<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>

# define ll long long

bool backtrack(vector<int>& jobs, vector<int>& dp, int ind, int& k, int& n, int& target){
    // for(auto&& x: dp) cout<<x.first<<' '<<x.second<<"; ";
    // cout<<'\n';
    if(ind == n) return true;
    set<int> seen;
    for(auto&& i: dp){
        if(seen.find(i) != seen.end()) continue;
        else if(i + jobs[ind] >= target) continue;
        seen.insert(i);
        i += jobs[ind];
        if(backtrack(jobs, dp, ind + 1, k, n, target)) return true;
        i -= jobs[ind];
    }
    return false;
}

class Solution {
public:
    int minimumTimeRequired(vector<int>& jobs, int k) {
        int n = jobs.size(), mx = 0, tot = 0;
        for(int i = 0; i < n; i++) mx = max(mx, jobs[i]), tot += jobs[i];
        int l = mx, r = tot;
        while(l <= r){
            vector<int> dp(k);
            int target = (l + r) / 2;
            if(backtrack(jobs, dp, 0, k, n, target)) r = target - 1;
            else l = target + 1;
        }
        return r;
    }
};
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