Algorithm
Problem Name: 640. Solve the Equation
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'.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public String solveEquation(String equation) {
int[] equationFirst = parseEquation(equation.split("=")[0]);
int[] equationSecond = parseEquation(equation.split("=")[1]);
int constantPart = equationSecond[1] - equationFirst[1];
int variablePart = equationFirst[0] - equationSecond[0];
if (variablePart == 0) {
return constantPart == 0 ? "Infinite solutions" : "No solution";
}
int sign = (variablePart < 0 && constantPart < 0) || (variablePart > 0 && constantPart > 0) ? 1 : -1;
if (Math.abs(constantPart) % Math.abs(variablePart) == 0) {
constantPart = Math.abs(constantPart) / Math.abs(variablePart);
variablePart = 1;
}
return (variablePart > 1 ? variablePart : "") + "x=" + sign * constantPart;
}
private int[] parseEquation(String s) {
int variablePart = 0;
int constantPart = 0;
int sign = 1;
int n = s.length();
for (int idx = 0; idx < n;) {
if (s.charAt(idx) == 'x') {
variablePart += sign;
idx++;
sign = 1;
} else if (Character.isDigit(s.charAt(idx))) {
int num = 0;
while (idx < n && Character.isDigit(s.charAt(idx))) {
num = num * 10 + Character.getNumericValue(s.charAt(idx++));
}
if (idx != n && s.charAt(idx) == 'x') {
variablePart += sign * num;
idx++;
} else {
constantPart += sign * num;
}
sign = 1;
} else {
sign = s.charAt(idx++) == '+' ? 1 : -1;
}
}
return new int[]{variablePart, constantPart};
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
{
num = 1
} else {
num = +tmp
}
return num
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def solveEquation(self, equation):
def calc(eq):
smX = smNum = 0
add, num = True, ""
for c in eq + "+":
if c.isdigit():
num += c
elif c == "x":
smX += int(num) if add and num else -int(num) if num else 1 if add else -1
num = ""
else:
smNum += int(num) if add and num else -int(num) if num else 0
num, add = "", c == "+"
return smX, smNum
eq = equation.split("=")
lX, lNum, rX, rNum = calc(eq[0]) + calc(eq[1])
if lX == rX:
return "No solution" if lNum != rNum else "Infinite solutions"
return "x=" + str((lNum - rNum) // (rX - lX))
Input
Output