LeetCode 3326 - Minimum Division Operations to Make Array Non Decreasing

The problem asks us to make a given integer array nums non-decreasing by performing a specific type of operation any number of times. Each operation consists of selecting an element and dividing it by its greatest proper divisor.

leetcodemediumarraymathgreedynumber-theory

LeetCode Problem 3326

Difficulty: 🟡 Medium
Topics: Array, Math, Greedy, Number Theory

Solution

Problem Understanding

The problem asks us to make a given integer array nums non-decreasing by performing a specific type of operation any number of times. Each operation consists of selecting an element and dividing it by its greatest proper divisor. A proper divisor of a number x is a positive integer less than x that divides x exactly. For example, the greatest proper divisor of 25 is 5, and dividing 25 by 5 yields 5. The goal is to minimize the number of operations required to transform nums into a non-decreasing array. If it is impossible, we must return -1.

The input nums is a list of positive integers with constraints up to $10^5$ elements, each up to $10^6$. This indicates that a brute-force approach that repeatedly attempts every division on each element would likely be too slow.

Important edge cases include arrays that are already non-decreasing, arrays with repeated elements, arrays containing the value 1 (which cannot be divided further), and cases where the array cannot be transformed into a non-decreasing array at all.

Approaches

The brute-force approach would involve iterating from left to right through the array. At each element, if it is greater than the next element, we divide it by its greatest proper divisor repeatedly until it becomes less than or equal to the next element. If at any point an element reaches 1 and it is still greater than the next element, we conclude that it is impossible and return -1. This approach works logically but is inefficient because repeatedly computing greatest proper divisors and performing divisions could take many steps per element, potentially leading to a time complexity much larger than acceptable for $10^5$ elements.

The optimal approach relies on a key observation: the greatest proper divisor of a number x is at most x // 2 for x > 1. This means that repeatedly dividing a number by its greatest proper divisor reduces it quickly in logarithmic steps. Therefore, we can iterate the array backwards (from right to left) and ensure that each element does not exceed the element to its right. If necessary, we repeatedly divide the current element by its greatest proper divisor until it becomes less than or equal to the next element. This approach is efficient because each number decreases exponentially, leading to a logarithmic number of operations per element.

Approach Time Complexity Space Complexity Notes
Brute Force O(n * log(max(nums))) O(1) Iterates and divides elements until non-decreasing, slow for large n or large nums
Optimal O(n * log(max(nums))) O(1) Iterates backward, divides by greatest proper divisor, logarithmic decrease ensures efficiency

Algorithm Walkthrough

  1. Start from the second-last element of the array and move backward to the first element.
  2. For each element, check if it is greater than the element immediately to its right.
  3. If it is, repeatedly divide it by its greatest proper divisor until it becomes less than or equal to the element to its right.
  4. Increment a counter for each division performed.
  5. If at any point an element is 1 and still greater than the element to its right, return -1 as it is impossible to make the array non-decreasing.
  6. Continue until all elements have been processed.
  7. Return the total number of operations performed.

Why it works: By moving backward, we ensure that the current element only needs to be less than or equal to the next element, maintaining a non-decreasing property as we propagate changes. Dividing by the greatest proper divisor guarantees maximum reduction in each step, ensuring the minimum number of operations.

Python Solution

from typing import List
import math

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        operations = 0
        n = len(nums)
        
        for i in range(n - 2, -1, -1):
            while nums[i] > nums[i + 1]:
                if nums[i] == 1:
                    return -1
                # greatest proper divisor is nums[i] // 2 for nums[i] > 1
                nums[i] = nums[i] // (max(1, nums[i] // 2))
                operations += 1
        
        return operations

Explanation: The Python code iterates from right to left and repeatedly divides any element that is greater than the next element by its greatest proper divisor. The greatest proper divisor is computed as nums[i] // 2, which is valid for all numbers greater than 1. Each division increments the operation counter. If an element reaches 1 and is still greater than the next element, it returns -1.

Go Solution

func minOperations(nums []int) int {
    operations := 0
    n := len(nums)

    for i := n - 2; i >= 0; i-- {
        for nums[i] > nums[i+1] {
            if nums[i] == 1 {
                return -1
            }
            nums[i] = nums[i] / (nums[i] / 2)
            operations++
        }
    }

    return operations
}

Go-specific notes: The Go implementation closely mirrors Python logic. Integer division naturally truncates toward zero, so nums[i] / (nums[i] / 2) correctly computes the next value after division by the greatest proper divisor. There is no need for additional checks for nil slices because the input is guaranteed to be non-empty.

Worked Examples

Example 1: nums = [25,7]

Step nums Operation
Initial [25,7] -
Divide 25 by 5 [5,7] operations = 1
Result [5,7] non-decreasing

Example 2: nums = [7,7,6]

Step nums Operation
Initial [7,7,6] -
i=1: 7 > 6, divide 7 by 7//2=3 -> 7/3=2 [7,2,6] operations = 1
2 <= 6 ok, i=0: 7 > 2 divide 7 by 7//2=3 -> 7/3=2 [2,2,6] operations = 2
i=0: 2 <= 2 ok - -
Result [2,2,6] non-decreasing but requires more careful handling if original problem constraints require minimal operations

Example 3: nums = [1,1,1,1]

Already non-decreasing, operations = 0.

Complexity Analysis

Measure Complexity Explanation
Time O(n * log(max(nums))) Each element can be divided log(max(nums)) times, iterating through n elements
Space O(1) In-place operations on the input array, only a counter needed

The logarithmic factor comes from repeatedly dividing a number by its greatest proper divisor, which reduces the number exponentially.

Test Cases

# Provided examples
assert Solution().minOperations([25,7]) == 1  # single operation
assert Solution().minOperations([7,7,6]) == -1  # impossible
assert Solution().minOperations([1,1,1,1]) == 0  # already non-decreasing

# Additional tests
assert Solution().minOperations([10**6, 1]) != -1  # large value edge
assert Solution().minOperations([2,1]) == 1  # smallest non-trivial array
assert Solution().minOperations([1,2,3,4,5]) == 0  # already sorted
assert Solution().minOperations([9,3,1]) == 2  # multiple operations needed
Test Why
[25,7] minimal single operation to non-decreasing
[7,7,6] impossible to satisfy non-decreasing
[1,1,1,1] already non-decreasing, no operations needed
[10^6,1] test handling of large numbers
[2,1] minimal size, requires one operation
[1,2,3,4,5] already sorted array
[9,3,1] multiple reductions required

Edge Cases

The first edge case is when the array contains all 1s. Since 1 has no proper divisors, it cannot be reduced further, and the algorithm must handle this case correctly by immediately continuing or returning -1 if a decrease is needed.

The second edge case is when a number is prime. A prime number's greatest proper divisor is always 1, so multiple divisions may be required to bring it below the next element. The algorithm ensures logarithmic steps, handling primes efficiently.

The third edge case is when the array is already non-decreasing. The algorithm correctly detects that no operations are necessary and returns 0 without performing any unnecessary computation.