LeetCode in Kotlin

1726. Tuple with Same Product

Medium

Given an array nums of distinct positive integers, return the number of tuples (a, b, c, d) such that a * b = c * d where a, b, c, and d are elements of nums, and a != b != c != d.

Example 1:

Input: nums = [2,3,4,6]

Output: 8

Explanation: There are 8 valid tuples:

(2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3)

(3,4,2,6) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)

Example 2:

Input: nums = [1,2,4,5,10]

Output: 16

Explanation: There are 16 valid tuples:

(1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2)

(2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1)

(2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,5,4)

(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 104
• All elements in nums are distinct.

Solution

class Solution {
    fun tupleSameProduct(nums: IntArray): Int {
        val ab = HashMap<Int, Int>()
        for (i in nums.indices) {
            for (j in i + 1 until nums.size) {
                ab[nums[i] * nums[j]] = ab.getOrDefault(nums[i] * nums[j], 0) + 1
            }
        }
        var count = 0
        for (entry: Map.Entry<Int, Int> in ab.entries) {
            val `val`: Int = entry.value
            count += `val` * (`val` - 1) / 2
        }
        return count * 8
    }
}

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