<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>

int dp[1 << 12][13];
class Solution {
    int n;
    int solve(vector<int>& jobs, int mask, int k) {
        if (mask == (1 << n) - 1)
            return 0;
        if (k == 0)
            return 1e9;

        if (dp[mask][k] != -1)
            return dp[mask][k];

        int ans = 1e9;
        vector<pair<int, int>> rem_jobs;

        for (int i = 0; i < n; i++)
            if (!(mask & (1 << i)))
                rem_jobs.push_back({jobs[i], i});

        int r = rem_jobs.size();
        for (int w = 0; w < (1 << r); w++) {
            int sum = 0, next_mask = mask;
            for (int i = 0; i < r; i++)
                if (w & (1 << i)) {
                    sum += rem_jobs[i].first;
                    next_mask |= (1 << rem_jobs[i].second);
                }
            ans = min(ans, max(sum, solve(jobs, next_mask, k - 1)));
        }

        return dp[mask][k] = ans;
    }

public:
    int minimumTimeRequired(vector<int>& jobs, int k) {
        n = jobs.size();
        for (int i = 0; i < (1 << n); i++)
            for (int j = 1; j <= k; j++)
                dp[i][j] = -1;
        return solve(jobs, 0, k);
    }
};
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