<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:
    #define ll long long
    unordered_map<ll,ll>mp;
    ll dp[13][((1ll<<12))];
    ll f(ll i,ll mask){
        // cout<<i<<' '<<mask<<endl;
        if(i<0)return 1e9;
        if(mask == 0){
            if(i==0)return 0;
            else return 1e9;
        }
        if(dp[i][mask]!=-1)return dp[i][mask];
        ll cur = mask;
        ll ans = 1e9;
        while(cur>0){
            ans = min(max(mp[cur],f(i-1,mask^cur)),ans);
            cur = (cur-1) & mask;
        }
        return dp[i][mask]=ans;
    }

    int minimumTimeRequired(vector<int>&v, int k) {
        ll n = v.size();
        for(ll mask=0;mask<=((1ll<<n)-1);mask++){
            for(int j=0;j<n;j++){
                if((mask&(1ll<<j))){
                    mp[mask]+=v[j];
                }
            }
        }
        memset(dp,-1,sizeof(dp));
        // for(auto it:mp)cout<<it.first<< ' '<<it.second<<endl;
        return f(k,((1ll<<n)-1));
    }
};
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