Successful Pairs of Spells and Potions.cpp

/*

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


*/


//SOLUTION

class Solution {
public:
    vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
        
        vector<int> res;
        
        sort(potions.begin(),potions.end());
        int n = potions.size();
        
        for(int i = 0; i < spells.size(); i++)
        {
            int pairs = 0;
            int idx = -1;
            
            long long low  = 0;
            long long high = potions.size()-1;
            
            while(low <= high)
            {
                long long mid = low + (high-low)/2;
                
               // long long int temp = spells[i] * potions[mid];
                if((long long int)spells[i] * (long long int)potions[mid] >= success)
                {
                    idx = mid;
                    high = mid-1;
                }
                
                else
                    low = mid + 1;
                
            }
            
            if(idx == -1)
                res.push_back(0);
            
            else
                res.push_back(n-idx);
        }
        
        return res;
        
    }
};