Prime Subtraction Operation

#834 Most AskedMedium

Prime Subtraction Operation #834 Most Asked

Medium

Bloomberg LP

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Topics:

ArraysGreedy Algorithms

You are given a 0-indexed integer array nums of length n.

You can perform the following operation as many times as you want:

Pick an index i that you haven’t picked before, and pick a prime p strictly less than nums[i], then subtract p from nums[i].

Return true if you can make nums a strictly increasing array using the above operation and false otherwise.

A strictly increasing array is an array whose each element is strictly greater than its preceding element.

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers in the input array `nums`?
Can the input array `nums` contain zero or negative numbers?
Is the list of primes guaranteed to be precomputed and readily available, or do I need to generate it myself?
If it is impossible to make the array strictly increasing using prime subtraction, what should I return (e.g., `true` or `false`)?
Are there any constraints on the size of the input array `nums`?

Brute Force Solution

Approach

The brute force method involves checking every single possible way to modify the numbers in the list, ensuring they become smaller as we go. We'll subtract different prime numbers from each number in the list, if possible, to make the entire list strictly increasing.

Here's how the algorithm would work step-by-step:

Start with the first number in the list.
Consider every prime number smaller than the current number, one at a time.
Subtract each prime number from the current number.
If, after subtracting a prime, the current number is now bigger than the number before it (or if it's the first number), move on to the next number in the list.
If subtracting any prime doesn't work (meaning it's still smaller than the number before it or we've tried all primes), move on and try subtracting 0 (do nothing to this number), and keep going.
Repeat this process for every number in the list.
If, after trying every possibility, we find at least one way to make the whole list strictly increasing, then it is possible to do, otherwise, it's impossible.

Code Implementation

def prime_subtraction_operation(numbers):
    def is_strictly_increasing(current_numbers):
        for index in range(1, len(current_numbers)):
            if current_numbers[index] <= current_numbers[index - 1]:
                return False
        return True

    def find_primes_less_than(number):
        primes = []
        for current_number in range(2, number):
            is_prime = True
            for divisor in range(2, int(current_number**0.5) + 1):
                if current_number % divisor == 0:
                    is_prime = False
                    break
            if is_prime:
                primes.append(current_number)
        return primes

    def solve(index, current_numbers):
        if index == len(numbers):
            return is_strictly_increasing(current_numbers)

        primes = find_primes_less_than(numbers[index])

        # We must try subtracting every possible prime
        for prime in primes:
            new_number = numbers[index] - prime

            # Skip if it violates the strictly increasing condition
            if index > 0 and new_number <= current_numbers[index - 1]:
                continue

            if solve(index + 1, current_numbers + [new_number]):
                return True

        # Try not subtracting any prime
        if index > 0 and numbers[index] <= current_numbers[index - 1]:
            return False

        if solve(index + 1, current_numbers + [numbers[index]]):
            return True

        return False

    return solve(0, [])

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through each of the n numbers in the input list nums. For each number, it iterates through prime numbers less than that number, up to a maximum of m primes in the worst case (where m is dependent on the value of the largest number in nums). The algorithm uses backtracking and needs to potentially re-evaluate earlier decisions each time, making the effective number of prime subtractions per number close to m. Thus the complexity is approximately n multiplied by m resulting in O(n*m), with m bound by the magnitude of numbers in the array, and potentially reaching n in the worst case.

Space Complexity

O(N) – The described brute force method appears to be implemented using recursion to explore all possible combinations of prime subtractions. In the worst-case scenario, the recursion depth could reach N, where N is the number of elements in the input list. Each recursive call requires space on the call stack for function arguments and local variables, leading to a maximum auxiliary space usage proportional to N. Thus, the space complexity is O(N).

Optimal Solution

Approach

The problem involves making each number in a list smaller than the number before it by subtracting prime numbers. The optimal approach is to go through the numbers one by one, and for each number, subtract the smallest possible prime to make it smaller than the previous number.

Here's how the algorithm would work step-by-step:

Start by checking if the first number is already smaller than any previous number (since there are no previous numbers, this check is skipped the first time).
For each subsequent number, compare it to the number right before it.
If the current number is already smaller, move on to the next number.
If the current number is equal to or bigger than the number before it, find the smallest prime number that you can subtract from the current number to make it strictly smaller than the number before it.
Subtract this prime number from the current number.
If you can't find such a prime number (meaning you can't make the current number smaller than the one before it), then you can't achieve the desired outcome, and the answer is 'no'.
Keep going through the list, making each number smaller than the number before it.
If you make it through the entire list without ever failing to find a suitable prime to subtract, then you've achieved the desired outcome, and the answer is 'yes'.

Code Implementation

def prime_subtraction_operation(numbers) -> bool:
    previous_number = float('inf')

    def is_prime(number_to_check):
        if number_to_check <= 1:
            return False
        for i in range(2, int(number_to_check**0.5) + 1):
            if number_to_check % i == 0:
                return False
        return True

    for current_number in numbers:
        # Check if the current number needs adjustment.
        if current_number >= previous_number:
            found_prime = False
            for prime_candidate in range(2, current_number + 1):
                if is_prime(prime_candidate) and current_number - prime_candidate < previous_number:

                    # Subtract smallest prime to meet the condition.
                    current_number -= prime_candidate
                    found_prime = True

                    break

            # If no suitable prime is found, it's impossible.
            if not found_prime:
                return False

        previous_number = current_number

    return True

Big(O) Analysis

Time Complexity

O(n * sqrt(M)) – The outer loop iterates through each of the n elements in the input array nums. For each element nums[i], we potentially need to find a suitable prime number to subtract. Finding the smallest prime involves iterating up to a maximum value determined by the current element (call this M, the maximum possible value in nums). The primality test itself has a time complexity of approximately sqrt(M). Therefore, in the worst case, for each of the n elements, we might perform a primality check up to M, giving us a time complexity of O(n * sqrt(M)).

Space Complexity

O(sqrt(M)) – The space complexity is dominated by the prime number generation if it's not precomputed. If a prime generation function such as the Sieve of Eratosthenes is used up to a certain limit M where M is the maximum value in the input array, then the space used is proportional to sqrt(M) due to optimization techniques or inherent sieve properties. The main algorithm itself only uses a few variables to keep track of the current and previous numbers which constitutes constant space, but is dwarfed by the space used for prime generation.

Edge Cases

Case	How to Handle
Null or empty input array	Return true if the array is null or empty, as it's already considered non-decreasing.
Array with one element	Return true if the array has only one element because it is already considered non-decreasing.
Array is already sorted in non-decreasing order	The algorithm should still execute correctly and return true without modifying the array.
Array with all elements equal	The algorithm should attempt to subtract primes but ultimately return true as it will remain sorted.
Large array with very large numbers that could lead to integer overflow during subtraction	Use appropriate data types (e.g., long) to avoid potential integer overflow during subtraction and comparison.
An element is smaller than the next, and no prime number can be subtracted from it to make it smaller or equal.	In such a case, return false because a non-decreasing order cannot be achieved.
The smallest element in array is less than the smallest prime number (2)	Handle this case gracefully, ensuring the subtraction doesn't result in negative numbers or incorrect comparisons.
Array with a very large number of elements, impacting time complexity for prime number generation.	Optimize prime generation by pre-calculating primes up to the maximum value in the array or using an efficient prime sieve.

Null or empty input array

How to Handle:

Return true if the array is null or empty, as it's already considered non-decreasing.

Array with one element

How to Handle:

Return true if the array has only one element because it is already considered non-decreasing.

Array is already sorted in non-decreasing order

How to Handle:

The algorithm should still execute correctly and return true without modifying the array.

Array with all elements equal

How to Handle:

The algorithm should attempt to subtract primes but ultimately return true as it will remain sorted.

Large array with very large numbers that could lead to integer overflow during subtraction

How to Handle:

Use appropriate data types (e.g., long) to avoid potential integer overflow during subtraction and comparison.

An element is smaller than the next, and no prime number can be subtracted from it to make it smaller or equal.

How to Handle:

In such a case, return false because a non-decreasing order cannot be achieved.

The smallest element in array is less than the smallest prime number (2)

How to Handle:

Handle this case gracefully, ensuring the subtraction doesn't result in negative numbers or incorrect comparisons.

Array with a very large number of elements, impacting time complexity for prime number generation.

How to Handle:

Optimize prime generation by pre-calculating primes up to the maximum value in the array or using an efficient prime sieve.