<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, num: str, target: int) -> List[str]:
        ## RC ##
		## APPROACH : BACKTRACKING ##
		## LOGIC ##
		#	1. Draw backtracking tree starting with "" and 1st level being number for num 123: its 1 or 12 or 123 (attach *, -, + and send for dfs)
		#	2. When you are the last digit, donot send for dfs, compute the result
        #   3. Check out for some interesting edge cases, interviewer is looking for those you to solve.
		# edge cases : 105 5, 100009 9, 569 45
        
        ## TIME COMPLEXICITY : O(4^N) ## (each level you are creating 4 branches)
		## SPACE COMPLEXICITY : O(4^N) ##
        
        def dfs( curr, num ):
            if curr in visited:
                return
            visited.add(curr)
            for i, ch in enumerate(num):
                if i == len(num) - 1:     # when you are at last digit, calculate the result
                    if eval(curr + num) == target:
                        result.append(curr + num)
                else:
                    dfs( curr + num[:i+1] + "*", num[i+1:])
                    dfs( curr + num[:i+1] + "-", num[i+1:])
                    dfs( curr + num[:i+1] + "+", num[i+1:])
                
                if num[:i+1] == "0":     # edge case, in the number you are sending in if the first digit is 0, donot send further. ex: 105, send 0 but not 05.
                    break
        result = []
        visited = set()
        dfs( "", num )
        return result
155000