What is the time complexity of Minimum Operations to Make Median of Array Equal to K?

The optimal solution for Minimum Operations to Make Median of Array Equal to K runs in O(n log n) time and uses O(1) space. The optimal solution has a time complexity of O(n log n) due to sorting the array, and a space complexity of O(1) since we are modifying the array in place.

What is the brute force solution for Minimum Operations to Make Median of Array Equal to K?

The brute force approach for Minimum Operations to Make Median of Array Equal to K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible combination of operations on the array to see how many operations it takes to make the median equal to k. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Operations to Make Median of Array Equal to K?

Minimum Operations to Make Median of Array Equal to K uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Sorting, Greedy.

What are the interview tips for Minimum Operations to Make Median of Array Equal to K?

Always clarify the definition of median and how it applies to the problem. Discuss your thought process and approach before jumping into coding. Consider edge cases, such as when the array is already optimal.

What are common mistakes when solving Minimum Operations to Make Median of Array Equal to K?

Not sorting the array before trying to find the median. Forgetting to consider both sides of the median when calculating operations.

#3107

Minimum Operations to Make Median of Array Equal to K

Medium

ArrayGreedySortingSortingGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(1)

The optimal approach focuses on the properties of the median and how it can be adjusted efficiently without brute-forcing through all elements. By sorting the array and adjusting only the necessary elements, we minimize operations.

⚙️

Algorithm

4 steps

1Step 1: Sort the array nums.
2Step 2: Identify the median index based on the length of the array.
3Step 3: If the current median is less than k, increase elements to k until the median is k.
4Step 4: If the current median is greater than k, decrease elements to k until the median is k.

solution.py20 lines

1# Full working Python code
2
3def min_operations(nums, k):
4    nums.sort()
5    n = len(nums)
6    median_index = (n - 1) // 2
7    current_median = nums[median_index]
8    operations = 0
9    if current_median < k:
10        for i in range(median_index, n):
11            if nums[i] < k:
12                operations += k - nums[i]
13    else:
14        for i in range(median_index + 1):
15            if nums[i] > k:
16                operations += nums[i] - k
17    return operations
18
19# Example usage
20print(min_operations([2,5,6,8,5], 4))  # Output: 2

ℹ

Complexity note: The optimal solution has a time complexity of O(n log n) due to sorting the array, and a space complexity of O(1) since we are modifying the array in place.

1Sorting the array is crucial to finding the median efficiently.
2Only elements around the median need to be adjusted to minimize operations.

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