# 2300. Successful Pairs of Spells and Potions

## 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 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 <= 105
• 1 <= spells[i], potions[i] <= 105
• 1 <= success <= 1010

## Solutions

Solution 1: Sorting + Binary Search

We can sort the potion array, then traverse the spell array. For each spell $v$, we use binary search to find the first potion that is greater than or equal to $\frac{success}{v}$. We mark its index as $i$. The length of the potion array minus $i$ is the number of potions that can successfully combine with this spell.

The time complexity is $O((m + n) \times \log m)$, and the space complexity is $O(\log n)$. Here, $m$ and $n$ are the lengths of the potion array and the spell array, respectively.

• class Solution {
public int[] successfulPairs(int[] spells, int[] potions, long success) {
Arrays.sort(potions);
int n = spells.length, m = potions.length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
int left = 0, right = m;
while (left < right) {
int mid = (left + right) >> 1;
if ((long) spells[i] * potions[mid] >= success) {
right = mid;
} else {
left = mid + 1;
}
}
ans[i] = m - left;
}
return ans;
}
}

• class Solution {
public:
vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
sort(potions.begin(), potions.end());
vector<int> ans;
int m = potions.size();
for (int& v : spells) {
int i = lower_bound(potions.begin(), potions.end(), success * 1.0 / v) - potions.begin();
ans.push_back(m - i);
}
return ans;
}
};

• class Solution:
def successfulPairs(
self, spells: List[int], potions: List[int], success: int
) -> List[int]:
potions.sort()
m = len(potions)
return [m - bisect_left(potions, success / v) for v in spells]


• func successfulPairs(spells []int, potions []int, success int64) (ans []int) {
sort.Ints(potions)
m := len(potions)
for _, v := range spells {
i := sort.Search(m, func(i int) bool { return int64(potions[i]*v) >= success })
ans = append(ans, m-i)
}
return ans
}

• function successfulPairs(spells: number[], potions: number[], success: number): number[] {
potions.sort((a, b) => a - b);
const m = potions.length;
const ans: number[] = [];
for (const v of spells) {
let left = 0;
let right = m;
while (left < right) {
const mid = (left + right) >> 1;
if (v * potions[mid] >= success) {
right = mid;
} else {
left = mid + 1;
}
}
ans.push(m - left);
}
return ans;
}