<p>A binary watch has 4 LEDs on the top to represent the hours (0-11), and 6 LEDs on the bottom to represent&nbsp;the minutes (0-59). Each LED represents a zero or one, with the least significant bit on the right.</p>

<ul>
	<li>For example, the below binary watch reads <code>&quot;4:51&quot;</code>.</li>
</ul>

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/binarywatch.jpg" style="width: 500px; height: 500px;" /></p>

<p>Given an integer <code>turnedOn</code> which represents the number of LEDs that are currently on (ignoring the PM), return <em>all possible times the watch could represent</em>. You may return the answer in <strong>any order</strong>.</p>

<p>The hour must not contain a leading zero.</p>

<ul>
	<li>For example, <code>&quot;01:00&quot;</code> is not valid. It should be <code>&quot;1:00&quot;</code>.</li>
</ul>

<p>The minute must&nbsp;consist of two digits and may contain a leading zero.</p>

<ul>
	<li>For example, <code>&quot;10:2&quot;</code> is not valid. It should be <code>&quot;10:02&quot;</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> turnedOn = 1
<strong>Output:</strong> ["0:01","0:02","0:04","0:08","0:16","0:32","1:00","2:00","4:00","8:00"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> turnedOn = 9
<strong>Output:</strong> []
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= turnedOn &lt;= 10</code></li>
</ul>

class Solution {
public:
    vector<string> readBinaryWatch(int turnedOn) {
        dfs(turnedOn, -1);
        return res;
    }

private:
    vector<string> res;
    bool pos[10] = {false};

    // DFS function to explore valid times
    void dfs(int count, int idx) {
        if(count == 0){
            // Reconstruct the time
            int h = 0, m = 0;
            for(int i = 0; i < 4; ++i){
                if(pos[i]) h += pow(2, i);
            }
            for(int i = 4; i < 10; ++i){
                if(pos[i]) m += pow(2, i - 4);
            }
            string hstr, mstr, temp;
            if (h >= 12) return;
            else hstr = to_string(h);

            if(m >= 60) return;
            else mstr = (m < 10? "0" + to_string(m) : to_string(m)); 

            temp = hstr + ":" + mstr;
            res.push_back(temp);
            return;
        }

        // Backtrack using for loop structure
        for(int i = idx + 1; i < 10; ++i){
            pos[i] = true;
            dfs(count - 1, i);
            pos[i] = false;
        }
    }
};
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