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Minimum Replacements to Sort the Array

Hard

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ArraysGreedy Algorithms

You are given a 0-indexed integer array nums. In one operation you can replace any element of the array with any two elements that sum to it.

For example, consider nums = [5,6,7]. In one operation, we can replace nums[1] with 2 and 4 and convert nums to [5,2,4,7].

Return the minimum number of operations to make an array that is sorted in non-decreasing order.

Example 1:

Input: nums = [3,9,3]
Output: 2
Explanation: Here are the steps to sort the array in non-decreasing order:
- From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3]
- From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3]
There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 0
Explanation: The array is already in non-decreasing order. Therefore, we return 0.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Minimum Replacements to Sort the Array

Hard

PayPal

1 view

Topics:

ArraysGreedy Algorithms

You are given a 0-indexed integer array nums. In one operation you can replace any element of the array with any two elements that sum to it.

For example, consider nums = [5,6,7]. In one operation, we can replace nums[1] with 2 and 4 and convert nums to [5,2,4,7].

Return the minimum number of operations to make an array that is sorted in non-decreasing order.

Example 1:

Input: nums = [3,9,3]
Output: 2
Explanation: Here are the steps to sort the array in non-decreasing order:
- From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3]
- From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3]
There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 0
Explanation: The array is already in non-decreasing order. Therefore, we return 0.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

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 the array?
Can the input array `nums` contain negative integers or zero?
If the array is already sorted in non-decreasing order, should I return 0?
Could you clarify the desired output if the input `nums` is null or empty?
When replacing an element, is there a limit to the number of elements I can replace it with (aside from the sum constraint)?

Brute Force Solution

Approach

The brute force approach to sorting with replacements involves trying every possible combination of breaking down numbers in the sequence. We explore all ways to make the sequence non-decreasing by substituting elements, counting how many replacements each way takes.

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

Start by considering the last number in the sequence; it's already 'sorted' by itself.
Now, look at the second to last number. Try all the possible ways to break it down into smaller numbers, making sure that the largest of the smaller numbers is less than or equal to the last number we already considered.
For each of these breakdowns of the second to last number, remember how many replacements were needed.
Move one step back, to the third to last number. Now, for each way you broke down the second to last number, try all possible ways to break down the third to last number, ensuring the largest of the new numbers is less than or equal to the smallest number from the previous breakdown.
Again, keep track of how many new replacements you are making.
Continue this process all the way to the beginning of the sequence, trying all possible breakdowns for each number and keeping track of the total replacements.
After considering all the possible breakdowns for all the numbers, choose the breakdown that results in the smallest number of replacements overall.

Code Implementation

def minimum_replacements_brute_force(numbers):
    def solve(index, previous_minimum):
        if index == -1:
            return 0

        current_number = numbers[index]
        minimum_replacements = float('inf')

        # Iterate through possible number of splits
        for number_of_splits in range(1, current_number + 1):
            largest_value = current_number // number_of_splits
            if current_number % number_of_splits != 0:
                largest_value += 1

            # Check if the largest value is less than or equal to previous minimum.
            if largest_value <= previous_minimum:

                replacements = number_of_splits - 1

                # Recursively find the minimum replacements for the rest of the array
                recursive_replacements = solve(index - 1, current_number // number_of_splits)

                if current_number % number_of_splits != 0:
                    recursive_replacements = solve(index - 1, largest_value - 1)

                minimum_replacements = min(minimum_replacements, replacements + recursive_replacements)

        return minimum_replacements

    # Start the recursion with the last element as the initial minimum.
    return solve(len(numbers) - 1, float('inf'))

Big(O) Analysis

Time Complexity

O(k^n) – The algorithm explores all possible ways to break down each number in the input array of size n. For each number, it tries multiple breakdowns (up to k possible values where k depends on the number's value), leading to exponential branching. Specifically, for each element, we potentially explore k different possibilities. Since we have to do this for n elements the total number of combinations grows exponentially. Thus the overall time complexity is O(k^n) where k is some factor depending on the magnitude of numbers in the input array and can be at most the value of the largest number in the input array.

Space Complexity

O(N^K) – The brute force approach explores all possible breakdowns of each number in the input array. The algorithm uses recursion (or an equivalent iterative approach that mimics recursion) to explore these possibilities. In the worst case, for each element, we might consider a number of possible breakdowns dependent on the number itself and the elements adjacent to it. The number of possible breakdowns escalates factorially as the algorithm traverses the input array. This is due to the exponential number of ways an integer can be partitioned subject to the non-decreasing constraint. Specifically, the space complexity explodes to O(N^K) where N is the input size and K is related to the magnitude of elements in the array, as storing and managing each breakdown for an element becomes significant, leading to the construction and retention of many intermediate sequences and replacement counts. This high space cost is largely driven by storing all possible sequences and replacement counts as the algorithm iterates through the numbers.

Optimal Solution

Approach

The key idea is to work backwards from the end of the sequence to make locally optimal choices. We focus on transforming the sequence such that each element is at most the element to its right, minimizing the required replacements.

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

Start from the rightmost element of the sequence, as this will be the basis for determining our replacements.
Consider the element directly to the left of the last element. Our goal is to make sure this element is no bigger than the element to its right.
If the element is already smaller or equal, we move on to the next element to the left. No replacement is needed here.
If the element is bigger, we need to split it into multiple smaller elements. The number of smaller elements we create is determined by how many pieces are needed so that the largest of those pieces does not exceed the value of the element to the right.
We perform the splitting such that each piece is of roughly equal size.
After splitting, we update the current element with the largest element we just created (which is less than or equal to its neighbor to the right), and we keep track of the number of replacements made. Now we move to the element just to the left of the split elements and repeat the process.
Continue this process moving from right to left, until you reach the beginning of the sequence.
The total number of splits is the answer.

Code Implementation

def minimum_replacements(numbers):
    number_of_replacements = 0
    last_element = numbers[-1]

    for i in range(len(numbers) - 2, -1, -1):
        current_element = numbers[i]

        # If sorted, move to the next element.
        if current_element <= last_element:
            last_element = current_element
            continue

        # Determine the number of pieces to split into.
        number_of_pieces = (current_element + last_element - 1) // last_element

        # This is number of operations required.
        number_of_replacements += number_of_pieces - 1

        # Update last_element to satisfy the sorted condition.
        last_element = current_element // number_of_pieces

    return number_of_replacements

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n from right to left. Inside the loop, the number of splits for each element is calculated based on a division operation. This division and comparison operation takes constant time. Therefore, the dominant operation is the single pass through the array, resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm operates in place, modifying the input array directly. It uses a few constant space variables to track the current element and its right neighbor, along with the number of replacements. No additional data structures like lists, hash maps, or recursion stacks are used that scale with the input size N (the number of elements in the array). Therefore, the auxiliary space used is constant regardless of the input array's size.

Edge Cases

Case	How to Handle
Empty array	Return 0 as no operations are needed for an empty array.
Array with one element	Return 0 as a single-element array is already sorted.
Array already sorted in non-decreasing order	Return 0 as no replacements are needed.
Array with two elements, needs one replacement	This serves as a minimal valid test case to check base logic.
Large integer values in the array that could lead to integer overflow when summing.	Use long data type to prevent integer overflow when performing calculations.
Array with all identical values	Return 0 since the array is already sorted.
Array with values sorted in strictly decreasing order	This represents the worst-case scenario requiring maximum replacements.
Array with very large size (e.g., 10^5 elements)	Ensure the algorithm's time complexity is efficient enough to handle large inputs (e.g., O(n) or O(n log n)).

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 the array?
Can the input array `nums` contain negative integers or zero?
If the array is already sorted in non-decreasing order, should I return 0?
Could you clarify the desired output if the input `nums` is null or empty?
When replacing an element, is there a limit to the number of elements I can replace it with (aside from the sum constraint)?

Brute Force Solution

Approach

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

Start by considering the last number in the sequence; it's already 'sorted' by itself.
Now, look at the second to last number. Try all the possible ways to break it down into smaller numbers, making sure that the largest of the smaller numbers is less than or equal to the last number we already considered.
For each of these breakdowns of the second to last number, remember how many replacements were needed.
Move one step back, to the third to last number. Now, for each way you broke down the second to last number, try all possible ways to break down the third to last number, ensuring the largest of the new numbers is less than or equal to the smallest number from the previous breakdown.
Again, keep track of how many new replacements you are making.
Continue this process all the way to the beginning of the sequence, trying all possible breakdowns for each number and keeping track of the total replacements.
After considering all the possible breakdowns for all the numbers, choose the breakdown that results in the smallest number of replacements overall.

Code Implementation

def minimum_replacements_brute_force(numbers):
    def solve(index, previous_minimum):
        if index == -1:
            return 0

        current_number = numbers[index]
        minimum_replacements = float('inf')

        # Iterate through possible number of splits
        for number_of_splits in range(1, current_number + 1):
            largest_value = current_number // number_of_splits
            if current_number % number_of_splits != 0:
                largest_value += 1

            # Check if the largest value is less than or equal to previous minimum.
            if largest_value <= previous_minimum:

                replacements = number_of_splits - 1

                # Recursively find the minimum replacements for the rest of the array
                recursive_replacements = solve(index - 1, current_number // number_of_splits)

                if current_number % number_of_splits != 0:
                    recursive_replacements = solve(index - 1, largest_value - 1)

                minimum_replacements = min(minimum_replacements, replacements + recursive_replacements)

        return minimum_replacements

    # Start the recursion with the last element as the initial minimum.
    return solve(len(numbers) - 1, float('inf'))

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start from the rightmost element of the sequence, as this will be the basis for determining our replacements.
Consider the element directly to the left of the last element. Our goal is to make sure this element is no bigger than the element to its right.
If the element is already smaller or equal, we move on to the next element to the left. No replacement is needed here.
If the element is bigger, we need to split it into multiple smaller elements. The number of smaller elements we create is determined by how many pieces are needed so that the largest of those pieces does not exceed the value of the element to the right.
We perform the splitting such that each piece is of roughly equal size.
After splitting, we update the current element with the largest element we just created (which is less than or equal to its neighbor to the right), and we keep track of the number of replacements made. Now we move to the element just to the left of the split elements and repeat the process.
Continue this process moving from right to left, until you reach the beginning of the sequence.
The total number of splits is the answer.

Code Implementation

def minimum_replacements(numbers):
    number_of_replacements = 0
    last_element = numbers[-1]

    for i in range(len(numbers) - 2, -1, -1):
        current_element = numbers[i]

        # If sorted, move to the next element.
        if current_element <= last_element:
            last_element = current_element
            continue

        # Determine the number of pieces to split into.
        number_of_pieces = (current_element + last_element - 1) // last_element

        # This is number of operations required.
        number_of_replacements += number_of_pieces - 1

        # Update last_element to satisfy the sorted condition.
        last_element = current_element // number_of_pieces

    return number_of_replacements

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty array	Return 0 as no operations are needed for an empty array.
Array with one element	Return 0 as a single-element array is already sorted.
Array already sorted in non-decreasing order	Return 0 as no replacements are needed.
Array with two elements, needs one replacement	This serves as a minimal valid test case to check base logic.
Large integer values in the array that could lead to integer overflow when summing.	Use long data type to prevent integer overflow when performing calculations.
Array with all identical values	Return 0 since the array is already sorted.
Array with values sorted in strictly decreasing order	This represents the worst-case scenario requiring maximum replacements.
Array with very large size (e.g., 10^5 elements)	Ensure the algorithm's time complexity is efficient enough to handle large inputs (e.g., O(n) or O(n log n)).