LeetCode in Kotlin

2300. Successful Pairs of Spells and Potions

Medium

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:

• 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

Solution

class Solution {
    fun successfulPairs(spells: IntArray, potions: IntArray, success: Long): IntArray {
        potions.sort()
        for (i in spells.indices) {
            var l = 0
            var r = potions.size
            while (l < r) {
                val m = l + (r - l) / 2
                if (spells[i].toLong() * potions[m] >= success) {
                    r = m
                } else {
                    l = m + 1
                }
            }
            spells[i] = potions.size - l
        }
        return spells
    }
}

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