2300.Successful Pairs of Spells and Potions

tags: Medium,Array,Two Pointers,Binary Search,Sorting

2300. Successful Pairs of Spells and Potions

題目描述

You are given two positive integer arrays spells and potions, of length n and m respectively, where spells[i] represents the strength of the i^th spell and potions[j] represents the strength of the j^th 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 i^th spell.

範例

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

解答

C++

思路:

1. 題目: spell強度*potion強度 >= success算成功的pair, 計算每個spell對於每個potion成功的pair有幾個
2. 每個spell其對應的potion不變, 故以potion為主要思考點, 每個potion需要多少spell強度才算成功
3. 排序potion, 因 spell*potion >= success才算成功, 移項得potion >= success/spell, 所以可以binary search potion找小於success/spell的值都是不合格的

vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
        int size = potions.size();
        vector<int> res;
        sort(begin(potions), end(potions));
        for(long long sp : spells)
        {
            int invalid_size = lower_bound(begin(potions), end(potions), ceil((double)success/(double)sp))-potions.begin();
            res.push_back(size-invalid_size);
        }
        return res;
    }

Time:

$O(nlog(n))$
Extra Space:

$O(n)$

XD Apr 4, 2023

Javascript

function successfulPairs(spells, potions, success) {
  const sortedSpells = [];
  for (let i = 0; i < spells.length; i++) {
    sortedSpells.push([spells[i], i]);
  }

  const result = new Array(spells.length).fill(0);
  sortedSpells.sort((a, b) => a[0] - b[0]);
  potions.sort((a, b) => a - b);
  let right = potions.length - 1;

  for (const [spell, index] of sortedSpells) {
    while (right >= 0 && spell * potions[right] >= success) {
      right--;
    }

    result[index] = potions.length - right - 1;
  }

  return result;
}

Sorting + Two Pointers
錯蠻多次的…只想到sorting後去做但還是一直TLE，樓上太神啦
Marsgoat Apr 21, 2023

Reference

回到題目列表