How do you solve Prime Subtraction Operation on LeetCode?

Prime Subtraction Operation (LeetCode #2601) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the properties of prime numbers can help optimize the solution.

What is the brute force solution for Prime Subtraction Operation?

The brute force approach for Prime Subtraction Operation is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try to make the array strictly increasing by checking each element and attempting to subtract every prime less than that element. This approach is simple but inefficient as it checks all possibilities. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Prime Subtraction Operation?

Prime Subtraction Operation uses the following concepts: Array, Math, Binary Search, Greedy, Number Theory. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Prime Subtraction Operation?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as already sorted arrays or arrays with repeated elements. Explain your thought process clearly while coding, as it helps interviewers understand your approach.

What are common mistakes when solving Prime Subtraction Operation?

Not checking if the adjusted number is strictly greater than the previous one. Failing to generate or utilize prime numbers effectively.

#2601

Prime Subtraction Operation

Medium

ArrayMathBinary SearchGreedyNumber TheoryGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of checking all primes for each element, we can directly calculate the largest prime we can subtract to ensure the array is strictly increasing. This reduces unnecessary checks and speeds up the process.

⚙️

Algorithm

5 steps

1Step 1: Generate a list of all prime numbers less than 1000.
2Step 2: Iterate through the nums array from the second element to the last.
3Step 3: For each element, calculate the minimum value it must be greater than (the previous element).
4Step 4: Find the largest prime less than the current element that can be subtracted to meet the condition.
5Step 5: If no such prime exists, return false. If all elements can be adjusted successfully, return true.

solution.py15 lines

1# Full working Python code
2from sympy import primerange
3
4
5def can_make_increasing(nums):
6    primes = list(primerange(1, 1000))
7    for i in range(1, len(nums)):
8        required = nums[i - 1] + 1
9        for p in reversed(primes):
10            if nums[i] - p >= required:
11                nums[i] -= p
12                break
13        else:
14            return False
15    return True

ℹ

Complexity note: The time complexity is O(n) because we iterate through the nums array once and check primes in a more efficient manner, reducing unnecessary checks.

1Understanding the properties of prime numbers can help optimize the solution.
2The order of operations matters; always consider the previous element when adjusting the current one.

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