<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
class Solution {
public:
    ll func(int x,vector<int>&a){
        ll n=a.size();
        vector<pair<int,int>> dp((1<<n));
        dp[0]={1,0};
        for(int i=1;i<(1<<n); i++){
            dp[i]={n+1,0};
            for(int j=0; j<n; j++){
                if(i&(1<<j)){
                    auto cur=dp[i^(1<<j)];
                    if(cur.second+a[j]<=x) cur.second+=a[j];
                    else{
                        cur.first++;
                        cur.second=a[j];
                    }
                    dp[i]=min(dp[i],cur);
                }
            }
        }
        return dp[(1<<n)-1].first;
    }
    int minimumTimeRequired(vector<int>& a, int k) {
        int sum=accumulate(a.begin(),a.end(),0);
        int l=*max_element(a.begin(),a.end()),h=sum,ans=sum;
        while(l<=h){
            int mid=(l+(h-l)/2);
            if(func(mid,a)<=k){
                ans=mid;
                h=mid-1;
            }
            else l=mid+1;
        }
        return ans;
    }
};
26168