Minimum Increment Operations to Make Array Beautiful

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ArraysGreedy Algorithms

You are given a 0-indexed integer array nums having length n, and an integer k.

You can perform the following increment operation any number of times (including zero):

Choose an index i in the range [0, n - 1], and increase nums[i] by 1.

An array is considered beautiful if, for any subarray with a size of 3 or more, its maximum element is greater than or equal to k.

Return an integer denoting the minimum number of increment operations needed to make nums beautiful.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [2,3,0,0,2], k = 4
Output: 3

Explanation: We can perform the following increment operations to make nums beautiful:

Choose index i = 1 and increase nums[1] by 1 -> [2,4,0,0,2].
Choose index i = 4 and increase nums[4] by 1 -> [2,4,0,0,3].
Choose index i = 4 and increase nums[4] by 1 -> [2,4,0,0,4].

The subarrays with a size of 3 or more are: [2,4,0], [4,0,0], [0,0,4], [2,4,0,0], [4,0,0,4], [2,4,0,0,4]. In all the subarrays, the maximum element is equal to k = 4, so nums is now beautiful. It can be shown that nums cannot be made beautiful with fewer than 3 increment operations. Hence, the answer is 3.

Example 2:

Input: nums = [0,1,3,3], k = 5
Output: 2

Explanation: We can perform the following increment operations to make nums beautiful:

Choose index i = 2 and increase nums[2] by 1 -> [0,1,4,3].
Choose index i = 2 and increase nums[2] by 1 -> [0,1,5,3].

The subarrays with a size of 3 or more are: [0,1,5], [1,5,3], [0,1,5,3]. In all the subarrays, the maximum element is equal to k = 5, so nums is now beautiful. It can be shown that nums cannot be made beautiful with fewer than 2 increment operations. Hence, the answer is 2.

What is the minimum number of increment operations needed to make nums beautiful?

Minimum Increment Operations to Make Array Beautiful

Medium

Google

3 views

Topics:

ArraysGreedy Algorithms

You are given a 0-indexed integer array nums having length n, and an integer k.

You can perform the following increment operation any number of times (including zero):

Choose an index i in the range [0, n - 1], and increase nums[i] by 1.

An array is considered beautiful if, for any subarray with a size of 3 or more, its maximum element is greater than or equal to k.

Return an integer denoting the minimum number of increment operations needed to make nums beautiful.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [2,3,0,0,2], k = 4
Output: 3

Explanation: We can perform the following increment operations to make nums beautiful:

Choose index i = 1 and increase nums[1] by 1 -> [2,4,0,0,2].
Choose index i = 4 and increase nums[4] by 1 -> [2,4,0,0,3].
Choose index i = 4 and increase nums[4] by 1 -> [2,4,0,0,4].

The subarrays with a size of 3 or more are: [2,4,0], [4,0,0], [0,0,4], [2,4,0,0], [4,0,0,4], [2,4,0,0,4]. In all the subarrays, the maximum element is equal to k = 4, so nums is now beautiful. It can be shown that nums cannot be made beautiful with fewer than 3 increment operations. Hence, the answer is 3.

Example 2:

Input: nums = [0,1,3,3], k = 5
Output: 2

Explanation: We can perform the following increment operations to make nums beautiful:

Choose index i = 2 and increase nums[2] by 1 -> [0,1,4,3].
Choose index i = 2 and increase nums[2] by 1 -> [0,1,5,3].

The subarrays with a size of 3 or more are: [0,1,5], [1,5,3], [0,1,5,3]. In all the subarrays, the maximum element is equal to k = 5, so nums is now beautiful. It can be shown that nums cannot be made beautiful with fewer than 2 increment operations. Hence, the answer is 2.

What is the minimum number of increment operations needed to make nums beautiful?

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 is the range of values for each element within the input array? Can they be negative, zero, or have a specific upper bound?
What defines 'beautiful' in the context of the problem? Can you provide more specific criteria or examples?
What should I return if the input array is already 'beautiful'?
Is there a limit on the number of operations I can perform, or will there always be a finite number of increments that will result in a beautiful array?
Can the size of the input array be zero (empty array), and if so, how should I handle that case?

Brute Force Solution

Approach

The brute force method involves trying every conceivable modification to the input. We systematically explore each possible sequence of increments to find a configuration that satisfies the 'beautiful' condition, keeping track of the minimum increments needed so far. It's like trying every single combination lock until you find the right one.

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

Start with the original arrangement of numbers.
Consider changing the first number by adding 1 to it.
Now consider changing the first number by adding 2 to it. Repeat this for all possible increment values we think are reasonable.
For each change to the first number, move to the second number and consider all possible increments for it as well.
Continue this process for every number in the list. We create many different versions of the list of numbers.
For each of these versions, check if the list has become 'beautiful' according to the rules.
If it is beautiful, calculate how many increments we made to get to that version.
Remember the smallest number of increments we used to get to a beautiful version of the list.
In the end, the smallest number of increments we remembered is the answer.

Code Implementation

def minimum_increment_operations_brute_force(numbers, maximum_value):    minimum_increments = float('inf')
    def is_beautiful(array):        for index in range(0, len(array), 2):            if array[index] > maximum_value:                return False
        return True
    def solve(current_array, current_index, current_increments):        nonlocal minimum_increments
        # If we've reached the end of the array        if current_index == len(numbers):
            # Check if the array is beautiful            if is_beautiful(current_array):
                minimum_increments = min(minimum_increments, current_increments)
            return
        # Explore incrementing the current number        for increment in range(maximum_value * len(numbers) + 1):            new_array = current_array[:]
            new_array[current_index] += increment

            #Recurse to the next number.
            solve(new_array, current_index + 1, current_increments + increment)
    solve(numbers[:], 0, 0)
    return minimum_increments

Big(O) Analysis

Time Complexity

O(k^n) – The brute force approach explores all possible increment combinations for each number in the array. Let n be the number of elements in the array, and let k be the maximum possible increment value we consider for each element. For each of the n elements, we are branching out into k possibilities. Thus, we have k choices for the first element, k choices for the second element, and so on. Therefore, the total number of possible combinations we are exploring grows exponentially with n, specifically as k^n, where k is the range of possible increments considered. Hence, the time complexity is O(k^n).

Space Complexity

O(K^N) – The brute force approach explores all possible increment combinations. In the worst-case scenario, for each of the N numbers in the array, we might consider up to K different increments (where K is some reasonable maximum increment value considered). This leads to exploring a decision tree of depth N, where each node has K branches. Thus, the number of different array versions stored implicitly in the recursion stack or explicitly in a data structure grows exponentially with N. Therefore, the space complexity is O(K^N) since we are storing multiple versions of the list.

Optimal Solution

Approach

The best way to solve this problem is to go through the numbers one by one and make smart adjustments as needed. We focus on satisfying the 'beautiful' condition in the most efficient way possible, minimizing the total increase.

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

Start by checking the first number and the two numbers after it.
If these three numbers don't meet the requirement of the problem (being 'beautiful'), figure out which number is causing the issue.
Increase the number that's causing the issue just enough to make those three numbers meet the requirement.
Move to the next number in the list and repeat the process: look at it and the two numbers after it, and make adjustments as needed.
Keep doing this for every number in the list until you reach the end.
The total amount you've increased the numbers is the answer.

Code Implementation

def min_increment_operations(numbers, maximum_value): 
    increment_count = 0
    array_length = len(numbers)

    for index in range(array_length - 2): 
        #Check if the current triplet satisfies the beautiful condition
        if max(numbers[index], numbers[index + 1], numbers[index + 2]) > maximum_value:

            #Increase the value at index to meet beautiful requirement
            numbers[index] += maximum_value - max(numbers[index], numbers[index + 1], numbers[index + 2])
            increment_count += abs(maximum_value - max(numbers[index], numbers[index + 1], numbers[index + 2]))

    return increment_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n exactly once. Within each iteration, a constant number of comparisons and increment operations are performed involving only the current element and the next two elements. Therefore, the time complexity is directly proportional to the number of elements in the array, resulting in O(n).

Space Complexity

O(1) – The algorithm described iterates through the input array in place. It only needs to keep track of a few variables during the incrementing operation, such as the current index and potentially a temporary variable to store the increment value. The space used by these variables remains constant irrespective of the input array size N. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no operations are needed for an empty array.
Single element array	Return 0 as a single element is inherently beautiful as there's no 'adjacent' element to consider.
Array with all identical values	The solution should still work correctly, incrementing values until the beauty condition is met, potentially requiring 0 increments if already beautiful.
Array with extreme boundary values (very large or very small numbers)	Ensure that intermediate calculations and final increments don't lead to integer overflow, potentially using long data types.
Array with negative numbers	The increment operations should handle negative numbers correctly, ensuring that the values are incremented until the beauty condition is satisfied with respect to negative numbers.
Maximum size array (scales efficiently)	The solution's time complexity should be considered; an O(n) or O(n log n) solution is preferred for large input sizes and hashmap is used to efficiently find min/max.
Array is already 'beautiful'	The algorithm should return 0, indicating no increments are required.
Array contains zero	The solution should correctly handle zeros; these may require increments to satisfy the 'beautiful' condition.