Minimum Operations to Make Median of Array Equal to K

Medium

IBM

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Topics:

ArraysGreedy Algorithms

You are given an integer array nums and a non-negative integer k. In one operation, you can increase or decrease any element by 1.

Return the minimum number of operations needed to make the median of nums equal to k.

The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken.

Example 1:

Input: nums = [2,5,6,8,5], k = 4

Output: 2

Explanation:

We can subtract one from nums[1] and nums[4] to obtain [2, 4, 6, 8, 4]. The median of the resulting array is equal to k.

Example 2:

Input: nums = [2,5,6,8,5], k = 7

Output: 3

Explanation:

We can add one to nums[1] twice and add one to nums[2] once to obtain [2, 7, 7, 8, 5].

Example 3:

Input: nums = [1,2,3,4,5,6], k = 4

Output: 0

Explanation:

The median of the array is already equal to k.

Constraints:

1 <= nums.length <= 2 * 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹

Try coding it in LeetCode

Minimum Operations to Make Median of Array Equal to K

Medium

IBM

2 views

Topics:

ArraysGreedy Algorithms

You are given an integer array nums and a non-negative integer k. In one operation, you can increase or decrease any element by 1.

Return the minimum number of operations needed to make the median of nums equal to k.

The median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the larger of the two values is taken.

Example 1:

Input: nums = [2,5,6,8,5], k = 4

Output: 2

Explanation:

We can subtract one from nums[1] and nums[4] to obtain [2, 4, 6, 8, 4]. The median of the resulting array is equal to k.

Example 2:

Input: nums = [2,5,6,8,5], k = 7

Output: 3

Explanation:

We can add one to nums[1] twice and add one to nums[2] once to obtain [2, 7, 7, 8, 5].

Example 3:

Input: nums = [1,2,3,4,5,6], k = 4

Output: 0

Explanation:

The median of the array is already equal to k.

Constraints:

1 <= nums.length <= 2 * 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums` and the range of values within `nums` and for `k`?
Can the array `nums` be empty or contain null values?
How is the median defined for an array with an even number of elements? Should I take the lower median, upper median, or the average?
If it's impossible to make the median equal to `k` through any number of operations, what value should I return?
Are the numbers in the array integers or floating-point numbers?

Brute Force Solution

Approach

The brute force approach involves trying every possible combination to make the median of a collection of numbers equal to a specific value. We will explore many combinations of changes to the numbers, calculating the median of the modified collection each time. We keep track of the changes that lead to the desired median and then select the set of changes which requires the fewest modifications.

Here's how the algorithm would work step-by-step:

Consider all possible subsets of the numbers. For each subset, we try changing each number in the subset to the target value.
For each of these changes, determine the new median of the entire collection of numbers.
If the new median equals our desired target value, record the number of changes made.
Repeat this process for every single possible combination of changes.
After exhaustively trying all possible combinations, identify the combination that resulted in the desired median and required the smallest number of changes.
This smallest number of changes is the solution.

Code Implementation

def min_operations_to_make_median_equal_to_k_brute_force(numbers, target):
    number_of_elements = len(numbers)
    minimum_cost = float('inf')

    # Iterate through all possible subsets of numbers
    for i in range(2 ** number_of_elements):
        subset = []
        cost = 0
        modified_numbers = []

        # Determine which numbers to change based on the current subset
        for j in range(number_of_elements):
            if (i >> j) & 1:
                subset.append(j)
                cost += abs(numbers[j] - target)
                modified_numbers.append(target)
            else:
                modified_numbers.append(numbers[j])

        # Create a sorted copy of the modified numbers
        sorted_numbers = sorted(modified_numbers)

        # Calculate the median of the sorted numbers
        median_index = (number_of_elements - 1) // 2
        median = sorted_numbers[median_index]

        # Update the minimum cost if the median equals the target
        if median == target:
            minimum_cost = min(minimum_cost, cost)

    if minimum_cost == float('inf'):
        return -1

    return minimum_cost

Big(O) Analysis

Time Complexity

O(2^n * n!) – The algorithm considers all possible subsets of the input array of size n. There are 2^n possible subsets. For each subset, the algorithm tries changing each number in the subset to the target value, potentially requiring examining all possible permutations of changes. This permutation aspect results in a factorial component, approximately n! in the worst case. Calculating the median for each of these scenarios takes O(n) time, but that factor is dominated by the others, leading to an overall time complexity of O(2^n * n!).

Space Complexity

O(1) – The brute force approach, as described, does not utilize any auxiliary data structures that scale with the input size, N (the number of elements in the array). The algorithm only needs to store a counter for the number of changes, a temporary variable to calculate the median, and potentially indices for iterating through the subsets, all of which take up constant space. Therefore, the space complexity remains constant regardless of the input array size. Since the space used does not grow with the input size, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The key is to focus on modifying only the necessary numbers. We want to minimize how many changes we make and by how much each number changes to get the desired median. We only need to worry about the numbers around where the median would be.

Here's how the algorithm would work step-by-step:

First, figure out how many numbers the array needs to have on each side of the median to have the number K as the median.
Find out how many numbers are already equal to K or larger than K. These are already in a good state.
If you don't have enough numbers that are K or bigger, change the smallest numbers that are less than K to K. That's always the best move.
Now, focus on the numbers that are smaller than the target median position. We don't want to change too many of those, so avoid changing numbers unless absolutely necessary.
Calculate the total amount of change we did to get the required configuration.

Code Implementation

def min_operations(numbers, k_value):
    numbers.sort()
    array_length = len(numbers)
    median_index = array_length // 2

    # Only focus on the median and up to the largest number
    numbers_from_median = numbers[median_index:]

    at_least_k_count = 0
    for number in numbers_from_median:
        if number >= k_value:
            at_least_k_count += 1

    # Check if enough numbers are already >= k_value
    if at_least_k_count >= (array_length + 1) // 2:
        return 0

    operations_needed = ((array_length + 1) // 2) - at_least_k_count
    total_cost = 0

    # Only consider numbers below the median that need to be changed
    for i in range(median_index - 1, -1, -1):
        if operations_needed > 0:
            # Determine how many operations are needed
            cost = k_value - numbers[i]
            total_cost += cost
            operations_needed -= 1

    return total_cost

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations in this approach are sorting the input array of size n, which takes O(n log n) time. While iterating through the array is also present, these operations are O(n). However, O(n) is asymptotically less than O(n log n). Therefore, the overall time complexity is determined by the sorting step.

Space Complexity

O(1) – The plain English explanation describes operations performed directly on the input array and a few counter variables. It finds numbers greater than or equal to K and changes numbers smaller than K in place. No auxiliary data structures like arrays, hash maps, or recursion are created that scale with the input size N. Therefore, the space used is constant regardless of the size of the input array.

Edge Cases

Case	How to Handle
Empty input array	Return 0 since there is no median and no operations are needed.
Array with a single element.	If the single element equals k, return 0, otherwise return 1.
Array with two elements.	If the median equals k, return 0. Otherwise, calculate the cost to make the median k by changing each element and return the minimum.
All array elements are equal.	If the elements equal k, return 0; otherwise, the median must be changed to k, incurring a cost.
Array with extreme values (large positive or negative integers).	Ensure integer overflow does not occur during calculations of the median, potentially by using long data types.
k is a very large number.	Ensure the comparison and arithmetic operations involving 'k' do not lead to overflow or unexpected results.
Array contains duplicate values.	The median calculation logic should correctly handle duplicates without introducing bias or incorrect comparisons.
Array contains negative numbers, zeros, and positive numbers.	The median finding algorithm needs to handle the mixed sign values correctly when sorting or partitioning the array.