How do you solve Minimum Operations to Collect Elements on LeetCode?

Minimum Operations to Collect Elements (LeetCode #2869) 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: Using a set allows for efficient tracking of collected elements.

What is the time complexity of Minimum Operations to Collect Elements?

The optimal solution for Minimum Operations to Collect Elements runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only iterate through the array once, and checking the size of the set is O(1).

What is the brute force solution for Minimum Operations to Collect Elements?

The brute force approach for Minimum Operations to Collect Elements is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we will simulate the process of removing elements from the end of the array until we collect all required elements from 1 to k. 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 Operations to Collect Elements?

Minimum Operations to Collect Elements uses the following concepts: Array, Hash Table, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Collect Elements?

Always consider edge cases, such as when k equals the length of the array. Explain your thought process clearly while coding. Optimize your solution after implementing the brute force approach.

What are common mistakes when solving Minimum Operations to Collect Elements?

Not checking the size of the set correctly. Overcomplicating the logic by using nested loops.

#2869

Minimum Operations to Collect Elements

Easy

ArrayHash TableBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution leverages a set to keep track of collected elements while iterating through the array in reverse. This allows us to efficiently check if we have collected all required elements without unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Initialize a set to keep track of collected elements.
2Step 2: Iterate through the array in reverse order.
3Step 3: Add each element to the set and increment the operation count.
4Step 4: Stop when the size of the set equals k, as we have collected all required elements.

solution.py9 lines

1def min_operations(nums, k):
2    collected = set()
3    operations = 0
4    for i in range(len(nums) - 1, -1, -1):
5        collected.add(nums[i])
6        operations += 1
7        if len(collected) == k:
8            return operations
9    return operations

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the array once, and checking the size of the set is O(1).

1Using a set allows for efficient tracking of collected elements.
2Iterating in reverse helps to minimize unnecessary checks.

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