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This approach involves rearranging the equation to separate the terms involving 'x' from the constant terms. We ensure that all terms involving 'x' are on one side of the equation and all constant terms on the other side. This allows us to solve for 'x' by simplifying both sides to isolated 'x' terms versus numerical constants.
Time Complexity: O(n), where n is the length of the equation string. Space Complexity: O(1), as we use a constant amount of extra space.
1def solveEquation(equation):
2 def parse(s):
3 x_coef = 0
4 const = 0
5 num = ''
6 sign = 1
7 i = 0
8
9 while i < len(s):
10 ch = s[i]
11 if ch == '+':
12 sign = 1
13 elif ch == '-':
14 sign = -1
15 elif ch.isdigit():
16 num = ch
17 while i + 1 < len(s) and s[i + 1].isdigit():
18 i += 1
19 num += s[i]
20 const += sign * int(num)
21 num = ''
22 elif ch == 'x':
23 if num == '':
24 num = '1'
25 x_coef += sign * int(num)
26 num = ''
27 i += 1
28
29 return x_coef, const
30
31 left, right = equation.split('=')
32 left_x_coef, left_const = parse(left)
33 right_x_coef, right_const = parse(right)
34
35 total_x_coef = left_x_coef - right_x_coef
36 total_const = right_const - left_const
37
38 if total_x_coef == 0:
39 if total_const == 0:
40 return 'Infinite solutions'
41 else:
42 return 'No solution'
43 else:
44 return f'x={total_const // total_x_coef}'
We split the equation at the equal sign and process both sides to calculate the total coefficients for 'x' and any constant. By comparing these totals, we determine if the equation has no solution, infinite solutions, or exactly one solution. Parsing involves iterating through each character of the string, summing coefficients and constants.
This approach focuses on scanning the equation and simplifying terms by combining all instances of the variable 'x' and the constant terms separately. After combining, we analyze the coefficients to deduce the solution.
Time Complexity: O(n), where n represents the number of characters in the input string. Space Complexity: O(1), since only fixed-space variables are utilized.
1public class Solution {
2 public string SolveEquation(string equation) {
3 (int xCoefL, int constL) = ParseEquation(equation.Split('=')[0]);
4 (int xCoefR, int constR) = ParseEquation(equation.Split('=')[1]);
5
int xCoef = xCoefL - xCoefR;
int constant = constR - constL;
if (xCoef == 0) {
return constant == 0 ? "Infinite solutions" : "No solution";
} else {
return "x=" + (constant / xCoef).ToString();
}
}
private (int, int) ParseEquation(string side) {
int xCoef = 0, constant = 0;
int sign = 1;
int i = 0;
while (i < side.Length) {
if (side[i] == '+') {
sign = 1;
i++;
} else if (side[i] == '-') {
sign = -1;
i++;
}
int val = 0;
bool isX = false;
if (i < side.Length && char.IsDigit(side[i])) {
while (i < side.Length && char.IsDigit(side[i])) {
val = val * 10 + (side[i] - '0');
i++;
}
} else {
val = 1;
}
if (i < side.Length && side[i] == 'x') {
isX = true;
i++;
}
if (isX) {
xCoef += sign * val;
} else {
constant += sign * val;
}
}
return (xCoef, constant);
}
}
In this C# solution, we further enhance parsing by leveraging tuples for clean coefficient and constant extraction. This method helps standardize processing by treating everything as terms to be added or subtracted from a total, revealing a systematic deduction path for solutions or infinite cases.