<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 minimumTimeRequired(vector<int>& a, int k) {
        int n = (int)a.size();
        int l = 0;
        int r = 1e9;
        int mx = *max_element(a.begin(),a.end());
        while(r > l + 1) {
            int m = (l+r) >> 1;
            if(m < mx) {
                l = m;
                continue;
            }
            vector<pair<int,int>>dp(1 << n);
            for(int i = 0 ; i < (1 << n) ; i++) {
                dp[i] = {2e9,2e9};
            }
            dp[0] = {1,0};
            for(int i = 0 ; i < (1 << n) ; i++) {
                int currW = dp[i].first;
                int hrs = dp[i].second;
                for(int j = 0 ; j < n ; j++) {
                    if(!(i & (1 << j))) {
                        if(hrs + a[j] <= m) {
                            dp[i|(1<<j)] = min(dp[i|(1<<j)],{currW,a[j] + hrs});
                        } else {
                            dp[i|(1<<j)] = min(dp[i|(1<<j)],{currW+1,a[j]});
                        }
                    }
                }
            }
            if(dp[(1<<n)-1].first <= k) {
                r = m;
            } else {
                l = m;
            }
        }
        return r;
    }
};
31101