Minimum Number of Increments on Subarrays to Form a Target Array

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ArraysGreedy Algorithms

You are given an integer array target. You have an integer array initial of the same size as target with all elements initially zeros.

In one operation you can choose any subarray from initial and increment each value by one.

Return the minimum number of operations to form a target array from initial.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: target = [1,2,3,2,1]
Output: 3
Explanation: We need at least 3 operations to form the target array from the initial array.
[0,0,0,0,0] increment 1 from index 0 to 4 (inclusive).
[1,1,1,1,1] increment 1 from index 1 to 3 (inclusive).
[1,2,2,2,1] increment 1 at index 2.
[1,2,3,2,1] target array is formed.

Example 2:

Input: target = [3,1,1,2]
Output: 4
Explanation: [0,0,0,0] -> [1,1,1,1] -> [1,1,1,2] -> [2,1,1,2] -> [3,1,1,2]

Example 3:

Input: target = [3,1,5,4,2]
Output: 7
Explanation: [0,0,0,0,0] -> [1,1,1,1,1] -> [2,1,1,1,1] -> [3,1,1,1,1] -> [3,1,2,2,2] -> [3,1,3,3,2] -> [3,1,4,4,2] -> [3,1,5,4,2].

Constraints:

1 <= target.length <= 10⁵
1 <= target[i] <= 10⁵

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 possible ranges for the integers in the `target` array? Can they be negative, zero, or only positive?
What is the maximum size of the `target` array? Should I be concerned about integer overflow when calculating the minimum increments?
If the `target` array is empty, what should I return?
Could you clarify with an example what constitutes an 'increment on a subarray'? Does it mean adding the same value to all elements within a contiguous range?
If the target array is already all zeros, what should the result be?

Brute Force Solution

Approach

The basic idea is to try absolutely everything. We consider every possible combination of operations to make the starting array match the target. We then pick the best solution from all the possibilities we tried.

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

Start with an initial array filled with zeros.
Try incrementing every single possible sub-portion of the initial array by every possible value.
After each set of increment operations, compare the resulting array to the target array.
If the resulting array matches the target array, record the total number of increments used.
Repeat steps 2-4 for every conceivable set of increment operations, ensuring we explore all options.
Finally, choose the set of increment operations that produced the target array with the fewest total increments.

Code Implementation

def min_increments_brute_force(target_array):
    array_length = len(target_array)
    min_increments = float('inf')

    def solve(current_array, increments):
        nonlocal min_increments

        if current_array == target_array:
            min_increments = min(min_increments, increments)
            return

        # Optimization: If current increments already exceed min, stop.
        if increments >= min_increments:
            return

        # Try every possible subarray and increment value.
        for start_index in range(array_length):
            for end_index in range(start_index, array_length):
                for increment_value in range(1, max(target_array) + 2):
                    new_array = current_array[:]

                    # Increment the subarray.
                    for index in range(start_index, end_index + 1):
                        new_array[index] += increment_value

                    # Explore this possibility.
                    solve(new_array, increments + increment_value)

    # Start with an initial array of zeros.
    initial_array = [0] * array_length
    solve(initial_array, 0)

    # If no solution was found, return 0
    if min_increments == float('inf'):
        return 0

    return min_increments

Big(O) Analysis

Time Complexity

O(exponential) – The given algorithm attempts every possible increment on every possible subarray. Consider an array of size n. There are O(n^2) possible subarrays. For each subarray, the algorithm tries incrementing by different values. The number of possible increment values is unbounded, but even if we limit it to the maximum value in the target array (call this m), we have m choices for each subarray. Since we are considering all combinations of these increments across all subarrays, the total number of possibilities explored grows exponentially with both n and m. Therefore, the time complexity is O(m^(n^2)) which simplifies to exponential.

Space Complexity

O(N^N) – The provided explanation describes an exhaustive search of all possible increment operations on all possible subarrays. This implies storing all possible combinations of increments and the resulting arrays in memory to compare them with the target. In the worst-case scenario, we are trying all combinations of incrementing subarrays from an initial array of size N, leading to a space complexity that grows factorially with N. Since the number of increment values can also vary depending on the target array values, the number of increment operations being stored becomes N^N where N is the length of the target array. This creates a space complexity of approximately O(N^N).

Optimal Solution

Approach

The core idea is to count only the necessary increases. Instead of directly comparing the numbers or checking every subarray, we cleverly track the ups and downs between neighboring elements.

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

Begin by assuming the first number in your target requires that many operations to reach it from zero.
Then, go through the target one number at a time and compare it to the number before it.
If the current number is bigger than the previous one, you'll need to perform more operations: specifically, the difference between the current number and the previous number.
If the current number is smaller or the same as the previous number, you don't need to do anything extra because the previous operations already cover the current number's requirement.
Finally, after going through all the numbers, add up all the operation counts you found. This total will be the minimum number of increments.

Code Implementation

def min_increments(target_array):
    number_of_operations = 0
    previous_number = 0

    # First increment needs to get to the first number.
    if target_array:
        number_of_operations = target_array[0]
        previous_number = target_array[0]

    for i in range(1, len(target_array)):
        current_number = target_array[i]

        # Only increment if current is greater than previous.
        if current_number > previous_number:
            number_of_operations += current_number - previous_number

        previous_number = current_number

    return number_of_operations

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the target array once. In each iteration, it performs a constant-time comparison between the current element and the previous element. The number of iterations is directly proportional to the size of the input array, n. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The algorithm iterates through the input array, comparing adjacent elements. It only uses a few variables to store intermediate sums (the operation counts) and indices. The amount of extra space does not scale with the input array's size (N). Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty target array	Return 0 immediately as no operations are needed to achieve an empty target.
Target array with one element	Return the value of the single element as the minimum increments required.
Target array with all elements equal to zero	Return 0 because no increments are required.
Target array is sorted in non-decreasing order	The number of increments will be target[n-1] - target[0] if target[0] > 0, or target[n-1] otherwise.
Target array with large values that could cause integer overflow	Use a data type that can accommodate large values, like long, to avoid overflow errors.
Target array with negative numbers	Since the problem states incrementing subarrays, negative numbers are not expected and should result in throwing an exception.
Target array with alternating increasing and decreasing values	The algorithm should correctly sum the differences between adjacent elements when the current element is greater than the previous.
Very large target array to assess time complexity	The solution should be O(n) to efficiently handle large arrays; avoid solutions with nested loops.

Null or empty target array

How to Handle:

Return 0 immediately as no operations are needed to achieve an empty target.

Target array with one element

How to Handle:

Return the value of the single element as the minimum increments required.

Target array with all elements equal to zero

How to Handle:

Return 0 because no increments are required.

Target array is sorted in non-decreasing order

How to Handle:

The number of increments will be target[n-1] - target[0] if target[0] > 0, or target[n-1] otherwise.

Target array with large values that could cause integer overflow

How to Handle:

Use a data type that can accommodate large values, like long, to avoid overflow errors.

Target array with negative numbers

How to Handle:

Since the problem states incrementing subarrays, negative numbers are not expected and should result in throwing an exception.

Target array with alternating increasing and decreasing values

How to Handle:

The algorithm should correctly sum the differences between adjacent elements when the current element is greater than the previous.

Very large target array to assess time complexity

How to Handle:

The solution should be O(n) to efficiently handle large arrays; avoid solutions with nested loops.