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Formatted question description: https://leetcode.ca/all/2300.html

2300. Successful Pairs of Spells and Potions

  • Difficulty: Medium.
  • Related Topics: Array, Two Pointers, Binary Search, Sorting.
  • Similar Questions: Most Profit Assigning Work, Longest Subsequence With Limited Sum.

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

Solution (Java, C++, Python)

  • class Solution {
        public int[] successfulPairs(int[] spells, int[] potions, long success) {
            Arrays.sort(potions);
            for (int i = 0; i < spells.length; i++) {
                int l = 0;
                int r = potions.length;
                while (l < r) {
                    int m = l + (r - l) / 2;
                    if ((long) spells[i] * potions[m] >= success) {
                        r = m;
                    } else {
                        l = m + 1;
                    }
                }
                spells[i] = potions.length - l;
            }
            return spells;
        }
    }
    
    ############
    
    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:
        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]
    
    ############
    
    # 2300. Successful Pairs of Spells and Potions
    # https://leetcode.com/problems/successful-pairs-of-spells-and-potions/
    
    class Solution:
        def successfulPairs(self, spells: List[int], potions: List[int], success: int) -> List[int]:
            n = len(potions)
            res = []
            potions.sort()
            
            for spell in spells:
                index = bisect_left(potions, success / spell)
                res.append(n - index)
            
            return res
            
            
    
    
  • 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;
        }
    };
    
  • 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;
    }
    
    

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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