What is the time complexity of Minimum Operations to Collect Elements?

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

What data structures are used to solve Minimum Operations to Collect Elements?

This problem uses Array, Hash Table, Bit Manipulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Operations to Collect Elements?

Insight 1: Processing the array from the end is crucial because each operation removes the last element. We are interested in the minimum number of operations, so we should start checking from the end. Insight 2: Using a set (or similar data structure like a boolean array) to track collected numbers efficiently checks if a number has already been collected. This avoids redundant checks.

Which companies ask LeetCode 2869. Minimum Operations to Collect Elements in interviews?

This problem is frequently asked at Deutsche Bank.

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Minimum Operations to Collect Elements - LeetCode 2869 Solution

Minimum Operations to Collect Elements - Complete Solution Guide

Minimum Operations to Collect Elements is LeetCode problem 2869, a Easy 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 of positive integers and an integer k . In one operation, you can remove the last element of the array and add it to your collection. Return the minimum number of operations needed to collect elements 1, 2, ..., k . Example 1: Input: nums = [3,1,5,4,2], k = 2 Output: 4 Explanation: After 4 operations, we collect elements 2, 4, 5, and 1, in this order. Our collection contains elements 1 and 2. Hence, the answer is 4. Example 2: Input: nums = [3,1,5,4,2], k = 5 Output:

Detailed Explanation

The problem asks you to find the minimum number of operations required to collect a set of consecutive integers (from 1 to k) from the end of an array `nums`. In each operation, you remove the last element of the array and add it to your collection. The goal is to determine how many elements you need to process (remove from the end) until your collection contains all integers from 1 to k.

Solution Approach

The solution iterates through the input array `nums` from the end. It maintains a set (`needed` in Python, `collected` in Java, `found` in C++) to keep track of the numbers (from 1 to k) that have not yet been collected. For each element, it checks if the element is among the numbers needed. If it is, the element is marked as collected. The process continues until all numbers from 1 to k are collected. The number of iterations (operations) needed is the solution.

Step-by-Step Algorithm

Step 1: Create a set (`needed`) containing integers from 1 to k. This set will store the numbers we still need to collect.
Step 2: Initialize a `count` variable to 0. This will track the number of operations performed.
Step 3: Iterate through the `nums` array from the end using a for loop.
Step 4: Increment `count` for each iteration (operation).
Step 5: Check if the current element is present in the `needed` set. If it is, remove it from the set.
Step 6: Check if the `needed` set is empty. If it is, all numbers from 1 to k have been collected; return `count`.
Step 7: After the loop, return `count`. All numbers should have been collected by then (guaranteed by problem statement).

Key Insights

Insight 1: Processing the array from the end is crucial because each operation removes the last element. We are interested in the minimum number of operations, so we should start checking from the end.
Insight 2: Using a set (or similar data structure like a boolean array) to track collected numbers efficiently checks if a number has already been collected. This avoids redundant checks.
Insight 3: The problem guarantees that it's always possible to collect the numbers from 1 to k. This simplifies the solution as we don't need to handle cases where it's not possible.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Hash Table, Bit Manipulation.

Companies

Asked at: Deutsche Bank.

Minimum Operations to Collect Elements | LeetCode 2869

Easy

Array Hash Table Bit Manipulation

Time: O(n)Space: O(k)

Solution Code

Available in:

import java.util.List;
import java.util.HashSet;
import java.util.Set;

class Solution {
    public int minOperations(List<Integer> nums, int k) {
        Set<Integer> collected = new HashSet<>();
        int operations = 0;
        for (int i = nums.size() - 1; i >= 0; i--) {
            operations++;
            int num = nums.get(i);
            if (num >= 1 && num <= k) {
                collected.add(num);
            }
            if (collected.size() == k) {
                return operations;
            }
        }
        return operations;
    }
}

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Problem Info

Description

Problem 2869: Minimum Operations to Collect Elements - Detailed explanation and solution approaches.

Topics

Array Hash Table Bit Manipulation

Companies

Deutsche Bank

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2869

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

7 steps

1
Step 1: Create a set (`needed`) containing integers from 1 to k. This set will store the numbers we still need to collect.
2
Step 2: Initialize a `count` variable to 0. This will track the number of operations performed.
3
Step 3: Iterate through the `nums` array from the end using a for loop.
4
Step 4: Increment `count` for each iteration (operation).
5
Step 5: Check if the current element is present in the `needed` set. If it is, remove it from the set.
6
Step 6: Check if the `needed` set is empty. If it is, all numbers from 1 to k have been collected; return `count`.
7
Step 7: After the loop, return `count`. All numbers should have been collected by then (guaranteed by problem statement).

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(k)

📝

Why?

The time complexity is O(n) because we iterate through the array once. The space complexity is O(k) because we store at most k elements in the `needed` set (or equivalent structure).

📖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 boolean array

Instead of a set, use a boolean array of size k+1 to track whether each number from 1 to k has been collected.

Time: O(n)Space: O(k)

Trade-offs: The space complexity is similar. A boolean array might be slightly faster in some implementations than a set due to simpler data structure overhead.

✨

Key Insights

1Insight 1: Processing the array from the end is crucial because each operation removes the last element. We are interested in the minimum number of operations, so we should start checking from the end.
2Insight 2: Using a set (or similar data structure like a boolean array) to track collected numbers efficiently checks if a number has already been collected. This avoids redundant checks.
3Insight 3: The problem guarantees that it's always possible to collect the numbers from 1 to k. This simplifies the solution as we don't need to handle cases where it's not possible.

📚

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!

Edge Cases

Edge case 1: `k` is equal to `nums.length`. The algorithm correctly handles this case, iterating through the whole array.
Edge case 2: All numbers from 1 to k are at the beginning of the array. The algorithm still correctly identifies that it needs to iterate through the whole array to collect them.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Iterating from the beginning of the array instead of the end. This would not find the minimum number of operations.

✗Common mistake 2: Not efficiently tracking which numbers have been collected. Using a linear search for each number would lead to O(n*k) time complexity instead of O(n).

✅

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: Imagine a system monitoring sensor readings. You need to collect at least the last k readings (e.g., temperature values) to determine a stable average. This algorithm efficiently determines how many readings to process to obtain those k readings.

🏦Real-world scenario 2: In data logging, you might need to collect the most recent complete set of specific data points (e.g., the last k transaction IDs). This algorithm helps in determining the amount of data to process for retrieval.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: November 17, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Minimum Operations to Collect Elements"?

For this easy-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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