This problem requires iterating backward through the array and greedily dividing elements by their greatest proper divisors. The goal is to ensure each number does not exceed the following element, maintaining a non-decreasing sequence. The solution focuses on minimizing operations while validating the invariant that no element surpasses its right neighbor after each division.
Problem Statement
Given an integer array nums, you can perform any number of operations where you select one element and divide it by its greatest proper divisor. A proper divisor of a number x is any positive integer less than x that divides x evenly.
Your task is to find the minimum number of such division operations needed to transform nums into a non-decreasing array. If it is impossible to achieve a non-decreasing sequence, return -1.
Examples
Example 1
Input: nums = [25,7]
Output: 1
Using a single operation, 25 gets divided by 5 and nums becomes [5, 7] .
Example 2
Input: nums = [7,7,6]
Output: -1
Example details omitted.
Example 3
Input: nums = [1,1,1,1]
Output: 0
Example details omitted.
Constraints
- 1 <= nums.length <= 105
- 1 <= nums[i] <= 106
Solution Approach
Iterate Backwards with Greedy Division
Start from the end of the array and move left, ensuring each element is less than or equal to the next. If an element is larger, divide it by its largest proper divisor repeatedly until the invariant holds.
Count Operations Efficiently
Maintain a counter for each division performed. Since each division reduces the number, ensure no unnecessary divisions are applied beyond what is needed to satisfy the non-decreasing condition.
Check for Impossible Cases
If an element becomes 1 and is still larger than its next neighbor, the sequence cannot be fixed. Immediately return -1 to handle these failure cases efficiently.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity depends on the number of division operations applied to each element, potentially O(n log m) where n is array length and m is the maximum element. Space complexity is O(1) extra beyond the input array.
What Interviewers Usually Probe
- Are you considering the largest proper divisor in each step?
- Can you explain why iterating backward simplifies maintaining the invariant?
- How do you handle elements that cannot be reduced further but still violate the non-decreasing property?
Common Pitfalls or Variants
Common pitfalls
- Dividing elements unnecessarily instead of only when they violate the non-decreasing condition.
- Iterating forward instead of backward, which complicates invariant validation.
- Forgetting to return -1 when an element cannot be reduced enough to satisfy the sequence.
Follow-up variants
- Compute minimum multiplication operations to make an array non-increasing.
- Allow division by any proper divisor rather than only the greatest and track minimum operations.
- Apply similar greedy operations on 2D arrays row by row while maintaining non-decreasing order.
How GhostInterview Helps
- Guides you through selecting the correct element and divisor for each operation to minimize steps.
- Visualizes the backward iteration and invariant checks to prevent unnecessary operations.
- Highlights impossible cases early, reducing trial-and-error during coding.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the key greedy strategy for this problem?
Always divide elements from right to left using the largest proper divisor to maintain the non-decreasing invariant.
How do I know if the array cannot be fixed?
If an element reaches 1 and is still larger than the next element, return -1 because no further valid operations exist.
Can I divide by divisors other than the greatest proper divisor?
The problem specifies using the greatest proper divisor, as smaller divisors may require more operations and break the greedy minimization.
Does this solution work efficiently for large arrays?
Yes, iterating backward and applying divisions only when necessary ensures the algorithm scales with array length and maximum element size.
Is there a variation involving multiplication instead of division?
Yes, a similar greedy approach can be applied in reverse to make arrays non-increasing using multiplication operations.
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