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1723. Find Minimum Time to Finish All Jobs

Description

You are given an integer array jobs, where jobs[i] is the amount of time it takes to complete the ith job.

There are k workers that you can assign jobs to. Each job should be assigned to exactly one worker. The working time 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 maximum working time of any worker is minimized.

Return the minimum possible maximum working time of any assignment.

 

Example 1:

Input: jobs = [3,2,3], k = 3
Output: 3
Explanation: By assigning each person one job, the maximum time is 3.

Example 2:

Input: jobs = [1,2,4,7,8], k = 2
Output: 11
Explanation: 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.

 

Constraints:

• 1 <= k <= jobs.length <= 12
• 1 <= jobs[i] <= 107

Solutions

• Java
• C++
• Python
• Go

• class Solution {
      private int[] cnt;
      private int ans;
      private int[] jobs;
      private int k;
  
      public int minimumTimeRequired(int[] jobs, int k) {
          this.k = k;
          Arrays.sort(jobs);
          for (int i = 0, j = jobs.length - 1; i < j; ++i, --j) {
              int t = jobs[i];
              jobs[i] = jobs[j];
              jobs[j] = t;
          }
          this.jobs = jobs;
          cnt = new int[k];
          ans = 0x3f3f3f3f;
          dfs(0);
          return ans;
      }
  
      private void dfs(int i) {
          if (i == jobs.length) {
              int mx = 0;
              for (int v : cnt) {
                  mx = Math.max(mx, v);
              }
              ans = Math.min(ans, mx);
              return;
          }
          for (int j = 0; j < k; ++j) {
              if (cnt[j] + jobs[i] >= ans) {
                  continue;
              }
              cnt[j] += jobs[i];
              dfs(i + 1);
              cnt[j] -= jobs[i];
              if (cnt[j] == 0) {
                  break;
              }
          }
      }
  }
  

• class Solution {
  public:
      int ans;
  
      int minimumTimeRequired(vector<int>& jobs, int k) {
          vector<int> cnt(k);
          ans = 0x3f3f3f3f;
          sort(jobs.begin(), jobs.end(), greater<int>());
          dfs(0, k, jobs, cnt);
          return ans;
      }
  
      void dfs(int i, int k, vector<int>& jobs, vector<int>& cnt) {
          if (i == jobs.size()) {
              ans = min(ans, *max_element(cnt.begin(), cnt.end()));
              return;
          }
          for (int j = 0; j < k; ++j) {
              if (cnt[j] + jobs[i] >= ans) continue;
              cnt[j] += jobs[i];
              dfs(i + 1, k, jobs, cnt);
              cnt[j] -= jobs[i];
              if (cnt[j] == 0) break;
          }
      }
  };
  

• class Solution:
      def minimumTimeRequired(self, jobs: List[int], k: int) -> int:
          def dfs(i):
              nonlocal ans
              if i == len(jobs):
                  ans = min(ans, max(cnt))
                  return
              for j in range(k):
                  if cnt[j] + jobs[i] >= ans:
                      continue
                  cnt[j] += jobs[i]
                  dfs(i + 1)
                  cnt[j] -= jobs[i]
                  if cnt[j] == 0:
                      break
  
          cnt = [0] * k
          jobs.sort(reverse=True)
          ans = inf
          dfs(0)
          return ans
  
  

• func minimumTimeRequired(jobs []int, k int) int {
  	cnt := make([]int, k)
  	ans := 0x3f3f3f3f
  	sort.Slice(jobs, func(i, j int) bool {
  		return jobs[i] > jobs[j]
  	})
  	var dfs func(int)
  	dfs = func(i int) {
  		if i == len(jobs) {
  			mx := slices.Max(cnt)
  			ans = min(ans, mx)
  			return
  		}
  		for j := 0; j < k; j++ {
  			if cnt[j]+jobs[i] >= ans {
  				continue
  			}
  			cnt[j] += jobs[i]
  			dfs(i + 1)
  			cnt[j] -= jobs[i]
  			if cnt[j] == 0 {
  				break
  			}
  		}
  	}
  	dfs(0)
  	return ans
  }
  

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