2300. Successful Pairs of Spells and Potions

1. Description

You are given two positive integer arrays spells and potions, of length n and m respectively, where spells[i] represents the strength of the $i^{th}$ spell and potions[j] represents the strength of the $j^{th}$ 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 $i^{th}$ spell.

2. Example

Example 1

Input: spells = [5,1,3], potions = [1,2,3,4,5], success = 7
Output: [4,0,3]
Explanation:

• $0^{th}$ spell: 5 * [1,2,3,4,5] = [5,10,15,20,25]. 4 pairs are successful.
• $1^{st}$ spell: 1 * [1,2,3,4,5] = [1,2,3,4,5]. 0 pairs are successful.
• $2^{nd}$ 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:

• $0^{th}$ spell: 3 * [8,5,8] = [24,15,24]. 2 pairs are successful.
• $1^{st}$ spell: 1 * [8,5,8] = [8,5,8]. 0 pairs are successful.
• $2^{nd}$ spell: 2 * [8,5,8] = [16,10,16]. 2 pairs are successful.

Thus, [2,0,2] is returned.

3. Constraints

• n == spells.length
• m == potions.length
• 1 <= n, m <= $10^5$
• 1 <= spells[i], potions[i] <= $10^5$
• 1 <= success <= $10^{10}$

4. Solutions

Binary Search (manual implementation)

m = spells.size(), n = potions.size()
Time complexity: O(nlogn + mlogn)
Space complexity: O(logn)

class Solution {
public:
    vector<int> successfulPairs(const vector<int> &spells, vector<int> potions, long long success) {
        sort(potions.begin(), potions.end());
        vector<int> valid_pairs(spells.size(), 0);
        for (int i = 0; i < spells.size(); ++i) {
            long long min_multiplier = (success + spells[i] - 1) /
                spells[i]; // compute ceil(success / spells[i]) using integer arithmetic
            int left = 0, right = potions.size();
            while (left < right) {
                int middle = left + (right - left) / 2;
                if (potions[middle] < min_multiplier) {
                    left = middle + 1;
                } else {
                    right = middle;
                }
            }

            valid_pairs[i] = potions.size() - left;
        }

        return valid_pairs;
    }
};

Binary Search (STL)

m = spells.size(), n = potions.size()
Time complexity: O(nlogn + mlogn)
Space complexity: O(logn)

class Solution {
public:
    vector<int> successfulPairs(const vector<int> &spells, vector<int> potions, long long success) {
        sort(potions.begin(), potions.end());
        vector<int> valid_pairs(spells.size(), 0);
        for (int i = 0; i < spells.size(); ++i) {
            long long min_multiplier = (success + spells[i] - 1) /
                spells[i]; // compute ceil(success / spells[i]) using integer arithmetic
            valid_pairs[i] =
                potions.end() - lower_bound(potions.begin(), potions.end(), min_multiplier);
        }

        return valid_pairs;
    }
};