<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>

class Solution {
public:
    int dp[5002][14];
    int func(int k,int mask, vector<int> v){
        int n=v.size();
        if(mask==(1<<n)-1) return 0;
        if(dp[mask][k]!=-1) return dp[mask][k];
        
        int ans=1e9,i;
        if(k==1){
            ans=0;
            for(i=0;i<v.size();i++){
                if(!(mask&(1<<i))){
                    ans+=v[i];
                }
            }
            return dp[mask][k]=ans;
        }
        vector<int> a;
        for(i=0;i<v.size();i++){
                if(!(mask&(1<<i))){
                    a.push_back(i);
                }
        }
        int m=a.size(),j;
        int p=(1<<m)-1,y=1e9+2;
        for(i=1;i<=p;i++){
            int nk=mask,an=0;
            for(j=0;j<m;j++){
                if(i&(1<<j)){
                    an+=v[a[j]];
                    nk|=(1<<a[j]);
                }
            }
            if(an>=y) continue;
            int x=func(k-1,nk,v);
            ans=min(ans,max(x,an));
            if(an>x){
                y=an;
            }
            
            
        }
        
        return dp[mask][k]=ans;
    }

    int minimumTimeRequired(vector<int>& v, int k) {
          memset(dp,-1,sizeof(dp));
          return func(k,0,v);
    }
};
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