LeetCode Problem Workspace
Prime Subtraction Operation
Determine if it's possible to make the array strictly increasing using prime subtractions.
5
Topics
5
Code langs
3
Related
Practice Focus
Medium · Binary search over the valid answer space
Answer-first summary
Determine if it's possible to make the array strictly increasing using prime subtractions.
Ace coding interviews with Interview AiBoxInterview AiBox guidance for Binary search over the valid answer space
This problem asks you to determine if you can make an array strictly increasing by subtracting primes. A binary search approach is useful to identify the most optimal prime to subtract from each element. The solution leverages mathematical concepts and binary search to efficiently solve the problem.
Problem Statement
You are given a 0-indexed integer array nums of length n. You can perform the following operation as many times as you want: for any i (0 <= i < n), you may pick a prime p such that nums[i] - p is still a valid number. The goal is to determine if it's possible to use these operations to make the array strictly increasing.
Return true if you can make nums a strictly increasing array using the above operation, and false otherwise. You should focus on using binary search over the valid primes that can be subtracted to achieve the increasing order.
Examples
Example 1
Input: nums = [4,9,6,10]
Output: true
In the first operation: Pick i = 0 and p = 3, and then subtract 3 from nums[0], so that nums becomes [1,9,6,10]. In the second operation: i = 1, p = 7, subtract 7 from nums[1], so nums becomes equal to [1,2,6,10]. After the second operation, nums is sorted in strictly increasing order, so the answer is true.
Example 2
Input: nums = [6,8,11,12]
Output: true
Initially nums is sorted in strictly increasing order, so we don't need to make any operations.
Example 3
Input: nums = [5,8,3]
Output: false
It can be proven that there is no way to perform operations to make nums sorted in strictly increasing order, so the answer is false.
Constraints
- 1 <= nums.length <= 1000
- 1 <= nums[i] <= 1000
- nums.length == n
Solution Approach
Prime Selection via Binary Search
A binary search can be used to efficiently find the largest prime number smaller than or equal to each element in nums. This allows for optimal prime subtraction at each step, improving the solution's efficiency.
Greedy Subtraction Strategy
By using a greedy approach, always subtract the largest valid prime at each index in nums to keep the array increasing. This ensures that the solution is both optimal and correct.
Mathematical Insights into Primes
Understanding prime numbers and their properties is essential. Leveraging prime tables and efficient prime checking algorithms is crucial to solving this problem within time constraints.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(n + m \log \log (m)) |
| Space | O(m) |
The time complexity is O(n + m \log \log (m)) where n is the length of the array and m is the range of values in nums. The space complexity is O(m) due to the prime number storage required for efficient subtraction.
What Interviewers Usually Probe
- Assess the candidate's understanding of prime number properties.
- Look for knowledge in binary search optimization.
- Check if the candidate is comfortable using greedy algorithms.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to subtract the correct prime at each index.
- Not applying binary search efficiently, leading to unnecessary computation.
- Confusing prime subtraction with other array manipulation techniques.
Follow-up variants
- Adjust the problem by limiting the available primes to smaller sets.
- Alter the array length to test scalability and performance.
- Change the problem to require the array to be non-decreasing instead of strictly increasing.
FAQ
What is the optimal approach to solving the Prime Subtraction Operation problem?
The optimal approach involves using binary search to efficiently select the largest valid prime to subtract from each element in the array.
How does binary search improve the solution for the Prime Subtraction Operation problem?
Binary search allows you to quickly find the largest prime smaller than or equal to each element, minimizing unnecessary operations and ensuring efficiency.
Can this problem be solved with a greedy algorithm?
Yes, a greedy algorithm is a key component of the solution, as you always want to subtract the largest possible prime at each step to make the array strictly increasing.
What is the time complexity of the Prime Subtraction Operation problem?
The time complexity is O(n + m \log \log (m)) where n is the length of the array and m is the range of values in nums.
Are there any common mistakes to avoid when solving the Prime Subtraction Operation problem?
Common mistakes include failing to select the correct prime at each index, inefficient prime searching, and misunderstanding the problem's greedy nature.
Solution
Solution 1: Preprocessing prime numbers + binary search
We first preprocess all the primes within $1000$ and record them in the array $p$.
class Solution:
def primeSubOperation(self, nums: List[int]) -> bool:
p = []
for i in range(2, max(nums)):
for j in p:
if i % j == 0:
break
else:
p.append(i)
n = len(nums)
for i in range(n - 2, -1, -1):
if nums[i] < nums[i + 1]:
continue
j = bisect_right(p, nums[i] - nums[i + 1])
if j == len(p) or p[j] >= nums[i]:
return False
nums[i] -= p[j]
return TrueSolution 2: Preprocessing prime numbers
#### TypeScript
class Solution:
def primeSubOperation(self, nums: List[int]) -> bool:
p = []
for i in range(2, max(nums)):
for j in p:
if i % j == 0:
break
else:
p.append(i)
n = len(nums)
for i in range(n - 2, -1, -1):
if nums[i] < nums[i + 1]:
continue
j = bisect_right(p, nums[i] - nums[i + 1])
if j == len(p) or p[j] >= nums[i]:
return False
nums[i] -= p[j]
return TrueContinue Topic
array
Practice more edge cases under the same topic.
arrow_forwardauto_awesomeContinue Pattern
Binary search over the valid answer space
Expand the same solving frame across more problems.
arrow_forwardsignal_cellular_altSame Difficulty Track
Medium
Stay on this level to stabilize interview delivery.
arrow_forward