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2300. 咒语和药水的成功对数

English Version

题目描述

给你两个正整数数组 spells 和 potions ，长度分别为 n 和 m ，其中 spells[i] 表示第 i 个咒语的能量强度，potions[j] 表示第 j 瓶药水的能量强度。

同时给你一个整数 success 。一个咒语和药水的能量强度 相乘 如果 大于等于 success ，那么它们视为一对 成功 的组合。

请你返回一个长度为 n 的整数数组 pairs，其中 pairs[i] 是能跟第 i 个咒语成功组合的 药水 数目。

 

示例 1：

输入：spells = [5,1,3], potions = [1,2,3,4,5], success = 7
输出：[4,0,3]
解释：
- 第 0 个咒语：5 * [1,2,3,4,5] = [5,10,15,20,25] 。总共 4 个成功组合。
- 第 1 个咒语：1 * [1,2,3,4,5] = [1,2,3,4,5] 。总共 0 个成功组合。
- 第 2 个咒语：3 * [1,2,3,4,5] = [3,6,9,12,15] 。总共 3 个成功组合。
所以返回 [4,0,3] 。

示例 2：

输入：spells = [3,1,2], potions = [8,5,8], success = 16
输出：[2,0,2]
解释：
- 第 0 个咒语：3 * [8,5,8] = [24,15,24] 。总共 2 个成功组合。
- 第 1 个咒语：1 * [8,5,8] = [8,5,8] 。总共 0 个成功组合。
- 第 2 个咒语：2 * [8,5,8] = [16,10,16] 。总共 2 个成功组合。
所以返回 [2,0,2] 。

 

提示：

  • n == spells.length
  • m == potions.length
  • 1 <= n, m <= 105
  • 1 <= spells[i], potions[i] <= 105
  • 1 <= success <= 1010

解法

方法一：二分查找

Python3

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, key=lambda x: s * x) for s in spells]

Java

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        Arrays.sort(potions);
        int m = potions.length, n = spells.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;
    }
}

C++

class Solution {
public:
    vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
        sort(potions.begin(), potions.end());
        int m = potions.size();
        vector<int> ans;
        for (int& s : spells) {
            int left = 0, right = m;
            while (left < right) {
                int mid = (left + right) >> 1;
                if (1ll * s * potions[mid] >= success)
                    right = mid;
                else
                    left = mid + 1;
            }
            ans.push_back(m - left);
        }
        return ans;
    }
};

Go

func successfulPairs(spells []int, potions []int, success int64) []int {
	sort.Ints(potions)
	m := len(potions)
	var ans []int
	for _, s := range spells {
		left, right := 0, m
		for left < right {
			mid := (left + right) >> 1
			if int64(s*potions[mid]) >= success {
				right = mid
			} else {
				left = mid + 1
			}
		}
		ans = append(ans, m-left)
	}
	return ans
}

TypeScript

function successfulPairs(
    spells: number[],
    potions: number[],
    success: number,
): number[] {
    const n = spells.length,
        m = potions.length;
    potions.sort((a, b) => a - b);
    let pairs = new Array(n);
    let hashMap = new Map();
    for (let i = 0; i < n; i++) {
        const target = Math.ceil(success / spells[i]);
        let idx = hashMap.get(target);
        if (!idx) {
            idx = searchLeft(potions, 0, m, target);
            hashMap.set(target, idx);
        }
        pairs[i] = m - idx;
    }
    return pairs;
}

function searchLeft(nums, left, right, target) {
    while (left < right) {
        let mid = (left + right) >> 1;
        if (nums[mid] >= target) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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