3326. Minimum Division Operations to Make Array Non Decreasing - Detailed Explanation

Problem Statement

You are given an array of positive integers. In one operation, you can choose any element and replace it with ⌊x/k⌋ (i.e. x divided by k and rounded down) for any integer k ≥ 2. Your goal is to make the array non-decreasing (i.e. each element is no smaller than the one before it) while minimizing the total number of operations. Return the minimum number of operations needed.

For example, consider an array:   nums = [a₀, a₁, ..., aₙ₋₁]

After applying some operations, the resulting array must satisfy:   a₀' ≤ a₁' ≤ ... ≤ aₙ₋₁'

where each aᵢ' is the value of the element aᵢ after some (possibly zero) division operations.

Hints

1. Process from Right to Left:
   Instead of trying to adjust the array from left to right, work backwards from the last element. The rightmost element sets an upper bound for the elements to its left.

2. Greedy Splitting with Division:
   For a given element that is too large compared to the allowed upper bound (coming from its right neighbor after operations), decide how many parts you should split it into (via division) so that the resulting value is small enough.

   • If an element x must be reduced so that its new value is at most cur (the allowed maximum from the right), choose the smallest integer k such that ⌊x/k⌋ ≤ cur.
   • Note that splitting x into k parts (i.e. applying one division operation that conceptually splits x into k equal pieces) takes (k - 1) operations.

3. Ceiling Division:
   To determine k, you can use the formula:
   [ k = \left\lceil \frac{x}{cur} \right\rceil = \frac{x + cur - 1}{cur} ] Then the new value for x becomes ⌊x/k⌋, and you add (k - 1) to your operation count.

Detailed Approach

1. Initialize:
   Set a variable cur equal to the last element of the array. This value is fixed (you cannot change it because there is nothing to its right that forces a reduction). Also, initialize an operation counter ops = 0.

2. Iterate Backwards:
   For each element from right-2 down to 0:

   • Let the current element be x = nums[i].
   • Case 1: If x ≤ cur, then it is already small enough, and update cur = x (since this element now becomes the new upper bound for all elements to its left).
   • Case 2: If x > cur, you must reduce x. Choose the minimum integer k such that
     [ \lfloor x/k \rfloor ≤ cur. ] This is equivalent to taking
     [ k = \left\lceil \frac{x}{cur} \right\rceil = \frac{x + cur - 1}{cur}. ] Add (k - 1) to your operation counter, and update cur to the new value, i.e., cur = x // k.

3. Result:
   After processing all elements, the accumulated count in ops is the minimum number of division operations needed to make the array non-decreasing.

Why This Greedy Strategy Works

• Right-to-Left Processing:
  By processing from right to left, you determine for each element the maximum allowed value it can have (which is the transformed value of the element immediately to its right). This ensures the non-decreasing property.

• Optimal Splitting:
  For any element x that is too high, splitting it into k parts minimizes the number of operations while ensuring the new value ⌊x/k⌋ is as high as possible but not greater than cur. Using the ceiling division guarantees that you pick the smallest k that achieves this condition.

Complexity Analysis

• Time Complexity: O(n)
  You iterate through the array once (from right to left) and perform O(1) work for each element.

• Space Complexity: O(1)
  You only use a few extra variables regardless of the input size.

Python Code