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
Companies: Goldman Sachs
Related Topics:
Array, Two Pointers, Binary Search, Sorting
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// OJ: https://leetcode.com/problems/successful-pairs-of-spells-and-potions
// Author: github.com/lzl124631x
// Time: O(AlogB)
// Space: O(1)
class Solution {
public:
vector<int> successfulPairs(vector<int>& A, vector<int>& B, long
long target) {
sort(begin(B), end(B));
vector<int> ans;
for (int n : A) {
ans.push_back((int)B.size() - (upper_bound(begin(B), end
(B), target, [&](long long target, int &val) {
return (long long)n * val >= target;
}) - begin(B)));
}
return ans;
}
};