3326. Minimum Division Operations to Make Array Non Decreasing LeetCode Solution
In this guide, you will get 3326. Minimum Division Operations to Make Array Non Decreasing LeetCode Solution with the best time and space complexity. The solution to Minimum Division Operations to Make Array Non Decreasing problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Division Operations to Make Array Non Decreasing
You are given an integer array nums.
Any positive divisor of a natural number x that is strictly less than x is called a proper divisor of x. For example, 2 is a proper divisor of 4, while 6 is not a proper divisor of 6.
You are allowed to perform an operation any number of times on nums, where in each operation you select any one element from nums and divide it by its greatest proper divisor.
Return the minimum number of operations required to make the array non-decreasing.
If it is not possible to make the array non-decreasing using any number of operations, return -1.
Example 1:
Input: nums = [25,7]
Output: 1
Explanation:
Using a single operation, 25 gets divided by 5 and nums becomes [5, 7].
Time Complexity: O(n \cdot \sqrt{(\max(\texttt{nums}))})
Space Complexity: O(1)
3326. Minimum Division Operations to Make Array Non Decreasing LeetCode Solution in C++
class Solution {
public:
int minOperations(vector<int>& nums) {
int ans = 0;
for (int i = nums.size() - 2; i >= 0; --i)
if (nums[i] > nums[i + 1]) {
const int minDivisor = getMinDivisor(nums[i]);
if (minDivisor > nums[i + 1])
return -1;
nums[i] = minDivisor;
++ans;
}
return ans;
}
private:
int getMinDivisor(int num) {
for (int divisor = 2; divisor <= sqrt(num); ++divisor)
if (num % divisor == 0)
return divisor;
return num;
}
}; /* code provided by PROGIEZ */
3326. Minimum Division Operations to Make Array Non Decreasing LeetCode Solution in Java
class Solution {
public int minOperations(int[] nums) {
int ans = 0;
for (int i = nums.length - 2; i >= 0; --i)
if (nums[i] > nums[i + 1]) {
final int minDivisor = getMinDivisor(nums[i]);
if (minDivisor > nums[i + 1])
return -1;
nums[i] = minDivisor;
++ans;
}
return ans;
}
private int getMinDivisor(int num) {
for (int divisor = 2; divisor <= Math.sqrt(num); ++divisor)
if (num % divisor == 0)
return divisor;
return num;
}
} // code provided by PROGIEZ
3326. Minimum Division Operations to Make Array Non Decreasing LeetCode Solution in Python
class Solution:
def minOperations(self, nums: list[int]) -> int:
ans = 0
for i in range(len(nums) - 2, -1, -1):
if nums[i] > nums[i + 1]:
minDivisor = self._getMinDivisor(nums[i])
if minDivisor > nums[i + 1]:
return -1
nums[i] = minDivisor
ans += 1
return ans
def _getMinDivisor(self, num: int) -> int:
for divisor in range(2, math.isqrt(num) + 1):
if num % divisor == 0:
return divisor
return num # code by PROGIEZ
