Verbal Arithmetic Puzzle
<p>Given an equation, represented by <code>words</code> on the left side and the <code>result</code> on the right side.</p>

<p>You need to check if the equation is solvable under the following rules:</p>

<ul>
	<li>Each character is decoded as one digit (0 - 9).</li>
	<li>No two characters can map to the same digit.</li>
	<li>Each <code>words[i]</code> and <code>result</code> are decoded as one number <strong>without</strong> leading zeros.</li>
	<li>Sum of numbers on the left side (<code>words</code>) will equal to the number on the right side (<code>result</code>).</li>
</ul>

<p>Return <code>true</code> <em>if the equation is solvable, otherwise return</em> <code>false</code>.</p>

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

<pre>
<strong>Input:</strong> words = [&quot;SEND&quot;,&quot;MORE&quot;], result = &quot;MONEY&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Map &#39;S&#39;-&gt; 9, &#39;E&#39;-&gt;5, &#39;N&#39;-&gt;6, &#39;D&#39;-&gt;7, &#39;M&#39;-&gt;1, &#39;O&#39;-&gt;0, &#39;R&#39;-&gt;8, &#39;Y&#39;-&gt;&#39;2&#39;
Such that: &quot;SEND&quot; + &quot;MORE&quot; = &quot;MONEY&quot; ,  9567 + 1085 = 10652</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;SIX&quot;,&quot;SEVEN&quot;,&quot;SEVEN&quot;], result = &quot;TWENTY&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Map &#39;S&#39;-&gt; 6, &#39;I&#39;-&gt;5, &#39;X&#39;-&gt;0, &#39;E&#39;-&gt;8, &#39;V&#39;-&gt;7, &#39;N&#39;-&gt;2, &#39;T&#39;-&gt;1, &#39;W&#39;-&gt;&#39;3&#39;, &#39;Y&#39;-&gt;4
Such that: &quot;SIX&quot; + &quot;SEVEN&quot; + &quot;SEVEN&quot; = &quot;TWENTY&quot; ,  650 + 68782 + 68782 = 138214</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;LEET&quot;,&quot;CODE&quot;], result = &quot;POINT&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no possible mapping to satisfy the equation, so we return false.
Note that two different characters cannot map to the same digit.
</pre>

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

<ul>
	<li><code>2 &lt;= words.length &lt;= 5</code></li>
	<li><code>1 &lt;= words[i].length, result.length &lt;= 7</code></li>
	<li><code>words[i], result</code> contain only uppercase English letters.</li>
	<li>The number of different characters used in the expression is at most <code>10</code>.</li>
</ul>
Array; Math; String; Backtracking
class Solution:
    def isAnyMapping(
        self, words, row, col, bal, letToDig, digToLet, totalRows, totalCols
    ):
        # If traversed all columns.
        if col == totalCols:
            return bal == 0

        # At the end of a particular column.
        if row == totalRows:
            return bal % 10 == 0 and self.isAnyMapping(
                words, 0, col + 1, bal // 10, letToDig, digToLet, totalRows, totalCols
            )

        w = words[row]

        # If the current string 'w' has no character in the ('col')th index.
        if col >= len(w):
            return self.isAnyMapping(
                words, row + 1, col, bal, letToDig, digToLet, totalRows, totalCols
            )

        # Take the current character in the variable letter.
        letter = w[len(w) - 1 - col]

        # Create a variable 'sign' to check whether we have to add it or subtract it.
        if row < totalRows - 1:
            sign = 1
        else:
            sign = -1

        # If we have a prior valid mapping, then use that mapping.
        # The second condition is for the leading zeros.
        if letter in letToDig and (
            letToDig[letter] != 0
            or (letToDig[letter] == 0 and len(w) == 1)
            or col != len(w) - 1
        ):

            return self.isAnyMapping(
                words,
                row + 1,
                col,
                bal + sign * letToDig[letter],
                letToDig,
                digToLet,
                totalRows,
                totalCols,
            )

        # Choose a new mapping.
        else:
            for i in range(10):
                # If 'i'th mapping is valid then select it.
                if digToLet[i] == "-" and (
                    i != 0 or (i == 0 and len(w) == 1) or col != len(w) - 1
                ):
                    digToLet[i] = letter
                    letToDig[letter] = i

                    # Call the function again with the new mapping.
                    if self.isAnyMapping(
                        words,
                        row + 1,
                        col,
                        bal + sign * letToDig[letter],
                        letToDig,
                        digToLet,
                        totalRows,
                        totalCols,
                    ):
                        return True

                    # Unselect the mapping.
                    digToLet[i] = "-"
                    if letter in letToDig:
                        del letToDig[letter]

        # If nothing is correct then just return false.
        return False

    def isSolvable(self, words, result):
        # Add the string 'result' in the list 'words'.
        words.append(result)

        # Initialize 'totalRows' with the size of the list.
        totalRows = len(words)

        # Find the longest string in the list and set 'totalCols' with the size of that string.
        totalCols = max(len(word) for word in words)

        # Create a HashMap for the letter to digit mapping.
        letToDig = {}

        # Create a list for the digit to letter mapping.
        digToLet = ["-"] * 10

        return self.isAnyMapping(
            words, 0, 0, 0, letToDig, digToLet, totalRows, totalCols
        )
