Minimum Operations to Make Array Values Equal to K

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Arrays

You are given an integer array nums and an integer k.

An integer h is called valid if all values in the array that are strictly greater than h are identical.

For example, if nums = [10, 8, 10, 8], a valid integer is h = 9 because all nums[i] > 9 are equal to 10, but 5 is not a valid integer.

You are allowed to perform the following operation on nums:

Select an integer h that is valid for the current values in nums.
For each index i where nums[i] > h, set nums[i] to h.

Return the minimum number of operations required to make every element in nums equal to k. If it is impossible to make all elements equal to k, return -1.

Example 1:

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

Output: 2

Explanation:

The operations can be performed in order using valid integers 4 and then 2.

Example 2:

Input: nums = [2,1,2], k = 2

Output: -1

Explanation:

It is impossible to make all the values equal to 2.

Example 3:

Input: nums = [9,7,5,3], k = 1

Output: 4

Explanation:

The operations can be performed using valid integers in the order 7, 5, 3, and 1.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 100

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 `nums` array and the range of values within the array and for `k`? Should I be concerned about integer overflow?
Can the array `nums` contain negative integers, zeros, or just positive integers?
Is the input array `nums` ever empty or null? What should I return in that case?
Are there any specific edge cases or boundary conditions I should be aware of for `k`? For example, could `k` be outside of the range of values in `nums`?
Is there a limit on the number of operations I can perform? If the minimum operations required exceeds a certain threshold, how should I handle that?

Brute Force Solution

Approach

The brute force method in this scenario involves checking every single possible way to change the numbers in the list to equal a target value. We will explore every combination of changes and then determine the solution that uses the least number of operations. This is a 'try everything' approach.

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

Consider all possible values to change the first number to, including changing it to the target value or leaving it as is.
For each of those possibilities, consider all possible values to change the second number to, including changing it to the target value or leaving it as is.
Continue this process for all numbers in the list.
For each complete set of changes you considered, count the number of operations (changes) you made.
Compare the operation count of all complete sets of changes and find the set that resulted in the fewest operations.
The fewest number of operations found is the answer.

Code Implementation

def min_operations_brute_force(numbers, target):
    min_operation_count = float('inf')

    def solve(current_index, current_operation_count):
        nonlocal min_operation_count

        # Base case: we have processed all numbers
        if current_index == len(numbers):
            min_operation_count = min(min_operation_count, current_operation_count)
            return

        # Option 1: Don't change the current number
        solve(current_index + 1, current_operation_count) 

        # Option 2: Change the current number to the target
        # This part of the code allows us to explore the effect of changing elements
        solve(current_index + 1, current_operation_count + 1)

    solve(0, 0)

    # Here is the returned result to the caller
    return min_operation_count

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force solution explores all possible combinations of changing each number in the input array to equal K or leaving it as is. For each of the n elements in the array, there are two choices: either change it to K or don't change it. This results in 2 * 2 * ... * 2 (n times) possible combinations, which equals 2^n. Therefore, the time complexity is O(2^n) since it examines all these possibilities.

Space Complexity

O(1) – The brute force approach described explores every possible combination of changes to the numbers in the input list. While conceptually this involves considering different branches of possibilities, the provided plain English explanation doesn't mention explicitly storing all these intermediate states simultaneously. Instead, it describes iterating through the list and counting operations. It implicitly overwrites the current state. Therefore, it does not create any auxiliary data structures that scale with the input size N (the number of elements in the list). The space required remains constant, regardless of the input array's length.

Optimal Solution

Approach

The key is to focus on what changes need to be made instead of recalculating everything repeatedly. We want to minimize the number of changes required to make each value in the array equal to a target value, focusing on values greater than the target.

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

First, find out how many values in the input are bigger than the target value.
Next, figure out the total number of times you need to divide values down to reduce values greater than the target.
Figure out how many times you can combine small values that are smaller than the target. Combining these values requires another operation.
The final answer is the combination of the divisions required on large numbers, plus the number of groups you need to perform additions to.

Code Implementation

def min_operations(numbers, target_value):
    larger_than_target_count = 0
    division_operations = 0
    smaller_than_target_sum = 0

    for number in numbers:
        if number > target_value:
            larger_than_target_count += 1
            # Count divisions to reduce to target
            while number > target_value:
                number //= 2
                division_operations += 1
        else:
            smaller_than_target_sum += number

    # Determine number of combination operations
    addition_operations = 0
    if smaller_than_target_sum < target_value and smaller_than_target_sum > 0:
        addition_operations = (target_value - 1) // smaller_than_target_sum

    return division_operations + addition_operations

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n to count values greater than k, calculate the total division operations, and determine the number of values less than k. Each of these steps takes O(n) time. Since these operations are performed sequentially, the overall time complexity is dominated by O(n).

Space Complexity

O(1) – The described algorithm primarily uses a fixed number of variables to count values greater than the target, track the number of divisions, and count the potential for combining smaller values. No data structures such as arrays, lists, or hash maps are created whose sizes depend on the input array size, N. Thus, the auxiliary space remains constant regardless of the input size.

Edge Cases

Case	How to Handle
Empty input array nums	Return 0 since no operations are needed on an empty array.
Null input array nums	Throw an IllegalArgumentException or return -1 to indicate invalid input.
Array with one element	Return the absolute difference between the single element and k.
Large input array with elements far from k	The solution should iterate through the array and calculate the sum of absolute differences, which handles large arrays efficiently with O(n) time complexity.
All elements are equal to k	Return 0, as no operations are needed when all elements match k.
Array contains negative numbers, zeros, and positive numbers	The absolute difference calculation handles both negative and positive numbers correctly.
Extreme boundary values in the array that could lead to integer overflow	Use long data type to store the intermediate sum of absolute differences to prevent integer overflow.
k is extremely large and far from the values in nums, leading to potential integer overflow during absolute difference calculation	Use long data type when calculating the absolute difference and accumulating the total operations to prevent integer overflow.

Empty input array nums

How to Handle:

Return 0 since no operations are needed on an empty array.

Null input array nums

How to Handle:

Throw an IllegalArgumentException or return -1 to indicate invalid input.

Array with one element

How to Handle:

Return the absolute difference between the single element and k.

Large input array with elements far from k

How to Handle:

The solution should iterate through the array and calculate the sum of absolute differences, which handles large arrays efficiently with O(n) time complexity.

All elements are equal to k

How to Handle:

Return 0, as no operations are needed when all elements match k.

Array contains negative numbers, zeros, and positive numbers

How to Handle:

The absolute difference calculation handles both negative and positive numbers correctly.

Extreme boundary values in the array that could lead to integer overflow

How to Handle:

Use long data type to store the intermediate sum of absolute differences to prevent integer overflow.

k is extremely large and far from the values in nums, leading to potential integer overflow during absolute difference calculation

How to Handle:

Use long data type when calculating the absolute difference and accumulating the total operations to prevent integer overflow.