Solve the Equation
Problem
Given a linear equation like '2x+3x-6x=x+2', return the solution as 'x=...', 'No solution', or 'Infinite solutions'.
equation = 'x+5-3+x=6+x-2''x=2'def solve_equation(eq):
def parse(s):
s = s.replace('-', '+-'); cx = 0; c = 0
for t in s.split('+'):
if not t: continue
if t.endswith('x'):
v = t[:-1]; cx += -1 if v == '-' else 1 if v in ('', '+') else int(v)
else: c += int(t)
return cx, c
l, r = eq.split('='); lx, lc = parse(l); rx, rc = parse(r)
a, b = lx - rx, rc - lc
if a == 0: return 'Infinite solutions' if b == 0 else 'No solution'
return f'x={b // a}'function solveEquation(eq) {
const parse = s => {
s = s.replace(/-/g, '+-'); let cx = 0, c = 0;
for (const t of s.split('+')) {
if (!t) continue;
if (t.endsWith('x')) { const v = t.slice(0, -1); cx += v === '-' ? -1 : (v === '' || v === '+') ? 1 : +v; }
else c += +t;
}
return [cx, c];
};
const [L, R] = eq.split('='); const [lx, lc] = parse(L); const [rx, rc] = parse(R);
const a = lx - rx, b = rc - lc;
if (a === 0) return b === 0 ? 'Infinite solutions' : 'No solution';
return 'x=' + Math.trunc(b / a);
}String solveEquation(String eq) {
int[] L = parse(eq.split("=")[0]), R = parse(eq.split("=")[1]);
int a = L[0] - R[0], b = R[1] - L[1];
if (a == 0) return b == 0 ? "Infinite solutions" : "No solution";
return "x=" + (b / a);
}
int[] parse(String s) {
s = s.replace("-", "+-"); int cx = 0, c = 0;
for (String t : s.split("\\+")) {
if (t.isEmpty()) continue;
if (t.endsWith("x")) { String v = t.substring(0, t.length()-1); cx += v.equals("-") ? -1 : (v.isEmpty() || v.equals("+") ? 1 : Integer.parseInt(v)); }
else c += Integer.parseInt(t);
}
return new int[]{cx, c};
}pair parse(string s) {
string r; for (char c : s) { if (c == '-') r += "+-"; else r += c; }
int cx = 0, c = 0; stringstream ss(r); string t;
while (getline(ss, t, '+')) {
if (t.empty()) continue;
if (t.back() == 'x') { string v = t.substr(0, t.size()-1); cx += v == "-" ? -1 : (v.empty() || v == "+" ? 1 : stoi(v)); }
else c += stoi(t);
}
return {cx, c};
}
string solveEquation(string eq) {
int eq_pos = eq.find('='); auto L = parse(eq.substr(0, eq_pos)), R = parse(eq.substr(eq_pos + 1));
int a = L.first - R.first, b = R.second - L.second;
if (a == 0) return b == 0 ? "Infinite solutions" : "No solution";
return "x=" + to_string(b / a);
} Explanation
A linear equation in x always boils down to two numbers: how many x's there are and what the constant is. So we parse each side into a pair (coef_x, constant) and then solve.
The parser turns every - into +- so splitting on + gives clean signed terms. For each term: if it ends in x we add its coefficient (where x means 1, -x means -1, 3x means 3); otherwise it's a plain number we add to the constant.
After parsing left and right, we move all x's to one side and all constants to the other: a = lx - rx and b = rc - lc, so the equation becomes a·x = b. If a != 0, the answer is x = b / a.
The special cases come from a == 0: if b is also 0 every x works (Infinite solutions); if b != 0 nothing works (No solution).
Example: x+5-3+x=6+x-2. Left is 2x + 2, right is 1x + 4. So a = 2-1 = 1, b = 4-2 = 2, giving x=2.