2601. Prime Subtraction Operation - Detailed Explanation

Problem Statement

You are given a 0-indexed array nums of positive integers. You can perform the following operation on each element at most once:

• Operation: Choose any prime number (p) such that (p < \text{nums}[i]) and subtract it from (\text{nums}[i]) (i.e. set (\text{nums}[i] = \text{nums}[i] - p)).

Return true if it is possible to make nums strictly increasing (that is, (\text{nums}[0] < \text{nums}[1] < \dots < \text{nums}[n-1])) after performing some (or none) of the operations; otherwise, return false.

Example 1:

• Input: nums = [5, 8, 9]
• Output: true
• Explanation:
  • For the first element (5):
    • You can subtract a prime from (5).
    • The allowed options are to subtract (2) or (3) (since both are primes less than (5)).
    • To minimize the resulting value, subtract the largest valid prime that still leaves a number greater than the previous element (for the first element, assume the “previous” value is (0)).
    • Here, subtracting (3) gives (5 - 3 = 2).
  • For the second element (8) (with previous value now (2)):
    • The difference (8 - 2 = 6) is greater than (2) so a subtraction may be possible.
    • The primes less than (6) are (2), (3), and (5); the largest is (5), and subtracting it gives (8 - 5 = 3).
  • For the third element (9) (with previous value now (3)):
    • The difference is (9 - 3 = 6); subtracting the largest prime less than (6) (which is (5)) gives (9 - 5 = 4).
  • The final transformed array becomes ([2, 3, 4]), which is strictly increasing.

Example 2:

• Input: nums = [4, 9, 6, 10]
• Output: true
• Explanation:
  • For (4): With a previous value (0), the difference is (4).
    • The valid primes less than (4) are (2) and (3); subtracting (3) gives (4 - 3 = 1).
  • For (9): With previous value (1), the difference is (9 - 1 = 8).
    • The valid primes less than (8) include (2), (3), (5), and (7); subtracting (7) yields (9 - 7 = 2).
  • For (6): With previous value (2), the difference is (6 - 2 = 4).
    • The valid primes less than (4) are (2) and (3); subtracting (3) gives (6 - 3 = 3).
  • For (10): With previous value (3), the difference is (10 - 3 = 7).
    • The valid primes less than (7) are (2), (3), and (5); subtracting (5) yields (10 - 5 = 5).
  • The resulting array is ([1, 2, 3, 5]), which is strictly increasing.

Example 3 (Failure Case):

• Input: nums = [5, 3, 2]
• Output: false
• Explanation:
  • For (5): With previous value (0), subtracting the largest prime (3) (from primes (2) and (3)) yields (5 - 3 = 2).
  • For (3): With previous value (2), the only option is to leave (3) unchanged (because (3 - p) for any valid prime (p) would be too small or not possible). So, the new value is (3), which is greater than (2).
  • For (2): Now, (2) is not greater than the previous value (3) and no operation can increase a number (only subtraction is allowed), so it is impossible to form a strictly increasing sequence.

Constraints:

• (1 \leq \text{nums.length} \leq \text{some reasonable limit}).
• (1 \leq \text{nums}[i] \leq \text{some upper bound}).
• Each element is a positive integer.

Hints Before Solving

• Hint 1: For each element, you have two choices: leave it unchanged or subtract a prime number (only once) to obtain a smaller value.
• Hint 2: To maximize your chances for future elements, it is beneficial to minimize the current element’s new value while still keeping it greater than the previous element.
• Hint 3: For a given number (x) and the current previous value (prev), notice that any subtraction gives you a candidate (x - p). You want the candidate to be as low as possible (to “save room” for later numbers) while satisfying (x - p > prev).

Approaches to Solve the Problem

Approach 1: Greedy with On-the-Fly Prime Checking

Idea:

Process the array from left to right and for each element (x), try to choose the smallest possible new value (by subtracting a prime) that is still greater than the previously chosen new value (denoted by (prev)). You have two options for each element:

• No Operation: Let the new value be (x) (if (x > prev)).
• Prime Subtraction: Choose a prime (p) with (p < x) such that (x - p > prev). To minimize (x - p), you want to subtract as much as possible (i.e. use the largest possible prime) while ensuring (x - p) remains above (prev).

For a given (x) and current (prev), define:

• (L = x - prev).
  You can only subtract a prime (p) if (p < L) (which guarantees (x - p > prev)).
  To minimize the new value (x - p), choose the largest prime (p) that is less than (L).

If no valid prime exists (for instance, if (L \leq 2) because the smallest prime is (2)), you are forced to leave (x) unchanged. If even that does not yield (x > prev) then it is impossible to form a strictly increasing array.

Process:

1. Initialize (prev = 0).
2. Iterate over each number (x) in nums:
   • If (x \leq prev), immediately return false (since you can only decrease numbers).
   • Compute (L = x - prev).
   • Option to Subtract:
     If (L > 2), find the largest prime (p) such that (p < L). Then a candidate new value is (x - p).
   • Default Option:
     Otherwise (or if no valid prime is found), the only option is to use (x) as the new value.
   • Validation:
     If the chosen candidate is ( \leq prev), then return false.
   • Update (prev) with the candidate.
3. After processing all elements, if each element could be transformed to a strictly increasing sequence, return true.

Approach 2: (Alternative Thought) Dynamic Programming

One could also think of this as a state‐transition problem where at each index you consider all possible new values obtainable from (x) (either (x) or (x - p) for valid primes (p)) that are greater than the previous value. Then, use dynamic programming (or backtracking with memoization) to decide if a valid sequence exists.

However, given the constraints and the natural greedy “minimize while keeping above prev” intuition, the greedy approach is more straightforward and efficient.

Code Implementations

Python Implementation