, where each pair contains an element from `nums` and its original index.
4
Step 4: Iterate through the `queries` array.
5
Step 5: For each query `(index, k)`, check if `marked[index]` is `false`. If it is, mark the element at `index` by setting `marked[index] = true` and subtract `nums[index]` from the `total_sum`.
6
Step 6: Iterate through the sorted list of pairs, marking the `k` smallest unmarked elements. Use a `ptr` to keep track of the current position in the sorted list. For each unmarked element encountered, mark it, subtract its value from `total_sum`, and increment a `marked_count`.
7
Step 7: Append the current `total_sum` to the answer array.
8
Step 8: Return the answer array.

💡

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(q*n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `n log n` comes from sorting the `nums` array. The outer loop runs `m` times. The inner loop can, in the worst case, iterate through all `n` elements of the sorted list to find `k` unmarked elements. The auxiliary space is dominated by the `marked` array and the sorted list.

📖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

Using a Heap (Priority Queue)

Instead of sorting the entire array initially, you could use a min-heap to store the unmarked elements. When marking 'k' elements, you can extract the minimum 'k' elements from the heap.

Time: O(n + m * (k log n + n)), where n is the length of the nums array, m is the number of queries, and k is the maximum value of k_i in queries. In the beginning we create heap of size n, it takes O(n) time. Also, for each query, extracting k min elements takes O(k log n), updating the marked array and calculating sum takes O(n).Space: O(n), for storing heap.

Trade-offs: The heap-based approach can be more efficient if 'k' is much smaller than 'n' for most queries. However, if 'k' is close to 'n' for many queries, the initial sorting approach can be faster due to avoiding repeated heap operations.

✨

Key Insights

1Insight 1: Sorting the array is crucial for efficiently finding the smallest unmarked elements.
2Insight 2: Using a boolean array (or similar) to track marked elements avoids repeated calculations and ensures elements are only marked once.
3Insight 3: The total sum needs to be maintained and updated as elements are marked to efficiently calculate the sum of unmarked elements after each query.

📚

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: `k` is larger than the number of unmarked elements. The code correctly handles this by marking all remaining unmarked elements and stopping the inner loop when `ptr` reaches the end of the array.
Edge case 2: The element at `index` in the query is already marked. The code correctly checks this condition using the `marked` array and skips marking the element if it's already marked, ensuring the correct `total_sum` is maintained.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not properly updating the `total_sum` after marking an element. This will lead to incorrect results.

✗Common mistake 2: Not considering the case where `k` is larger than the number of unmarked elements. The code needs to handle this gracefully by marking all remaining unmarked elements.

✗Common mistake 3: Incorrectly using the index while marking. One might use the index of the sorted array for marking instead of the original index of the number in the input nums array. It's important to use the original index to mark the correct elements.

✅

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. Imagine you have a list of tasks with different costs, and you need to prioritize and allocate resources to the least costly tasks first based on several requests (queries) from different departments. This problem's logic can be adapted to efficiently track allocated resources and available capacity.

🏦Real-world scenario 2: Online advertising. Consider a system that displays ads based on a bidding system. Each ad has a bid price, and the system needs to select the 'k' highest-bidding ads to display based on various criteria (queries). This problem's core algorithm can be modified to track which ads have been displayed and prioritize displaying the highest-bidding ads that haven't been shown recently.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: June 21, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Mark Elements on Array by Performing Queries"?

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.

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