What is the time complexity of Last Visited Integers?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Last Visited Integers?

This problem uses Array, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Last Visited Integers?

Using a deque (or list with `insert(0, num)`) for `seen` efficiently handles prepending new positive integers. The variable `k` tracks consecutive -1s, resetting when a positive integer is encountered. This is crucial for correctly identifying the 'last visited' integer.

Which companies ask LeetCode 2899. Last Visited Integers in interviews?

This problem is frequently asked at General Motors.

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Last Visited Integers - LeetCode 2899 Solution

Last Visited Integers - Complete Solution Guide

Last Visited Integers is LeetCode problem 2899, 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

Given an integer array nums where nums[i] is either a positive integer or -1 . We need to find for each -1 the respective positive integer, which we call the last visited integer. To achieve this goal, let's define two empty arrays: seen and ans . Start iterating from the beginning of the array nums . If a positive integer is encountered, prepend it to the front of seen . If -1 is encountered, let k be the number of consecutive -1 s seen so far (including the current -1 ), If k is less than or e

Detailed Explanation

The problem asks to process an integer array `nums`. Each element is either a positive integer or -1. For each -1 encountered, we need to find the corresponding 'last visited' positive integer. We maintain a list `seen` storing positive integers encountered so far in reverse order of appearance (most recently seen at the front). When a -1 is encountered, we check how many consecutive -1s have been seen. If this count `k` is less than or equal to the number of elements in `seen`, the k-th element in `seen` is the 'last visited' integer. Otherwise, if `k` exceeds the size of `seen`, the 'last visited' is -1. The output is a list `ans` containing these 'last visited' integers.

Solution Approach

The solution uses a `seen` list to store the positive integers encountered so far in reverse order of their appearance. A `k` variable counts consecutive -1s. The algorithm iterates through `nums`. If a positive integer is found, it's added to the beginning of `seen`, and `k` is reset. If a -1 is found, `k` is incremented. The k-th element of `seen` (or -1 if k exceeds `seen`'s size) is appended to the `ans` list.

Step-by-Step Algorithm

Initialize an empty list `seen` to store positive integers in reverse order of appearance and an empty list `ans` to store the results.
Initialize `k` to 0 to count consecutive -1s.
Iterate through the input array `nums`:
If the current element `num` is positive, prepend it to `seen` and reset `k` to 0.
If `num` is -1, increment `k`. If `k` is less than or equal to the length of `seen`, append the `k`-th element of `seen` to `ans`. Otherwise, append -1 to `ans`.
Return the `ans` list.

Key Insights

Using a deque (or list with `insert(0, num)`) for `seen` efficiently handles prepending new positive integers.
The variable `k` tracks consecutive -1s, resetting when a positive integer is encountered. This is crucial for correctly identifying the 'last visited' integer.
Handling the case where `k` exceeds `seen`'s length (no 'last visited' positive integer within the window of consecutive -1s) requires appending -1 to the result.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: General Motors.

Last Visited Integers | LeetCode 2899

Easy

Array Simulation

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

class Solution {
    public List<Integer> lastVisitedIntegers(int[] nums) {
        List<Integer> ans = new ArrayList<>();
        LinkedList<Integer> seen = new LinkedList<>();
        int k = 0;
        for (int num : nums) {
            if (num > 0) {
                seen.addFirst(num);
                k = 0;
            } else {
                k++;
                if (k <= seen.size()) {
                    ans.add(seen.get(k - 1));
                } else {
                    ans.add(-1);
                }
            }
        }
        return ans;
    }
}

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Problem Info

Description

Problem 2899: Last Visited Integers - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

General Motors

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2899

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

6 steps

1
Initialize an empty list `seen` to store positive integers in reverse order of appearance and an empty list `ans` to store the results.
2
Initialize `k` to 0 to count consecutive -1s.
3
Iterate through the input array `nums`:
4
If the current element `num` is positive, prepend it to `seen` and reset `k` to 0.
5
If `num` is -1, increment `k`. If `k` is less than or equal to the length of `seen`, append the `k`-th element of `seen` to `ans`. Otherwise, append -1 to `ans`.
6
Return the `ans` list.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through `nums` once (O(n)). Prepending to a list (or using a deque) takes O(1) amortized time. Accessing the k-th element of `seen` is O(1). Therefore, the overall time complexity is O(n). The space used by `seen` and `ans` can be at most O(n) in the worst case where all elements are positive or all elements are -1 after a positive integer.

📖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 stack

Instead of a list, a stack can be used for `seen`. Pushing positive numbers onto the stack and popping when needed for -1. This approach would achieve the same functionality.

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

Trade-offs: The stack-based approach might offer slightly better performance for prepending elements, especially in languages where prepending to a list has O(n) time complexity in the worst case. However, the difference is usually negligible for typical input sizes.

✨

Key Insights

1Using a deque (or list with `insert(0, num)`) for `seen` efficiently handles prepending new positive integers.
2The variable `k` tracks consecutive -1s, resetting when a positive integer is encountered. This is crucial for correctly identifying the 'last visited' integer.
3Handling the case where `k` exceeds `seen`'s length (no 'last visited' positive integer within the window of consecutive -1s) requires appending -1 to the result.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Empty input array: The solution handles this correctly, returning an empty `ans` list.
All -1s: The solution correctly appends -1 to `ans` for each -1 since `k` will always be greater than `seen`'s length.
All positive integers: The solution correctly adds all positive integers to `seen` and `ans` will be empty.

⚠️

Common Mistakes to Avoid

✗Incorrectly handling the consecutive -1s count: Failing to reset `k` when a positive integer is encountered. This would lead to incorrect 'last visited' integers being identified.

✗Inefficient list manipulation: Using inefficient list operations (e.g., repeatedly inserting at the beginning) could lead to O(n^2) 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?

💼Tracking recently accessed data: Similar logic can be used in caching systems to track the most recently accessed items. Replacing the least recently used item when the cache is full.

🏦Network packet processing: Keeping a record of the last seen valid packets before dropping corrupted or invalid packets can use this concept.

💡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: June 20, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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.

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 "Last Visited Integers"?

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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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