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.lengthm == potions.length1 <= n, m <= 1051 <= spells[i], potions[i] <= 1051 <= success <= 1010
class Solution {
public:
int potionCount(vector<int> &potions, long long spell, long long success)
{
long long start=0;
long long end=potions.size()-1;
long long mid=start+(end-start)/2;
long long ans=-1;
while(start<=end)
{
mid=start+(end-start)/2;
if(spell*potions[mid]>=success)
{
ans=mid;
end=mid-1;
}
else if(spell*potions[mid]<success)start=mid+1;
}
if(ans==-1)return 0;
return potions.size()-ans;
}
vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
vector<int> pairs;
sort(potions.begin(),potions.end());
for(auto i:spells)
{
int ans=potionCount(potions,i,success);
pairs.push_back(ans);
}
return pairs;
}
};
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