Successful Pairs of Spells and Potions

MEDIUM

Description

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.

Example 1:

Input: spells = [5,1,3], potions = [1,2,3,4,5], success = 7
Output: [4,0,3]
Explanation:
- 0^th spell: 5 * [1,2,3,4,5] = [5,10,15,20,25]. 4 pairs are successful.
- 1^st spell: 1 * [1,2,3,4,5] = [1,2,3,4,5]. 0 pairs are successful.
- 2^nd 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:
- 0^th spell: 3 * [8,5,8] = [24,15,24]. 2 pairs are successful.
- 1^st spell: 1 * [8,5,8] = [8,5,8]. 0 pairs are successful. 
- 2^nd 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

Approaches

Checkout 3 different approaches to solve Successful Pairs of Spells and Potions. Click on different approaches to view the approach and algorithm in detail.

Sorting Both Arrays with Two Pointers

This is a highly optimized approach that involves sorting both the spells and potions arrays and then using a two-pointer technique to find the counts efficiently. By processing spells in increasing order of strength, we can avoid re-scanning the potions array for each spell.

Algorithm

1. Create a 2D array sortedSpells of size n x 2 to store (spell, original_index).
1. Sort sortedSpells based on spell strength.
1. Sort the potions array.
1. Initialize potionIndex = m - 1 and the result array pairs.
1. Iterate i from 0 to n-1 through sortedSpells.
1. Let spell = sortedSpells[i][0] and originalIndex = sortedSpells[i][1].
1. While potionIndex >= 0 and (long) spell * potions[potionIndex] >= success:
1. ```
   Decrement `potionIndex`.
```
1. The count of successful potions is m - 1 - potionIndex.
1. Set pairs[originalIndex] to this count.
1. Return pairs.

The main challenge with sorting spells is that the final result must correspond to the original order of spells. To handle this, we create a 2D array to store each spell's value along with its original index. First, we populate this new structure with (spells[i], i). We then sort this array based on the spell strength and also sort the potions array. We initialize two pointers: one for the sorted spells (say i, from 0 to n-1) and one for the potions array (say j, from m-1 down to 0). We iterate through the sorted spells. The key insight is that as we move to a stronger spell, the required potion strength decreases. Therefore, the j pointer only needs to move from right to left across the potions array and never needs to be reset. For each spell, we move the j pointer to the left (j--) as long as the product is successful. After the inner loop, j points to the strongest potion that is not successful. The number of successful potions is m - 1 - j. We store this count in the result array at the spell's original index, which we saved earlier. Because the j pointer traverses the potions array at most once, the scanning part is very efficient.

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        int n = spells.length;
        int m = potions.length;
        
        int[][] sortedSpells = new int[n][2];
        for (int i = 0; i < n; i++) {
            sortedSpells[i][0] = spells[i];
            sortedSpells[i][1] = i;
        }
        
        Arrays.sort(sortedSpells, Comparator.comparingInt(a -> a[0]));
        Arrays.sort(potions);
        
        int[] pairs = new int[n];
        int potionIndex = m - 1;
        
        for (int i = 0; i < n; i++) {
            int spell = sortedSpells[i][0];
            int originalIndex = sortedSpells[i][1];
            
            while (potionIndex >= 0 && (long) spell * potions[potionIndex] >= success) {
                potionIndex--;
            }
            
            pairs[originalIndex] = m - 1 - potionIndex;
        }
        
        return pairs;
    }
}

Complexity Analysis

Time Complexity: O(n log n + m log m). O(n log n) for sorting spells, O(m log m) for sorting potions. The two-pointer scan takes O(n + m). The dominant part is the sorting.Space Complexity: O(n) to store the `(spell, index)` pairs and the result array.

Pros and Cons

Pros:

Generally the most efficient approach, especially when n and m are of similar magnitude.
The two-pointer scan is very fast in practice due to linear memory access and fewer complex operations per step compared to binary search.

Cons:

Requires extra space to store spell-index pairs.
The logic is the most complex of the three approaches.

Code Solutions

Checking out 3 solutions in different languages for Successful Pairs of Spells and Potions. Click on different languages to view the code.

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;
  }
}

Video Solution

Watch the video walkthrough for Successful Pairs of Spells and Potions

Algorithms:

Binary SearchSorting

Patterns:

Two Pointers

Data Structures:

Array

Companies:

Goldman Sachs |