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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;
}

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