Prime Subtraction Operation

Problem Statement

In this problem, you are provided with an array nums of integers, which is 0-indexed and has a length of n. The task is to determine if you can convert this array into a strictly increasing sequence through a series of operations. Each operation involves selecting an unused index i from the array, choosing a prime number p that is strictly less than the value nums[i] at that index, and then subtracting p from nums[i].

A key detail here is that each index can be used only once and each operation strictly reduces the value at that index by subtracting a prime number from it. The goal is to reorder or adjust the values such that every element in the array after the operations is strictly greater than its preceding element.

Examples

Example 1

Input:

nums = [4,9,6,10]

Output:

true

Explanation:

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

Explanation:

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

Explanation:

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

Approach and Intuition

Let's analyze and strategize how to approach the problem based on the given examples:

1. Examine if the array is already in strictly increasing order. If true, return true immediately as no operations are needed.
2. For an array not already in order:
   • Assess each element following the first one. If any element is not larger than its previous one, identify a suitable prime p and perform the subtraction.
   • When choosing a prime, it should be large enough to make a difference but also keep the modified number greater than its preceding number in the list where possible.
3. Ensure that once an index is used, it isn't considered again, meaning each i in nums can only be decreased once.
4. After attempting or performing operations, check if the adjusted sequence is strictly increasing. If it is, return true, otherwise, return false.

The above strategy must take into account:

• The set of prime numbers less than the smallest number in nums, as they would be the candidates for the subtraction operation.
• Efficient verification at each step to assess if the sequence is becoming strictly increasing.
• Managing the state of which indices have been used.

These aspects are fundamental to designing an efficient solution. Considerations on constraints highlight that with an array size up to 1000 and values up to 1000, the operations are computationally feasible within acceptable limits.

Solutions

class Solution {
public:
    bool isSubOperationPrime(vector<int>& elements) {
        int maximum = *max_element(elements.begin(), elements.end());
    
        // Prepare the prime check array.
        vector<int> primeCheck(maximum + 1, 1);
        primeCheck[1] = 0;
        for (int i = 2; i <= sqrt(maximum); i++) {
            if (primeCheck[i] == 1) {
                for (int j = i * i; j <= maximum; j += i) {
                    primeCheck[j] = 0;
                }
            }
        }
    
        // Initialize necessary values and index.
        int desiredValue = 1;
        int index = 0;
        while (index < elements.size()) {
            // Calculate the difference required to match the desiredValue.
            int requiredDifference = elements[index] - desiredValue;
    
            // Check if the requiredDifference is negative, implying we cannot match.
            if (requiredDifference < 0) {
                return false;
            }
    
            // If the requiredDifference is either prime or zero, proceed to update.
            if (primeCheck[requiredDifference] == 1 or requiredDifference == 0) {
                index++;
                desiredValue++;
            } else {
                // Advance to the next desiredValue if the current is unmatched.
                desiredValue++;
            }
        }
        return true;
    }
};