What is the time complexity of First Missing Positive?

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

What data structures are used to solve First Missing Positive?

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

What are the key insights for solving First Missing Positive?

Insight 1: The missing positive number must be between 1 and n+1, where n is the length of the array. This is because if all numbers from 1 to n are present, the answer is n+1. Insight 2: We can use the array itself as a hash table by rearranging the elements such that `nums[i]` should ideally be `i+1`. This allows us to check for the missing number in O(n) time without using extra space.

Which companies ask LeetCode 41. First Missing Positive in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Celigo and 35 more companies.

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First Missing Positive - LeetCode 41 Solution

First Missing Positive - Complete Solution Guide

First Missing Positive is LeetCode problem 41, a Hard 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 unsorted integer array nums . Return the smallest positive integer that is not present in nums . You must implement an algorithm that runs in O(n) time and uses O(1) auxiliary space. Example 1: Input: nums = [1,2,0] Output: 3 Explanation: The numbers in the range [1,2] are all in the array. Example 2: Input: nums = [3,4,-1,1] Output: 2 Explanation: 1 is in the array but 2 is missing. Example 3: Input: nums = [7,8,9,11,12] Output: 1 Explanation: The smallest positive integer 1 is missing

Detailed Explanation

The problem asks us to find the smallest positive integer that is missing from a given unsorted integer array `nums`. The solution must have a time complexity of O(n) and a space complexity of O(1). In simpler terms, given an array like [3, 4, -1, 1], we need to find the smallest positive number (1, 2, 3, ...) that's not in the array. In this case, it would be 2.

Solution Approach

The solution uses the array itself as a hash table. It iterates through the array and, for each element `nums[i]`, it attempts to place it at its correct index `nums[i] - 1`. If `nums[i]` is within the range [1, n] and `nums[i]` is not already at its correct position (i.e., `nums[i] != nums[nums[i] - 1]`), we swap `nums[i]` with the element at its correct index. This process is repeated until all elements are either in their correct positions or are out of range. Finally, we iterate through the rearranged array and return the first index `i` where `nums[i] != i + 1`. If no such index is found, it means all numbers from 1 to n are present, so the answer is n + 1.

Step-by-Step Algorithm

Step 1: Iterate through the input array `nums` from index 0 to n-1.
Step 2: For each element `nums[i]`, check if it falls within the range [1, n]. Also, check if `nums[i]` is already at its correct position i.e., check if `nums[i] == nums[nums[i] - 1]`.
Step 3: If `nums[i]` satisfies the above conditions, swap `nums[i]` with the element at index `nums[i] - 1`. This places `nums[i]` at its correct position.
Step 4: Repeat steps 2 and 3 until the current element `nums[i]` is either out of the valid range or is already at its correct position or it's equal to the value at its correct index.
Step 5: Increment `i` to move to the next element in the array.
Step 6: After the rearrangement, iterate through the array again from index 0 to n-1.
Step 7: For each element `nums[i]`, check if `nums[i]` is equal to `i + 1`. If not, it means the number `i + 1` is missing, so return `i + 1`.
Step 8: If the loop completes without finding a missing number, it means all numbers from 1 to n are present in the array. In this case, return `n + 1`.

Key Insights

Insight 1: The missing positive number must be between 1 and n+1, where n is the length of the array. This is because if all numbers from 1 to n are present, the answer is n+1.
Insight 2: We can use the array itself as a hash table by rearranging the elements such that `nums[i]` should ideally be `i+1`. This allows us to check for the missing number in O(n) time without using extra space.
Insight 3: Pay attention to boundary conditions and avoid infinite loops during the rearrangement process. Ensure that only numbers within the range [1, n] are moved to their correct positions.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Celigo, Databricks, EPAM Systems, Flipkart, Geico, General Motors, Goldman Sachs, Google, Licious, MakeMyTrip, Meta, Microsoft, Morgan Stanley, Myntra, Netflix, Nutanix, Oracle, PayPal, PhonePe, Roblox, Salesforce, Samsung, ServiceNow, Siemens, SoundHound, Sprinklr, Sumo Logic, Swiggy, Tesla, TikTok, Twilio, Uber, Walmart Labs, Zomato, athenahealth, eBay.

First Missing Positive | LeetCode 41

Hard

Array Hash Table

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int firstMissingPositive(int[] nums) {
        int n = nums.length;
        int i = 0;
        while (i < n) {
            int correct_idx = nums[i] - 1;
            if (1 <= nums[i] && nums[i] <= n && nums[i] != nums[correct_idx]) {
                int temp = nums[i];
                nums[i] = nums[correct_idx];
                nums[correct_idx] = temp;
            } else {
                i++;
            }
        }
        
        for (i = 0; i < n; i++) {
            if (nums[i] != i + 1) {
                return i + 1;
            }
        }
        
        return n + 1;
    }
}

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Problem Info

Description

Problem 41: First Missing Positive - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

AdobeAmazonAppleBloombergCeligoDatabricksEPAM SystemsFlipkart

Stats

Difficulty:Hard

Languages:4

Companies:40

Problem #:41

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: Iterate through the input array `nums` from index 0 to n-1.
2
Step 2: For each element `nums[i]`, check if it falls within the range [1, n]. Also, check if `nums[i]` is already at its correct position i.e., check if `nums[i] == nums[nums[i] - 1]`.
3
Step 3: If `nums[i]` satisfies the above conditions, swap `nums[i]` with the element at index `nums[i] - 1`. This places `nums[i]` at its correct position.
4
Step 4: Repeat steps 2 and 3 until the current element `nums[i]` is either out of the valid range or is already at its correct position or it's equal to the value at its correct index.
5
Step 5: Increment `i` to move to the next element in the array.
6
Step 6: After the rearrangement, iterate through the array again from index 0 to n-1.
7
Step 7: For each element `nums[i]`, check if `nums[i]` is equal to `i + 1`. If not, it means the number `i + 1` is missing, so return `i + 1`.
8
Step 8: If the loop completes without finding a missing number, it means all numbers from 1 to n are present in the array. In this case, return `n + 1`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because each element is visited and potentially swapped a constant number of times. The outer loop runs 'n' times. The inner while loop does not necessarily run 'n' times in total, but since each element is only swapped a maximum of once, the total iterations of the inner loop across all iterations of the outer loop is bounded by 'n'. The space complexity is O(1) because the solution modifies the input array in-place and uses a fixed number of extra variables.

📖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 Hash Set

Store all the numbers in a hash set. Iterate from 1 upwards and check if each number is present in the set. The first number not present is the answer.

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

Trade-offs: This approach is simpler to implement but requires O(n) extra space for the hash set. It's suitable when space complexity is not a major concern.

Sorting the array

Sort the array and then iterate through the sorted array to find the first missing positive integer.

Time: O(n log n)Space: O(1) or O(n) depending on the sorting algorithm

Trade-offs: This is less efficient than the in-place rearrangement as most efficient sorting algorithms have time complexity of O(n log n). Some sorting algorithms may use extra space too.

✨

Key Insights

1Insight 1: The missing positive number must be between 1 and n+1, where n is the length of the array. This is because if all numbers from 1 to n are present, the answer is n+1.
2Insight 2: We can use the array itself as a hash table by rearranging the elements such that `nums[i]` should ideally be `i+1`. This allows us to check for the missing number in O(n) time without using extra space.
3Insight 3: Pay attention to boundary conditions and avoid infinite loops during the rearrangement process. Ensure that only numbers within the range [1, n] are moved to their correct positions.

📚

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: Empty array: The code works correctly for an empty array. In this case, n will be 0, and the algorithm will immediately return 1, which is the correct answer.
Edge case 2: Array with only negative numbers or zeros: The algorithm will correctly rearrange the array, and in the second loop, it will find that `nums[0]` is not equal to 1 and returns 1.
Edge case 3: Array with duplicate positive numbers. The algorithm still works, placing numbers into their correct positions as much as possible. The duplicates that can't be placed won't affect the final result.
Edge case 4: Array with large positive numbers. These will be ignored in the rearrangement phase, and will not affect the result since the first missing positive is guaranteed to be less than or equal to n+1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case when all numbers from 1 to n are present in the array. The algorithm should return n + 1 in this case.

✗Common mistake 2: Infinite loop during the rearrangement process due to not checking for `nums[i] != nums[correct_idx]` before swapping. If this condition is not checked, the algorithm can get stuck in an infinite loop if `nums[i]` is already in the correct position.

✗Common mistake 3: Forgetting to check for the valid range `1 <= nums[i] <= n` during the rearrangement process. Swapping elements that are outside this range will corrupt the array and lead to incorrect results.

✅

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: Finding the smallest available resource ID in a system where resource IDs are positive integers.

🏦Real-world scenario 2: Data integrity checks: Identifying the first missing sequence number in a stream of data packets.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: July 12, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

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

@CodeNinjaJun 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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "First Missing Positive"?

For this hard-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?