花花酱 LeetCode 1723. Find Minimum Time to Finish All Jobs

zxi — Sun, 10 Jan 2021 19:30:53 +0000

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

Solution 1: All subsets

dp[i][t] := min of max working time by assigning a subset of jobs s to the first i workers.

dp[i][t] = min{max(dp[i – 1][s], cost[s ^ t])} where s is a subset of t.

Time complexity: O(k*3^n)
Space complexity: O(k*2^n)

C++

// Author: Huahua
// Author: Huahua
class Solution {
public:
  int minimumTimeRequired(vector& jobs, int k) {    
    const int n = jobs.size();
    int dp[13][4096];
    memset(dp, 0x7f, sizeof(dp));
    dp[0][0] = 0;
    
    vector cost(1 << n);
    for (int t = 0; t < 1 << n; ++t) {
      for (int j = 0; j < n; ++j)
        if (t >> j & 1) cost[t] += jobs[j];
      dp[1][t] = cost[t]; // do jobs t by one worker
    }
    
    for (int i = 2; i <= k; ++i)
      for (int t = 0; t < 1 << n; ++t)
        for (int s = t; s; s = (s - 1) & t)
          dp[i][t] = min(dp[i][t], max(dp[i - 1][s], cost[t ^ s]));
    return dp[k][(1 << n) - 1];
  }
};

Solution 2: Search + Pruning

Time complexity: O(k^n)
Space complexity: O(k*n)

C++

class Solution {
public:
  int minimumTimeRequired(vector& jobs, int k) {
    vector times(k);
    int ans = INT_MAX;
    function dfs = [&](int i, int cur) {
      if (cur >= ans) return;
      if (i == jobs.size()) {
        ans = cur;
        return;
      }
      unordered_set seen;
      for (int j = 0; j < k; ++j) {
        if (!seen.insert(times[j]).second) continue;
        times[j] += jobs[i];
        dfs(i + 1, max(cur, times[j]));
        times[j] -= jobs[i];
      }
    };
    sort(rbegin(jobs), rend(jobs));
    dfs(0, 0);
    return ans;
  }
};

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花花酱 LeetCode 1723. Find Minimum Time to Finish All Jobs

Solution 1: All subsets

C++

Solution 2: Search + Pruning

C++