LeetCode in Kotlin
640. Solve the Equation
Medium
Solve a given equation and return the value of 'x' in the form of a string "x=#value". The equation contains only '+', '-' operation, the variable 'x' and its coefficient. You should return "No solution" if there is no solution for the equation, or "Infinite solutions" if there are infinite solutions for the equation.
If there is exactly one solution for the equation, we ensure that the value of 'x' is an integer.
Example 1:
Input: equation = “x+5-3+x=6+x-2”
Output: “x=2”
Example 2:
Input: equation = “x=x”
Output: “Infinite solutions”
Example 3:
Input: equation = “2x=x”
Output: “x=0”
Constraints:
3 <= equation.length <= 1000equationhas exactly one'='.equationconsists of integers with an absolute value in the range[0, 100]without any leading zeros, and the variable'x'.
Solution
class Solution {
fun solveEquation(equation: String): String {
val eqs = equation.split("=".toRegex()).dropLastWhile { it.isEmpty() }.toTypedArray()
val arr1 = evaluate(eqs[0])
val arr2 = evaluate(eqs[1])
return if (arr1[0] == arr2[0] && arr1[1] == arr2[1]) {
"Infinite solutions"
} else if (arr1[0] == arr2[0]) {
"No solution"
} else {
"x=" + (arr2[1] - arr1[1]) / (arr1[0] - arr2[0])
}
}
private fun evaluate(eq: String): IntArray {
val arr = eq.toCharArray()
var f = false
var a = 0
var b = 0
var i = 0
if (arr[0] == '-') {
f = true
i++
}
while (i < arr.size) {
if (arr[i] == '-') {
f = true
i++
} else if (arr[i] == '+') {
i++
}
val sb = StringBuilder()
while (i < arr.size && Character.isDigit(arr[i])) {
sb.append(arr[i])
i++
}
val n = sb.toString()
if (i < arr.size && arr[i] == 'x') {
var number: Int
number = if (n == "") {
1
} else {
n.toInt()
}
if (f) {
number = -number
}
a += number
i++
} else {
var number = n.toInt()
if (f) {
number = -number
}
b += number
}
f = false
}
val op = IntArray(2)
op[0] = a
op[1] = b
return op
}
}