How do you solve Find Missing and Repeated Values on LeetCode?

Find Missing and Repeated Values (LeetCode #2965) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The problem can be solved using frequency counting or mathematical properties.

What is the time complexity of Find Missing and Repeated Values?

The optimal solution for Find Missing and Repeated Values runs in O(n²) time and uses O(n) space. The time complexity is O(n²) due to iterating through the n*n matrix, while the space complexity is O(n) for the set used to track seen numbers.

What is the brute force solution for Find Missing and Repeated Values?

The brute force approach for Find Missing and Repeated Values is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves checking each number in the expected range and counting its occurrences in the matrix. This helps us identify which number is repeated and which one is missing. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Find Missing and Repeated Values?

Find Missing and Repeated Values uses the following concepts: Array, Hash Table, Math, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Missing and Repeated Values?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process while coding; it shows your understanding. Consider both time and space complexity when discussing your approach.

What are common mistakes when solving Find Missing and Repeated Values?

Forgetting to check for the repeated number while iterating through the matrix. Not considering edge cases where the matrix is at its minimum size.

#2965

Find Missing and Repeated Values

Easy

ArrayHash TableMathMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal solution leverages mathematical properties to find the missing and repeated numbers without using extra space for counting. We can use the sum of the first n² natural numbers to find the missing number and the repeated number through a single pass.

⚙️

Algorithm

3 steps

1Step 1: Calculate the expected sum of numbers from 1 to n² using the formula n² * (n² + 1) / 2.
2Step 2: Calculate the actual sum of numbers in the matrix and identify the repeated number during this process.
3Step 3: Use the difference between the expected sum and actual sum to find the missing number.

solution.py16 lines

1# Full working Python code
2
3def findMissingAndRepeated(grid):
4    n = len(grid)
5    expected_sum = n * n * (n * n + 1) // 2
6    actual_sum = 0
7    seen = set()
8    repeated = -1
9    for row in grid:
10        for num in row:
11            actual_sum += num
12            if num in seen:
13                repeated = num
14            seen.add(num)
15    missing = expected_sum - (actual_sum - repeated)
16    return [repeated, missing]

ℹ

Complexity note: The time complexity is O(n²) due to iterating through the n*n matrix, while the space complexity is O(n) for the set used to track seen numbers.

1The problem can be solved using frequency counting or mathematical properties.
2Identifying the repeated number can be done while calculating the sum.

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