problem

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

submission

impl Solution {
    pub fn successful_pairs(
        spells: Vec<i32>,
        mut potions: Vec<i32>,
        success: i64,
    ) -> Vec<i32> {
        potions.sort();
        let lower_bound = |spell: i32| {
            let (mut left, mut right) = (0, potions.len());
            while left < right {
                let mid = left + (right - left) / 2;
                let strength: i64 = potions[mid] as i64 * spell as i64;
                if strength < success {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            (potions.len() - left) as i32
        };
        spells.into_iter().map(lower_bound).collect()
    }
}