2601. Prime Subtraction Operation

Problem Description

You are provided with an array of integers, and the goal is to determine if you can transform this array into a strictly increasing array through a specific operation. The operation allows you to pick an index in the array that has not been picked before and then choose a prime number less than the value at that index. You subtract this prime number from the value at the chosen index. An array is considered strictly increasing if every element is larger than the element before it. You are allowed to perform this operation as many times as needed, as long as each index is only picked once. The main challenge is to find out whether it is possible to make the array strictly increasing through these operations and return true if it is possible, or false if it is not.

Intuition

The key insight to solving this problem is understanding that each non-increasing pair of elements in the array needs to be dealt with by subtracting a prime number in such a way as to make the current element less than the next. To accomplish this, we must consider the primes smaller than the current element.

Because we are ensuring strictly increasing order, we need to find primes that are just small enough to decrease the current number, but not so much that we can't maintain the strictly increasing property. We approach this by starting from the second-to-last element and moving backward, checking if the current number needs adjustment with respect to its subsequent number.

To optimize prime number finding and selection, we first pre-generate a list of all prime numbers smaller than the maximum number in the original array. We then use binary search to find the smallest prime number that we can subtract from the current number to make it smaller than the subsequent number.

The algorithm is designed to find the optimal prime for each element from back to front. If, at any point, we cannot find a suitable prime, we know it isn't possible to achieve a strictly increasing array, and we return false. If, after processing all elements, we haven't encountered this issue, we return true, as we've successfully transformed the array into a strictly increasing one.

Learn more about Greedy, Math and Binary Search patterns.

Solution Approach

The implementation of the solution uses two main algorithms: prime number generation and binary search, and it employs the list data structure to store the prime numbers and to manipulate the nums array.

Prime Number Generation

The prime number generation part of the algorithm involves creating a list of prime numbers up to the largest element in the nums array. We initialize an empty list p to store the prime numbers. We then iterate through numbers from 2 to the maximum number found in nums. For each number i, we test whether it is divisible by any previously identified prime numbers in the p list. If i is divisible by any of these primes, it is not prime and we skip to the next number. If no divisors are found, i is prime and we append it to the p list.

Binary Search for Prime Subtraction

Once we have our list of prime numbers, we then iterate through the nums array in reverse order, starting from the second-to-last element. For each element nums[i], we need to find a prime number p[j] that, when subtracted from nums[i], would make nums[i] less than nums[i+1]. This ensures the strictly increasing property for these two elements.

We use bisect_right from Python's bisect module to perform a binary search for the proper prime number that can be subtracted. bisect_right finds the index j in the prime list such that all the primes up to index j-1 are strictly less than nums[i] - nums[i+1]. If j is equal to the length of p or p[j] is greater than or equal to nums[i], it indicates there is no suitable prime that can be subtracted from nums[i] without making it negative or non-increasing with respect to nums[i+1]. In this case, we return false because it is impossible to make the array strictly increasing.

After the loop, if we have successfully found primes for all required elements, we return true, indicating it is possible to form a strictly increasing array with the specified operations.

To summarize, this approach cleverly combines efficient prime generation with binary search to minimize the number of elements that need to be checked and to quickly find the smallest appropriate prime for each subtraction operation. This results in an efficient algorithm suitable for processing arrays to determine if they can be made strictly increasing.

Example Walkthrough

Let's illustrate the solution approach with a small example. Consider the array nums = [10, 5, 15]. We want to determine if we can subtract prime numbers from each element (only once per element) to make the array strictly increasing.

First, we generate the list of prime numbers up to the largest number in nums, which is 15. So our list of primes p up to 15 would be [2, 3, 5, 7, 11, 13].

Following the solution approach:

1. We start from the second-to-last element in nums, which is 5 in this case.

2. We see that 5 is not less than 15 (the subsequent element), so we need to check if we can subtract a prime number from the element [10] to make it less than 5.

3. From the list p, we search for the smallest prime number that would make the element at index 0 smaller than 5 (the next element) when subtracted. Here, we can subtract 7 from 10 (10 - 7 = 3), and 3 is less than 5.

4. The array now looks like [3, 5, 15].

5. Upon iteratively checking all elements, we find that no more operations are needed because the array is already strictly increasing.

6. Since we were able to apply the operations to the array to make it strictly increasing, the final return value would be true.

To summarize the example, we navigated through each element of the array against its subsequent element, using binary search to quickly find a suitable prime to subtract from nums[i] to achieve the strictly increasing property. This process demonstrated the algorithm's strategy of prime subtraction and how it can effectively transform an array while adhering to the constraints of the problem.

Solution Implementation

Time and Space Complexity

The time complexity of the given code can be analyzed in several parts:

1. Generating the prime numbers up to max(nums) involves checking each number against all previously found primes. In the worst case, if max(nums) is a prime number itself, this would generate O(max(nums) / \log(max(nums))) primes, since the density of primes is roughly 1 / \log(n) where n is the number being considered for primality. However, this check is done O(\sqrt{max(num)}) times for each number since we only need to check divisibility up to its square root. This gives us a complexity of O(max(nums) \sqrt{max(nums)} / \log(max(nums))) for the loop generating primes.

2. The main processing of nums array iterates over the elements once, which contributes O(n), where n is the length of nums.

3. The bisect_right function is used within the loop over nums. Assuming that the list of prime numbers is roughly m in size, where m is O(max(nums) / \log(max(nums))), the bisect_right operation contributes O(\log m) complexity each time it's called. It's called once for each of n iterations, so that part contributes O(n \log m).

4. The operation nums[i] -= p[j] is constant time, O(1).

Considering the time complexities together, the dominating term is from generating the primes and the iterative process with binary search. Since the prime generation complexity is less common in analysis, we'll assume that max(nums) is bounded by a polynomial size of n, which means that m is O(n / \log(n)). Hence, we focus on the O(n \log m) term for binary search within the loop. This suggests a time complexity of O(n(\log(n) - \log(\log(n)))), which simplifies to O(n \log n) under the assumption that max(nums) is not exponentially larger than n.

The space complexity of the algorithm is dominated by the list p that stores the primes. As calculated before, this will be about O(max(nums) / \log(max(nums))). Using the same bounding assumption as for the time complexity, we see that this simplifies to O(n), under the assumption that max(nums) is not exponentially larger than n.

So, in agreement with the reference answer, we can say the time complexity is O(n \log n) and the space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.