2601. Prime Subtraction Operation

[mah-shamim]

Topics: Array, Math, Binary Search, Greedy, Number Theory

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

Hint:

1. Think about if we have many primes to subtract from nums[i]. Which prime is more optimal?
2. The most optimal prime to subtract from nums[i] is the one that makes nums[i] the smallest as possible and greater than nums[i-1].

[mah-shamim]

We need to break down the algorithm and adapt it to PHP syntax and functionality. The solution primarily involves the following steps:

1. Generating Primes (Sieve of Eratosthenes): Generate a list of all primes up to the maximum possible value in nums (1000).
2. Prime Subtraction Operation: For each number in nums, check if we can subtract a prime to make the array strictly increasing.
3. Binary Search for Prime: Use a binary search to find the largest prime less than the current number that would still keep the sequence strictly increasing.

Let's implement this solution in PHP: 2601. Prime Subtraction Operation

<?php
class Solution {

    /**
     * @param Integer[] $nums
     * @return Boolean
     */
    function primeSubOperation($nums) {
        $maxNum = 1000;
        $primes = $this->sieveEratosthenes($maxNum);

        $prevNum = 0;
        foreach ($nums as &$num) {
            // Find the largest prime `p` that makes `num - p > prevNum`
            $prime = $this->findLargestPrimeLessThan($primes, $num - $prevNum);
            if ($prime !== null) {
                $num -= $prime;
            }

            // Check if the current number is greater than the previous number
            if ($num <= $prevNum) {
                return false;
            }

            $prevNum = $num;
        }

        return true;
    }

    /**
     * Helper function to generate all primes up to n using Sieve of Eratosthenes
     *
     * @param $n
     * @return array
     */
    private function sieveEratosthenes($n) {
        $isPrime = array_fill(0, $n + 1, true);
        $isPrime[0] = $isPrime[1] = false;
        $primes = [];

        for ($i = 2; $i <= $n; $i++) {
            if ($isPrime[$i]) {
                $primes[] = $i;
                for ($j = $i * $i; $j <= $n; $j += $i) {
                    $isPrime[$j] = false;
                }
            }
        }

        return $primes;
    }

    /**
     * Helper function to find the largest prime less than a given limit using binary search
     *
     * @param $primes
     * @param $limit
     * @return mixed|null
     */
    private function findLargestPrimeLessThan($primes, $limit) {
        $left = 0;
        $right = count($primes) - 1;

        while ($left <= $right) {
            $mid = $left + (int)(($right - $left) / 2);
            if ($primes[$mid] < $limit) {
                $left = $mid + 1;
            } else {
                $right = $mid - 1;
            }
        }

        return $right >= 0 ? $primes[$right] : null;
    }
}

// Example usage:
$solution = new Solution();
echo $solution->primeSubOperation([4, 9, 6, 10]) ? 'true' : 'false';  // Output: true
echo $solution->primeSubOperation([6, 8, 11, 12]) ? 'true' : 'false'; // Output: true
echo $solution->primeSubOperation([5, 8, 3]) ? 'true' : 'false';      // Output: false
?>

Explanation:

1. primeSubOperation: Loops through each element in nums and checks if we can make each element greater than the previous one by subtracting an appropriate prime.

   • We use $this->findLargestPrimeLessThan to find the largest prime less than num - prevNum.
   • If such a prime exists, we subtract it from the current num.
   • If after subtracting the prime, the current num is not greater than prevNum, we return false.
   • Otherwise, we update prevNum to the current num.

2. sieveEratosthenes: Generates all primes up to 1000 using the Sieve of Eratosthenes and returns them as an array.

3. findLargestPrimeLessThan: Uses binary search to find the largest prime less than a given limit, ensuring we find the optimal prime for subtraction.

Complexity Analysis

• Time Complexity: O(n . √m), where n is the length of nums and m is the maximum value of an element in nums (here m = 1000).
• Space Complexity: O(m), used to store the list of primes up to 1000.

This solution will return true or false based on whether it is possible to make nums strictly increasing by performing the described prime subtraction operations.

[basharul-siddike]

Thanks to solve this problem

[mah-shamim]

Thanks