How do you solve Set Mismatch on LeetCode?

Set Mismatch (LeetCode #645) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The sum of the first n natural numbers can be used to find the missing number.

What is the brute force solution for Set Mismatch?

The brute force approach for Set Mismatch is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every number in the array against all others to find duplicates and missing numbers. It's straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Set Mismatch?

Set Mismatch uses the following concepts: Array, Hash Table, Bit Manipulation, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Set Mismatch?

Always clarify the problem constraints and edge cases. Discuss your thought process and approach before jumping into coding. Consider the trade-offs of different approaches and explain them.

What are common mistakes when solving Set Mismatch?

Not accounting for the range of numbers correctly. Overlooking the need to differentiate between the duplicate and missing numbers.

#645

Set Mismatch

Easy

ArrayHash TableBit ManipulationSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages mathematical properties to find the duplicate and missing numbers in linear time. By using a single pass and simple arithmetic, we can achieve better performance.

⚙️

Algorithm

5 steps

1Step 1: Create a set to track numbers seen so far.
2Step 2: Iterate through the array to find the duplicate number.
3Step 3: Calculate the expected sum of numbers from 1 to n using the formula n*(n+1)/2.
4Step 4: Calculate the actual sum of the numbers in the set.
5Step 5: The missing number is the difference between the expected sum and the actual sum.

solution.py13 lines

1def findErrorNums(nums):
2    seen = set()
3    duplicate = -1
4    total = 0
5    n = len(nums)
6    expected_sum = n * (n + 1) // 2
7    for num in nums:
8        if num in seen:
9            duplicate = num
10        seen.add(num)
11        total += num
12    missing = expected_sum - (total - duplicate)
13    return [duplicate, missing]

ℹ

Complexity note: The time complexity is O(n) because we traverse the array only once. The space complexity is O(n) due to the additional set used to track seen numbers.

1The sum of the first n natural numbers can be used to find the missing number.
2Using a set allows for O(1) average time complexity for lookups.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.