What is the time complexity of Minimum Number of Operations to Have Distinct Elements?

The optimal solution for Minimum Number of Operations to Have Distinct Elements runs in O(n) time and uses O(n) space. The algorithm processes each element once and uses a set to track distinct values, leading to linear time complexity.

What is the brute force solution for Minimum Number of Operations to Have Distinct Elements?

The brute force approach for Minimum Number of Operations to Have Distinct Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. Repeatedly remove the first three elements until all remaining elements are distinct. This approach checks after each operation, leading to potential 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 Have Distinct Elements?

Minimum Number of Operations to Have Distinct Elements 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 Have Distinct Elements?

Explain your thought process clearly. Consider edge cases like all distinct or all duplicates. Optimize your solution before coding.

What are common mistakes when solving Minimum Number of Operations to Have Distinct Elements?

Not updating the pointer correctly after removals. Failing to check for duplicates after each operation.

#3779

Minimum Number of Operations to Have Distinct Elements

Medium

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)

Use a hash set to track distinct elements and a pointer to simulate the removal of elements. This approach efficiently checks for duplicates and counts operations.

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Algorithm

3 steps

1Step 1: Initialize a set to track distinct elements and a pointer to the start of the array.
2Step 2: While duplicates exist, move the pointer forward by three and increment the operation counter.
3Step 3: Return the operation count when all elements are distinct.

solution.py10 lines

1def min_operations(nums):
2    seen = set()
3    ops = 0
4    i = 0
5    while i < len(nums):
6        seen.update(nums[i:i+3])
7        i += 3
8        ops += 1
9        if len(seen) == len(nums): break
10    return ops

ℹ

Complexity note: The algorithm processes each element once and uses a set to track distinct values, leading to linear time complexity.

1Removing elements in groups can quickly reduce duplicates.
2Using a hash set allows for efficient duplicate checking.

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