2300. Successful Pairs of Spells and Potions

Problem

You are given two positive integer arrays spells and potions, of length n and m respectively, where spells[i] represents the strength of the ith spell and potions[j] represents the strength of the jth potion.

You are also given an integer success. A spell and potion pair is considered successful if the product of their strengths is at least success.

Return **an integer array *pairs* of length n where pairs[i] is the number of potions that will form a successful pair with the ith spell.**

  Example 1:

Input: spells = [5,1,3], potions = [1,2,3,4,5], success = 7
Output: [4,0,3]
Explanation:
- 0th spell: 5 * [1,2,3,4,5] = [5,10,15,20,25]. 4 pairs are successful.
- 1st spell: 1 * [1,2,3,4,5] = [1,2,3,4,5]. 0 pairs are successful.
- 2nd spell: 3 * [1,2,3,4,5] = [3,6,9,12,15]. 3 pairs are successful.
Thus, [4,0,3] is returned.

Example 2:

Input: spells = [3,1,2], potions = [8,5,8], success = 16
Output: [2,0,2]
Explanation:
- 0th spell: 3 * [8,5,8] = [24,15,24]. 2 pairs are successful.
- 1st spell: 1 * [8,5,8] = [8,5,8]. 0 pairs are successful. 
- 2nd spell: 2 * [8,5,8] = [16,10,16]. 2 pairs are successful. 
Thus, [2,0,2] is returned.

  Constraints:

• n == spells.length

• m == potions.length

• 1 <= n, m <= 10^5

• 1 <= spells[i], potions[i] <= 10^5

• 1 <= success <= 10^10

Solution (Java)

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        Arrays.sort(potions);
        for (int i = 0; i < spells.length; i++) {
            int l = 0;
            int r = potions.length;
            while (l < r) {
                int m = l + (r - l) / 2;
                if ((long) spells[i] * potions[m] >= success) {
                    r = m;
                } else {
                    l = m + 1;
                }
            }
            spells[i] = potions.length - l;
        }
        return spells;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).