How do you solve Successful Pairs of Spells and Potions on LeetCode?

Successful Pairs of Spells and Potions (LeetCode #2300) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + n log m) time complexity and O(1) space complexity. Key insight: If a spell can pair successfully with a potion, it can also pair successfully with all stronger potions.

What is the time complexity of Successful Pairs of Spells and Potions?

The optimal solution for Successful Pairs of Spells and Potions runs in O(n log n + n log m) time and uses O(1) space. The sorting of potions takes O(m log m), and for each spell, we perform a binary search in O(log m), leading to a total of O(n log m).

What is the brute force solution for Successful Pairs of Spells and Potions?

The brute force approach for Successful Pairs of Spells and Potions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every spell against every potion to see if their product meets the success threshold. This is straightforward but inefficient, especially with larger arrays. The optimal Optimal Solution approach improves this to O(n log n + n log m).

What algorithm or data structure is used to solve Successful Pairs of Spells and Potions?

Successful Pairs of Spells and Potions uses the following concepts: Array, Two Pointers, Binary Search, Sorting. The recommended patterns to study are: Binary Search, Sorting.

What are the interview tips for Successful Pairs of Spells and Potions?

Always discuss your thought process and approach before jumping into coding. Consider edge cases, such as minimum and maximum values for spells and potions. Practice explaining your code after writing it, as this is often part of the interview process.

What are common mistakes when solving Successful Pairs of Spells and Potions?

Not considering the case where the potion strength is less than the required strength for the spell. Failing to sort the potions before performing binary search.

#2300

Successful Pairs of Spells and Potions

Medium

ArrayTwo PointersBinary SearchSortingBinary SearchSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + n log m)
Space	O(1)	O(1)

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Intuition

Time O(n log n + n log m)Space O(1)

By sorting the potions and using binary search, we can efficiently find the number of successful potions for each spell. This reduces the time complexity significantly.

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Algorithm

4 steps

1Step 1: Sort the potions array.
2Step 2: For each spell, calculate the minimum required potion strength as success / spell.
3Step 3: Use binary search to find the first potion that meets or exceeds this minimum required strength.
4Step 4: The number of successful potions is the total potions minus the index found in the previous step.

solution.py9 lines

1def successfulPairs(spells, potions, success):
2    potions.sort()
3    result = []
4    from bisect import bisect_left
5    for spell in spells:
6        min_potion = success / spell
7        index = bisect_left(potions, min_potion)
8        result.append(len(potions) - index)
9    return result

ℹ

Complexity note: The sorting of potions takes O(m log m), and for each spell, we perform a binary search in O(log m), leading to a total of O(n log m).

1If a spell can pair successfully with a potion, it can also pair successfully with all stronger potions.
2Sorting allows us to efficiently find the first successful potion using binary search.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.