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 {
public:
    bool isAnyMapping(vector<string>& words, int row, int col, int bal, unordered_map<char, int>& letToDig,
        vector<char>& digToLet, int totalRows, int 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 && isAnyMapping(words, 0, col + 1, bal / 10, letToDig, digToLet, totalRows, totalCols));
        }

        string w = words[row];

        // If the current string 'W' has no character in the ('COL')th index.
        if (col >= w.length()) {
            return isAnyMapping(words, row + 1, col, bal, letToDig, digToLet, totalRows, totalCols);
        }

        // Take the current character in the variable letter.
        char letter = w[w.length() - 1 - col];

        // Create a variable 'SIGN' to check whether we have to add it or subtract it.
        int sign;

        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 (letToDig.count(letter) && (letToDig[letter] != 0 || (letToDig[letter] == 0 && w.length() == 1) || col != w.length() - 1)) {

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

        }
        // Choose a new mapping.
        else {
            for (int i = 0; i < 10; i++) {

                // If 'i'th mapping is valid then select it.
                if (digToLet[i] == '-' && (i != 0 || (i == 0 && w.length() == 1) || col != w.length() - 1)) {
                    digToLet[i] = letter;
                    letToDig[letter] = i;

                    // Call the function again with the new mapping.
                    bool x = isAnyMapping(words, row + 1, col, bal + sign * letToDig[letter],
                        letToDig, digToLet, totalRows, totalCols);

                    if (x == true) {
                        return true;
                    }

                    // Unselect the mapping.
                    digToLet[i] = '-';
                    if (letToDig.find(letter) != letToDig.end()) {
                        letToDig.erase(letter);
                    }
                }
            }
        }

        // If nothing is correct then just return false.
        return false;
    }

    bool isSolvable(vector<string>& words, string result) {
        // Add the string 'RESULT' in the vector 'WORDS'.
        words.push_back(result);

        int totalRows;
        int totalCols;

        // Initialize 'TOTALROWS' with the size of the vector.
        totalRows = words.size();

        // Find the longest string in the vector and set 'TOTALCOLS' with the size of that string.
        totalCols = 0;

        for (int i = 0; i < words.size(); i++) {

            // If the current string is the longest then update 'TOTALCOLS' with its length.
            if (totalCols < words[i].size()) {
                totalCols = words[i].size();
            }
        }

        // Create a HashMap for the letter to digit mapping.
        unordered_map<char, int> letToDig;

        // Create a vector for the digit to letter mapping.
        vector<char> digToLet(10, '-');

        return isAnyMapping(words, 0, 0, 0, letToDig, digToLet, totalRows, totalCols);
    }
};
