Introduction

In this blog post, we will explore how to solve the LeetCode problem "2300" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

• Binary Search: For each spell, use binary search on the sorted potions array to find the first potion that, when multiplied by the spell, is greater than or equal to success.
  • Sorting: Sort the potions array to enable efficient binary search.
  • Counting: The number of potions greater than or equal to the required value is the number of potions that form a successful pair.

• Runtime Complexity: O(n log m), where n is the length of spells and m is the length of potions. Storage Complexity: O(log m) due to sorting, or O(m) if a non-in-place sorting algorithm is used.

Code

    def successfulPairs(spells, potions, success):
    """
    Finds the number of successful pairs between spells and potions.

    Args:
        spells: A list of integers representing spell strengths.
        potions: A list of integers representing potion strengths.
        success: An integer representing the minimum product required for success.

    Returns:
        A list of integers where each element represents the number of successful pairs for the corresponding spell.
    """

    potions.sort()
    n = len(spells)
    m = len(potions)
    pairs = []

    for spell in spells:
        l, r = 0, m - 1
        count = 0
        while l <= r:
            mid = (l + r) // 2
            if spell * potions[mid] >= success:
                count = m - mid  # All potions from mid to the end are successful
                r = mid - 1  # Try to find an even smaller index that satisfies the condition
            else:
                l = mid + 1
        pairs.append(count)

    return pairs