What is the time complexity of Set Mismatch?

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

What data structures are used to solve Set Mismatch?

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

What are the key insights for solving Set Mismatch?

Insight 1: The expected sum of numbers from 1 to n is n * (n + 1) / 2. We can use this to calculate the missing number if we know the duplicate. Insight 2: We need to efficiently identify the duplicate number. A hash set or frequency counting array can help with this in O(n) time.

Which companies ask LeetCode 645. Set Mismatch in interviews?

This problem is frequently asked at Grammarly.

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Set Mismatch - LeetCode 645 Solution

Set Mismatch - Complete Solution Guide

Set Mismatch is LeetCode problem 645, 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 have a set of integers s , which originally contains all the numbers from 1 to n . Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number. You are given an integer array nums representing the data status of this set after the error. Find the number that occurs twice and the number that is missing and return them in the form of an array . Example 1: Input: nums = [1,2,2,4] Outp

Detailed Explanation

The problem states that we're given an array `nums` which is supposed to contain numbers from 1 to `n` (where `n` is the length of the array). However, one number is duplicated, and another number is missing. Our task is to find both the duplicated number and the missing number and return them in an array.

Solution Approach

The provided solutions use the following approach: First, they iterate through the input array to find the duplicate number. This is done using either a hash set or a frequency counting array. Simultaneously, they calculate the sum of all elements in the input array. Finally, they compute the expected sum of numbers from 1 to `n` and use the duplicate and the actual sum to deduce the missing number.

Step-by-Step Algorithm

Step 1: Initialize variables: `n` (length of the array), `duplicate` (to store the duplicated number, initialized to -1 or 0), `sum_nums` (to store the sum of numbers in the array, initialized to 0), and `expected_sum` (calculated as n * (n + 1) // 2). Also, initialize either a hash set or a frequency counter array.
Step 2: Iterate through the `nums` array. For each `num`:
Step 3: Add `num` to `sum_nums`.
Step 4: If using a hash set, check if `num` is already present in the set. If it is, then `num` is the duplicate. If using a frequency counter, increment the count for the number. If the count becomes 2, then the number is duplicate.
Step 5: If using a hash set, add `num` to the set.
Step 6: Calculate the missing number: `missing = expected_sum - sum_nums + duplicate`.
Step 7: Return an array containing the duplicate and the missing numbers in that order.

Key Insights

Insight 1: The expected sum of numbers from 1 to n is n * (n + 1) / 2. We can use this to calculate the missing number if we know the duplicate.
Insight 2: We need to efficiently identify the duplicate number. A hash set or frequency counting array can help with this in O(n) time.
Insight 3: Once we find the duplicate and the actual sum of the input numbers, we can deduce the missing number by comparing with the expected sum.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Grammarly.

Set Mismatch | LeetCode 645

Easy

Array Hash Table Bit Manipulation Sorting

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

Solution Code

Available in:

class Solution {
    public int[] findErrorNums(int[] nums) {
        int n = nums.length;
        int[] count = new int[n + 1];
        int duplicate = 0;
        int missing = 0;
        long sum = 0;
        long expectedSum = (long) n * (n + 1) / 2;

        for (int num : nums) {
            count[num]++;
            sum += num;
        }

        for (int i = 1; i <= n; i++) {
            if (count[i] == 2) {
                duplicate = i;
            }
            if (count[i] == 0) {
                missing = i;
            }
        }
        
        return new int[]{duplicate, missing};
    }
}

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Problem Info

Description

Problem 645: Set Mismatch - Detailed explanation and solution approaches.

Topics

Array Hash Table Bit Manipulation Sorting

Companies

Grammarly

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:645

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: Initialize variables: `n` (length of the array), `duplicate` (to store the duplicated number, initialized to -1 or 0), `sum_nums` (to store the sum of numbers in the array, initialized to 0), and `expected_sum` (calculated as n * (n + 1) // 2). Also, initialize either a hash set or a frequency counter array.
2
Step 2: Iterate through the `nums` array. For each `num`:
3
Step 3: Add `num` to `sum_nums`.
4
Step 4: If using a hash set, check if `num` is already present in the set. If it is, then `num` is the duplicate. If using a frequency counter, increment the count for the number. If the count becomes 2, then the number is duplicate.
5
Step 5: If using a hash set, add `num` to the set.
6
Step 6: Calculate the missing number: `missing = expected_sum - sum_nums + duplicate`.
7
Step 7: Return an array containing the duplicate and the missing numbers in that order.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the loop that iterates through the array. The space complexity depends on the auxiliary data structure (hash set or frequency array) used to find the duplicate. The calculation of expected_sum and missing takes constant time and does not affect the overall complexity.

📖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

Bit Manipulation

We can use bit manipulation (XOR) to solve this problem in O(n) time and O(1) space. XOR all numbers from 1 to n with the numbers in the array. The result will be the XOR of the missing and the duplicate. We can then find the rightmost set bit, which will be different in the missing and duplicate. Group numbers based on whether that bit is set, then find the missing and duplicate from these two groups.

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

Trade-offs: More complex to implement and understand, but has better space complexity.

Sorting

Sort the array. Then, iterate through the sorted array. The duplicate will be adjacent elements with the same value. Once the duplicate is found, iterate again to find the missing element by comparing the expected value at each position (1, 2, 3, etc.) with the actual value. If they are different, and the actual value is not the duplicate, then you found the missing.

Time: O(n log n) due to sorting.Space: O(1) if sorting is done in-place; otherwise, O(n) for the sorted copy.

Trade-offs: Sorting takes O(n log n) time, which is slower than the hash set or frequency counting approach. However, it potentially uses less space if in-place sorting is available.

✨

Key Insights

1Insight 1: The expected sum of numbers from 1 to n is n * (n + 1) / 2. We can use this to calculate the missing number if we know the duplicate.
2Insight 2: We need to efficiently identify the duplicate number. A hash set or frequency counting array can help with this in O(n) time.
3Insight 3: Once we find the duplicate and the actual sum of the input numbers, we can deduce the missing number by comparing with the expected sum.

📚

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: `nums` contains only the duplicate number (e.g., `[1, 1]`). The algorithm correctly identifies 1 as the duplicate, calculates the expected sum as 2, actual sum as 2, and correctly computes the missing number as 2. Returns [1,2]
Edge case 2: The duplicate is the last element in the array. The algorithm still iterates through the array and finds the duplicate correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider the case when calculating the missing number involves integer overflow, especially when 'n' is large. This can be addressed by using `long` data types, as seen in the provided solutions.

✗Common mistake 2: Incorrectly calculating the missing number formula. It's `expected_sum - sum_nums + duplicate`, not just `expected_sum - sum_nums`.

✅

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: Data validation in databases. When inserting a batch of records, this algorithm can be used to quickly identify duplicate entries and missing IDs.

🏦Real-world scenario 2: Error detection in communication systems. In a sequence of transmitted packets, this algorithm can help identify if any packets are lost (missing) or duplicated due to network issues.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: September 14, 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

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

@InterviewProJun 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 "Set Mismatch"?

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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