What is the time complexity of Minimum Number of Operations to Make Array Continuous?

The optimal solution for Minimum Number of Operations to Make Array Continuous runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n), and the sliding window runs in O(n), making this approach efficient for large inputs.

What is the brute force solution for Minimum Number of Operations to Make Array Continuous?

The brute force approach for Minimum Number of Operations to Make Array Continuous is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible combination of unique numbers that could form a continuous array. This involves replacing elements until the array meets the conditions for continuity, which can be inefficient. The optimal Optimal Solution approach improves this to O(n log n).

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

Minimum Number of Operations to Make Array Continuous uses the following concepts: Array, Hash Table, Binary Search, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Operations to Make Array Continuous?

Always consider edge cases, such as arrays with duplicates or already continuous arrays. Explain your thought process clearly; this shows your understanding of the problem. Practice identifying patterns in problems, as many can be solved using similar techniques.

What are common mistakes when solving Minimum Number of Operations to Make Array Continuous?

Failing to account for duplicates before processing the array. Not properly managing the sliding window pointers, leading to incorrect counts.

#2009

Minimum Number of Operations to Make Array Continuous

Hard

ArrayHash TableBinary SearchSliding WindowHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses sorting and a sliding window approach to efficiently find the minimum number of operations needed. By focusing on the range of unique elements, we can minimize replacements.

⚙️

Algorithm

3 steps

1Step 1: Sort the input array to arrange the numbers in increasing order.
2Step 2: Use a sliding window to find the longest continuous segment of unique numbers that can fit the criteria.
3Step 3: Calculate the number of operations needed as the difference between the total length and the length of the longest valid segment.

solution.py13 lines

1# Full working Python code
2
3def min_operations(nums):
4    nums = sorted(set(nums))  # Remove duplicates and sort
5    n = len(nums)
6    left = 0
7    min_operations = float('inf')
8    for right in range(n):
9        while nums[right] - nums[left] >= n:
10            left += 1
11        min_operations = min(min_operations, n - (right - left + 1))
12    return min_operations
13

ℹ

Complexity note: The sorting step takes O(n log n), and the sliding window runs in O(n), making this approach efficient for large inputs.

1Sorting the array helps in easily identifying the range of unique numbers.
2Using a sliding window allows us to efficiently find the longest valid segment without unnecessary checks.

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