<p>Solve a given equation and return the value of <code>&#39;x&#39;</code> in the form of a string <code>&quot;x=#value&quot;</code>. The equation contains only <code>&#39;+&#39;</code>, <code>&#39;-&#39;</code> operation, the variable <code>&#39;x&#39;</code> and its coefficient. You should return <code>&quot;No solution&quot;</code> if there is no solution for the equation, or <code>&quot;Infinite solutions&quot;</code> if there are infinite solutions for the equation.</p>

<p>If there is exactly one solution for the equation, we ensure that the value of <code>&#39;x&#39;</code> is an integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> equation = &quot;x+5-3+x=6+x-2&quot;
<strong>Output:</strong> &quot;x=2&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> equation = &quot;x=x&quot;
<strong>Output:</strong> &quot;Infinite solutions&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> equation = &quot;2x=x&quot;
<strong>Output:</strong> &quot;x=0&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= equation.length &lt;= 1000</code></li>
	<li><code>equation</code> has exactly one <code>&#39;=&#39;</code>.</li>
	<li><code>equation</code> consists of integers with an absolute value in the range <code>[0, 100]</code> without any leading zeros, and the variable <code>&#39;x&#39;</code>.</li>
	<li>The input is generated that if there is a single solution, it will be an integer.</li>
</ul>

class Solution:
    def solveEquation(self, equation: str) -> str:
        muls = {'+': 1, '-': -1}
        
        def gen(s):
            res = [0, 0]
            i, m = 0, 1
            c = '0'
            while i <= len(s):
                if i == len(s) or s[i] in muls: 
                    if c[-1] == 'x':
                        res[0] += m * int(c[:-1] if c[:-1] != '0' else '1')
                    else:
                        res[1] += m * int(c)
                    if i == len(s): break
                    m = muls[s[i]]
                    c = '0'
                else:
                    c += s[i]
                i += 1
            return res

        holds = [gen(x) for x in equation.split('=')]
        final = [holds[0][0] - holds[1][0], holds[0][1] - holds[1][1]]
        if final[0] == final[1] == 0:
            return 'Infinite solutions'
        elif final[0] == 0:
            return 'No solution'
        else:
            return f"x={-final[1] // final[0]}"



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