2366. Minimum Replacements to Sort the Array

Problem Description

You are provided with an array of integers called nums. The goal is to sort this array into non-decreasing order (where each number is less than or equal to the next) by performing a specific operation as many times as needed. The operation you can perform involves replacing a single element in the array with any two elements that sum up to the original element.

For instance, if you have an element 6 in the array, you can replace it with two elements 2 and 4 since 2 + 4 = 6. The challenge lies in figuring out the minimum number of such operations required to sort the entire array.

Intuition

The solution to this problem hinges on a key observation: replacing any number with two smaller numbers can potentially make sorting the array harder, as it introduces more elements that need to be in non-decreasing order. Therefore, we should aim to perform each replacement in a way that either maintains the current order or requires the least amount of subsequent operations.

To minimize operations, we work our way from the end of the array (the largest elements in a sorted array) to the beginning, adjusting each element so that it's not larger than the one already considered and positioned at the end.

As we move backwards:

1. We want the current element to be less than or equal to the element after it (since the array must be in non-decreasing order).
2. If the current element is already less than or equal to the next element, no operation is needed, and we move to the previous element.
3. If the current element is larger, we need to split it into smaller numbers in a way that requires the minimum number of splits while ensuring that each new number is not larger than the next element.

The algorithm keeps track of the maximum allowed value (mx) for each replacement, which initially is the value of the last element. It computes the minimum number of splits (k) needed for the current element to satisfy the non-decreasing order constraint, and updates the answer (ans) with the number of operations needed (which is k - 1, since replacing one number with two requires one operation). After this, it updates the maximum allowed value for the next element accordingly.

The process continues until we've considered all elements of the array, and the minimum number of operations required to achieve a non-decreasing array is returned.

Learn more about Greedy and Math patterns.

Solution Approach

To implement the solution, we follow a reverse iteration. We initiate by setting the variable mx to the value of the last element in the array since we aim to ensure that all preceding numbers are less than or equal to this value.

The main logic resides in a for-loop that starts from the second-to-last element and moves towards the first element (index 0). Here's a breakdown:

1. For-loop iteration: The loop starts at index n - 2 because we're comparing each element with the one after it. We decrement our index with every iteration, essentially moving from the end of the array to the start.

2. Conditional Operation: For each number, we check if it's less than or equal to mx. If it is, we're already maintaining the non-decreasing order, so we set mx to the current number (since this can be the new maximum for the upcoming previous elements) and continue to the next iteration without any further action.

   1if nums[i] <= mx:
   2    mx = nums[i]
   3    continue

3. Calculation of Splits (k): When the current number is greater than mx, we calculate k, the minimum number of parts we need to split the current number into. This must be done such that each part does not exceed the value of mx. The calculation of k is given by (nums[i] + mx - 1) // mx. We add mx - 1 before integer division to ensure that all parts will be as large as possible without exceeding mx.

4. Updating Answer: The number of new elements added (which is equal to the number of operations performed) will be k - 1. This value is added to ans, the variable tallying the minimum number of operations.

   1ans += k - 1

5. Updating mx for the next iteration: We update mx to the largest possible value that a part can take after the split, which is nums[i] // k. This new value of mx will now act as the upper limit for the next (actually the previous since we're iterating in reverse) number in the array.

6. Returning the result: Once the loop has been completed, the variable ans will hold the minimum number of operations needed to sort the array, which is returned as the result.

The algorithm does not require any additional data structures; it operates in-place, using only a few additional variables for keeping track of the state (ans, mx, k). It's a simple, yet efficient solution with a time complexity of O(n) as it requires a single iteration through the array.

Example Walkthrough

Let's consider an example to illustrate the solution approach. Suppose we are given the following array nums:

1nums = [10, 5, 13]

We need to sort this array into non-decreasing order by replacing elements into sums as needed. Let's walk through the procedure step by step.

1. We start by setting mx to the value of the last element in nums, which is 13. This is the maximum allowed value for its preceding elements.

2. We then begin iterating from the second-to-last element, which is 5 at index 1.

   • For nums[1] which is 5, since 5 ≤ mx (13), we do not need to perform any operations. We set mx to 5 (since 5 now becomes the maximum allowed value for the preceding element) and continue.

3. Next, we check nums[0], which is 10. Since 10 > mx (5 now), we need to perform operations. Calculating k gives us:

   1k = (nums[0] + mx - 1) // mx = (10 + 5 - 1) // 5 = 2

   Since k is 2, it means we need to split 10 into two parts, each not exceeding 5 (the current mx). We can split it into [5, 5]. The number of new elements added is k - 1 which is 1. Thus, we have 1 operation performed.

4. We update ans with the number of operations performed, so ans becomes 1.

5. We update mx to the largest value possible after the splits, which is nums[0] // k = 10 // 2 = 5.

6. Having iterated over all elements, we conclude that we needed a minimum of 1 operation to sort the array into non-decreasing order. So the answer (ans) is 1.

This is the result of using the approach described in the solution. The process is time-efficient as it iterates through the array just once, making the complexity O(n).

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code provided is O(n) where n is the length of the input list nums. This is because the algorithm iterates through the list once in reverse, beginning from the penultimate element to the first element. The operations within each iteration of the loop take constant time, such as comparison, arithmetic operations, and variable assignments. There are no nested loops or additional function calls that would change the linearity of the time complexity.

Space Complexity

The space complexity of the code provided is O(1). The algorithm uses a fixed amount of extra space regardless of the input size. Only a few single-value variables (ans, n, mx) are used for storage, and their space does not scale with the size of the input nums. No additional data structures that would grow with the size of the input are used, so the space usage remains constant.

Learn more about how to find time and space complexity quickly using problem constraints.