<p>Given a string <code>num</code> that contains only digits and an integer <code>target</code>, return <em><strong>all possibilities</strong> to insert the binary operators </em><code>&#39;+&#39;</code><em>, </em><code>&#39;-&#39;</code><em>, and/or </em><code>&#39;*&#39;</code><em> between the digits of </em><code>num</code><em> so that the resultant expression evaluates to the </em><code>target</code><em> value</em>.</p>

<p>Note that operands in the returned expressions <strong>should not</strong> contain leading zeros.</p>

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

<pre>
<strong>Input:</strong> num = &quot;123&quot;, target = 6
<strong>Output:</strong> [&quot;1*2*3&quot;,&quot;1+2+3&quot;]
<strong>Explanation:</strong> Both &quot;1*2*3&quot; and &quot;1+2+3&quot; evaluate to 6.
</pre>

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

<pre>
<strong>Input:</strong> num = &quot;232&quot;, target = 8
<strong>Output:</strong> [&quot;2*3+2&quot;,&quot;2+3*2&quot;]
<strong>Explanation:</strong> Both &quot;2*3+2&quot; and &quot;2+3*2&quot; evaluate to 8.
</pre>

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

<pre>
<strong>Input:</strong> num = &quot;3456237490&quot;, target = 9191
<strong>Output:</strong> []
<strong>Explanation:</strong> There are no expressions that can be created from &quot;3456237490&quot; to evaluate to 9191.
</pre>

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

<ul>
	<li><code>1 &lt;= num.length &lt;= 10</code></li>
	<li><code>num</code> consists of only digits.</li>
	<li><code>-2<sup>31</sup> &lt;= target &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution:
    def addOperators(self, s: str, target: int) -> List[str]:
        #TC is O(4^N*N)
        #4 -- 3 for +,-,* and 1 for skipping and proceeding i.e.
        #123 -> 1+ 1- 1* 12 --> 4^N
        #N for adding expre to final res
        #SC O(N)
        # need to traverse through s
        # backtrack each case of  start index and then +,*,-
        # need empty array ofcourse
        # need curidx = i
        # need the str path to append to the arr if my --> cur_num == target
        # need the prevNum access for multiplication
        # 1 2 3 4 5 --> 1 + 2 + 3 + 4 * 5
        #               ^ ^ ^ ^ ^ ^ ^ == 10 but we wanna do 4*5 so ---
        # just do 1 + 2 + 3 + 4 + (- 4) + (4 * 5)
        # return ans
        
        res = []

        def dfs(i, path, cur_num, prevNum):
            if i == len(s):
                if cur_num == target:
                    res.append(path)
                return
            
            for j in range(i, len(s)):
                # starting with zero?
                if j > i and s[i] == '0':
                    break
                num = int(s[i:j+1])

                # if cur index is 0 then simple add that number
                if i == 0:
                    dfs(j + 1, path + str(num), cur_num + num, num)
                else:
                    dfs(j + 1, path + "+" + str(num), cur_num + num, num)
                    dfs(j + 1, path + "-" + str(num), cur_num - num, -num)
                    dfs(j + 1, path + "*" + str(num), cur_num - prevNum + prevNum * num, prevNum * num)
        
        dfs(0, "", 0, 0)
        return res
        
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