Find Minimum Time to Finish All Jobs
<p>You are given an integer array <code>jobs</code>, where <code>jobs[i]</code> is the amount of time it takes to complete the <code>i<sup>th</sup></code> job.</p>

<p>There are <code>k</code> workers that you can assign jobs to. Each job should be assigned to <strong>exactly</strong> one worker. The <strong>working time</strong> of a worker is the sum of the time it takes to complete all jobs assigned to them. Your goal is to devise an optimal assignment such that the <strong>maximum working time</strong> of any worker is <strong>minimized</strong>.</p>

<p><em>Return the <strong>minimum</strong> possible <strong>maximum working time</strong> of any assignment. </em></p>

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

<pre>
<strong>Input:</strong> jobs = [3,2,3], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> By assigning each person one job, the maximum time is 3.
</pre>

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

<pre>
<strong>Input:</strong> jobs = [1,2,4,7,8], k = 2
<strong>Output:</strong> 11
<strong>Explanation:</strong> Assign the jobs the following way:
Worker 1: 1, 2, 8 (working time = 1 + 2 + 8 = 11)
Worker 2: 4, 7 (working time = 4 + 7 = 11)
The maximum working time is 11.</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= jobs.length &lt;= 12</code></li>
	<li><code>1 &lt;= jobs[i] &lt;= 10<sup>7</sup></code></li>
</ul>
Bit Manipulation; Array; Dynamic Programming; Backtracking; Bitmask
class Solution {
public:
    int ans;

    int minimumTimeRequired(vector<int>& jobs, int k) {
        vector<int> cnt(k);
        ans = 0x3f3f3f3f;
        sort(jobs.begin(), jobs.end(), greater<int>());
        dfs(0, k, jobs, cnt);
        return ans;
    }

    void dfs(int i, int k, vector<int>& jobs, vector<int>& cnt) {
        if (i == jobs.size()) {
            ans = min(ans, *max_element(cnt.begin(), cnt.end()));
            return;
        }
        for (int j = 0; j < k; ++j) {
            if (cnt[j] + jobs[i] >= ans) continue;
            cnt[j] += jobs[i];
            dfs(i + 1, k, jobs, cnt);
            cnt[j] -= jobs[i];
            if (cnt[j] == 0) break;
        }
    }
};