What is the time complexity of Minimum Number of Operations to Make Elements in Array Distinct?

The optimal solution for Minimum Number of Operations to Make Elements in Array Distinct runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the array once to build the frequency map and then iterate through the map, which is efficient given the constraints.

What is the brute force solution for Minimum Number of Operations to Make Elements in Array Distinct?

The brute force approach for Minimum Number of Operations to Make Elements in Array Distinct is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly removing the first three elements of the array until all elements are distinct. This method is straightforward but inefficient, as it may require multiple passes through the array. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Operations to Make Elements in Array Distinct?

Minimum Number of Operations to Make Elements in Array Distinct uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Operations to Make Elements in Array Distinct?

Clarify the problem requirements before coding. Think about edge cases, such as arrays with all distinct elements or all duplicates. Discuss your thought process and approach with the interviewer.

What are common mistakes when solving Minimum Number of Operations to Make Elements in Array Distinct?

Not considering the case where the array is already distinct. Failing to account for how many duplicates can be removed in one operation.

#3396

Minimum Number of Operations to Make Elements in Array Distinct

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach involves using a set to track distinct elements and counting how many duplicates we have. This allows us to calculate the minimum number of operations needed without repeatedly modifying the array.

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Algorithm

3 steps

1Step 1: Create a frequency map to count occurrences of each number.
2Step 2: Count how many numbers appear more than once.
3Step 3: For each duplicate, calculate how many operations are needed to remove them in groups of three.

solution.py8 lines

1def min_operations(nums):
2    from collections import Counter
3    freq = Counter(nums)
4    operations = 0
5    for count in freq.values():
6        if count > 1:
7            operations += (count - 1) // 3 + 1
8    return operations

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once to build the frequency map and then iterate through the map, which is efficient given the constraints.

1Using a frequency map helps in efficiently counting duplicates.
2Understanding how to group duplicates can minimize operations.

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