What is the time complexity of Minimum Division Operations to Make Array Non Decreasing?

The optimal solution for Minimum Division Operations to Make Array Non Decreasing runs in O(n log m) time and uses O(1) space. The complexity is O(n log m) where n is the number of elements and m is the maximum value in the array. This is due to the prime factorization process for each number.

What is the brute force solution for Minimum Division Operations to Make Array Non Decreasing?

The brute force approach for Minimum Division Operations to Make Array Non Decreasing is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible combination of operations to make the array non-decreasing. This means repeatedly dividing each number by its greatest proper divisor until we can no longer do so or until the array becomes non-decreasing. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Minimum Division Operations to Make Array Non Decreasing?

Minimum Division Operations to Make Array Non Decreasing uses the following concepts: Array, Math, Greedy, Number Theory. The recommended patterns to study are: Greedy, Math.

What are the interview tips for Minimum Division Operations to Make Array Non Decreasing?

Always start with a brute-force solution to demonstrate your understanding of the problem. Explain your thought process clearly, especially when transitioning from brute-force to optimal solutions. Practice dry runs of your code to ensure you understand how it works with different inputs.

What are common mistakes when solving Minimum Division Operations to Make Array Non Decreasing?

Failing to check if the array can ever be made non-decreasing and returning -1 prematurely. Not considering the smallest prime divisor for optimal reductions.

#3326

Minimum Division Operations to Make Array Non Decreasing

Medium

ArrayMathGreedyNumber TheoryGreedyMath

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log m)Space O(1)

The optimal approach involves iterating backward through the array and using the smallest prime divisor to reduce numbers efficiently. This allows us to minimize the number of operations needed to achieve a non-decreasing order.

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Algorithm

3 steps

1Step 1: Iterate through the array from the second last element to the first.
2Step 2: For each element, check if it is greater than the next element.
3Step 3: If it is, divide it by its smallest prime divisor until it is less than or equal to the next element, counting operations.

solution.py15 lines

1def min_operations(nums):
2    def smallest_prime_divisor(x):
3        for i in range(2, int(x**0.5) + 1):
4            if x % i == 0:
5                return i
6        return x
7
8    operations = 0
9    n = len(nums)
10    for i in range(n - 2, -1, -1):
11        while nums[i] > nums[i + 1]:
12            spd = smallest_prime_divisor(nums[i])
13            nums[i] //= spd
14            operations += 1
15    return operations

ℹ

Complexity note: The complexity is O(n log m) where n is the number of elements and m is the maximum value in the array. This is due to the prime factorization process for each number.

1Dividing by the greatest proper divisor or smallest prime divisor helps reduce numbers efficiently.
2Iterating backward allows us to ensure that each element is less than or equal to the next.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.