##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1723.html

# 1723. Find Minimum Time to Finish All Jobs

Hard

## Description

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] <= 10^7

## Solution

Use dynamic programming with compressed states. Let n be the length of jobs. Create a 2D array dp with k + 1 rows and 1 << n columns, where dp[i][j] represents the minimum time required starting from job i and mask j.

Initialize dp[k][(1 << n) - 1] = 0. Then for i from k - 1 to 0 and for mask from (1 << length) - 2 to 0, loop over all possible values curr from 1 to (1 << length) - 1. If (curr & mask) == 0, then update dp[i][mask] accordingly.

Finally, return dp[0][0].

• class Solution {
public int minimumTimeRequired(int[] jobs, int k) {
int length = jobs.length;
int[][] dp = new int[k + 1][1 << length];
for (int i = 0; i <= k; i++)
Arrays.fill(dp[i], Integer.MAX_VALUE);
int[] times = new int[1 << length];
for (int mask = 0; mask < 1 << length; mask++) {
int time = 0;
for (int i = 0; i < length; i++) {
if ((mask & (1 << i)) != 0)
time += jobs[i];
}
times[mask] = time;
}
dp[k][(1 << length) - 1] = 0;
for (int i = k - 1; i >= 0; i--) {
for (int mask = (1 << length) - 2; mask >= 0; mask--) {
for (int curr = 1; curr < 1 << length; curr++) {
if ((curr & mask) == 0) {
int time = Math.max(times[curr], dp[i + 1][curr | mask]);
dp[i][mask] = Math.min(dp[i][mask], time);
}
}
}
}
return dp[0][0];
}
}

############

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;
}
}
}
}

• // OJ: https://leetcode.com/problems/find-minimum-time-to-finish-all-jobs/
// Time: O(K^N) ~150ms as of 1/17/2021
// Space: O(NK)
// Ref: https://leetcode.com/problems/find-minimum-time-to-finish-all-jobs/discuss/1009817/One-branch-cutting-trick-to-solve-three-LeetCode-questions
class Solution {
int ans = INT_MAX;
vector<int> v;
void dfs(vector<int> &A, int i) {
if (i == A.size()) {
ans = min(ans, *max_element(begin(v), end(v)));
return;
}
unordered_set<int> seen;
for (int j = 0; j < v.size(); ++j) {
if (seen.count(v[j]) || v[j] + A[i] > ans) continue;
seen.insert(v[j]);
v[j] += A[i];
dfs(A, i + 1);
v[j] -= A[i];
}
}
public:
int minimumTimeRequired(vector<int>& A, int k) {
v.assign(k, 0);
sort(begin(A), end(A));
dfs(A, 0);
return ans;
}
};

• 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 := 0
for _, v := range cnt {
mx = max(mx, v)
}
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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

func min(a, b int) int {
if a < b {
return a
}
return b
}