LeetCode in Kotlin

399. Evaluate Division

Medium

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [A_i, B_i] and values[i] represent the equation A_i / B_i = values[i]. Each A_i or B_i is a string that represents a single variable.

You are also given some queries, where queries[j] = [C_j, D_j] represents the j^th query where you must find the answer for C_j / D_j = ?.

Return the answers to all queries. If a single answer cannot be determined, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

Example 1:

Input: equations = [[“a”,”b”],[“b”,”c”]], values = [2.0,3.0], queries = [[“a”,”c”],[“b”,”a”],[“a”,”e”],[“a”,”a”],[“x”,”x”]]

Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]

Explanation: Given: a / b = 2.0, b / c = 3.0 queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? return: [6.0, 0.5, -1.0, 1.0, -1.0 ]

Example 2:

Input: equations = [[“a”,”b”],[“b”,”c”],[“bc”,”cd”]], values = [1.5,2.5,5.0], queries = [[“a”,”c”],[“c”,”b”],[“bc”,”cd”],[“cd”,”bc”]]

Output: [3.75000,0.40000,5.00000,0.20000]

Example 3:

Input: equations = [[“a”,”b”]], values = [0.5], queries = [[“a”,”b”],[“b”,”a”],[“a”,”c”],[“x”,”y”]]

Output: [0.50000,2.00000,-1.00000,-1.00000]

Constraints:

1 <= equations.length <= 20
equations[i].length == 2
1 <= A_i.length, B_i.length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= C_j.length, D_j.length <= 5
A_i, B_i, C_j, D_j consist of lower case English letters and digits.

Solution

@Suppress("kotlin:S6518")
class Solution {
    private var root: MutableMap<String?, String?>? = null
    private var rate: MutableMap<String?, Double>? = null

    fun calcEquation(
        equations: List<List<String?>>,
        values: DoubleArray,
        queries: List<List<String?>>
    ): DoubleArray {
        root = HashMap()
        rate = HashMap()
        val n = equations.size
        for (equation in equations) {
            val x = equation[0]
            val y = equation[1]
            (root as HashMap<String?, String?>)[x] = x
            (root as HashMap<String?, String?>)[y] = y
            (rate as HashMap<String?, Double>)[x] = 1.0
            (rate as HashMap<String?, Double>)[y] = 1.0
        }
        for (i in 0 until n) {
            val x = equations[i][0]
            val y = equations[i][1]
            union(x, y, values[i])
        }
        val result = DoubleArray(queries.size)
        for (i in queries.indices) {
            val x = queries[i][0]
            val y = queries[i][1]
            if (!(root as HashMap<String?, String?>).containsKey(x) ||
                !(root as HashMap<String?, String?>).containsKey(y)
            ) {
                result[i] = -1.0
                continue
            }
            val rootX = findRoot(x, x, 1.0)
            val rootY = findRoot(y, y, 1.0)
            result[i] = if (rootX == rootY) (rate as HashMap<String?, Double>).get(x)!! /
                (rate as HashMap<String?, Double>).get(y)!! else -1.0
        }
        return result
    }

    private fun union(x: String?, y: String?, v: Double) {
        val rootX = findRoot(x, x, 1.0)
        val rootY = findRoot(y, y, 1.0)
        root!![rootX] = rootY
        val r1 = rate!![x]!!
        val r2 = rate!![y]!!
        rate!![rootX] = v * r2 / r1
    }

    private fun findRoot(originalX: String?, x: String?, r: Double): String? {
        if (root!![x] == x) {
            root!![originalX] = x
            rate!![originalX] = r * rate!![x]!!
            return x
        }
        return findRoot(originalX, root!![x], r * rate!![x]!!)
    }
}

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