Find Minimum Time to Finish All Jobs

• The core idea is a recursive function, backtrack(jobIndex, workerTimes), which tries to assign jobs[jobIndex] to one of the k workers.
• Algorithm Steps:
  1. Sort the jobs array in descending order. This helps in pruning more effectively, as assigning larger jobs first will quickly lead to exceeding any potential optimal maximum time.
  2. Initialize an array workerTimes of size k to all zeros, representing the current total time for each worker.
  3. Initialize a global variable minMaxTime to Integer.MAX_VALUE to store the minimum of the maximum worker times found so far.
  4. Start the recursion with backtrack(0, workerTimes).
  5. Inside backtrack(jobIndex, workerTimes):
     • Base Case: If jobIndex equals the number of jobs, all jobs have been assigned. We find the maximum time in workerTimes and update minMaxTime.
     • Pruning: Before assigning a job to a worker, check if their new time would exceed the current minMaxTime. If workerTimes[i] + jobs[jobIndex] >= minMaxTime, there's no need to explore this path further, so we skip this worker.
     • Recursive Step: Iterate through each worker i from 0 to k-1.
       • Assign jobs[jobIndex] to worker i.
       • Recursively call backtrack(jobIndex + 1, workerTimes).
       • Undo the assignment (backtrack) to explore other possibilities.
     • Symmetry Breaking: Since all workers are identical, assigning a job to any empty worker is equivalent. To avoid redundant computations, if we assign a job to a worker with 0 current time, we can stop trying other empty workers for this job. We break the loop after the recursive call for the first empty worker.

The core idea is a recursive function, backtrack(jobIndex, workerTimes), which tries to assign jobs[jobIndex] to one of the k workers. We apply several optimizations to make this feasible.

First, we sort the jobs array in descending order. This is a heuristic that often helps pruning work more effectively. Assigning larger jobs first means we are more likely to quickly create a high workerTime, which helps in pruning subsequent branches of the search tree that would exceed this time.

Second, we use pruning. We maintain a global variable minMaxTime that stores the minimum of the maximum worker times found so far. During the recursion, if assigning a job to a worker would make that worker's time greater than or equal to the current minMaxTime, we know this path cannot lead to a better solution, so we prune it.

Third, we break symmetry. Since the workers are indistinguishable, the assignment {W1:[j1], W2:[j2]} is the same as {W1:[j2], W2:[j1]}. We can avoid exploring these symmetric states by enforcing an order. A simple way is: when assigning a job, if we assign it to an empty worker (time = 0), we don't need to try assigning it to any other empty workers. This is because any assignment to another empty worker would lead to a state that is symmetric to one we are already exploring.

1class Solution {2    int minMaxTime = Integer.MAX_VALUE;3 4    public int minimumTimeRequired(int[] jobs, int k) {5        // Sort jobs in descending order for better pruning6        Arrays.sort(jobs);7        int[] reversedJobs = new int[jobs.length];8        for (int i = 0; i < jobs.length; i++) {9            reversedJobs[i] = jobs[jobs.length - 1 - i];10        }11 12        backtrack(reversedJobs, 0, new int[k]);13        return minMaxTime;14    }15 16    private void backtrack(int[] jobs, int jobIndex, int[] workerTimes) {17        if (jobIndex == jobs.length) {18            int currentMax = 0;19            for (int time : workerTimes) {20                currentMax = Math.max(currentMax, time);21            }22            minMaxTime = Math.min(minMaxTime, currentMax);23            return;24        }25        26        for (int i = 0; i < workerTimes.length; i++) {27            if (workerTimes[i] + jobs[jobIndex] >= minMaxTime) {28                continue;29            }30            31            workerTimes[i] += jobs[jobIndex];32            backtrack(jobs, jobIndex + 1, workerTimes);33            workerTimes[i] -= jobs[jobIndex];34 35            // Symmetry breaking: if this worker was empty,36            // no need to try other empty workers.37            if (workerTimes[i] == 0) {38                break;39            }40        }41    }42}

• The possible values for the minimum maximum working time are within a specific range. The lower bound is the largest single job time, and the upper bound is the sum of all job times.
• We perform a binary search on this range of possible answers. For each mid value, we check if it's possible to assign all jobs such that no worker's time exceeds mid.
• Algorithm Steps:
  1. Define the search range: low = max(jobs), high = sum(jobs).
  2. Perform binary search:
     • While low <= high, pick mid = low + (high - low) / 2.
     • Call a helper function isPossible(jobs, k, mid) to check if an assignment is feasible with mid as the time limit.
     • If isPossible is true, it means mid is a potential answer. We try for a better (smaller) one: ans = mid, high = mid - 1.
     • If isPossible is false, mid is too small: low = mid + 1.
  3. The isPossible function is implemented using a backtracking/DFS approach, similar to the first method.
• isPossible(jobs, k, limit) function:
  1. Sort jobs descendingly for better pruning.
  2. Use a recursive function canAssign(jobIndex, workerTimes, limit).
  3. Base Case: If jobIndex reaches the end, all jobs are assigned within the limit. Return true.
  4. Recursive Step: For the current jobs[jobIndex], try to assign it to each worker i. If workerTimes[i] + jobs[jobIndex] <= limit, assign the job and recurse. If the recursive call returns true, we've found a valid assignment, so return true up the call stack. Backtrack if it fails.
  5. Apply the same symmetry-breaking optimization as in the pure backtracking approach.

This approach reframes the problem from an optimization problem ('find the minimum maximum') to a decision problem ('is it possible to achieve a maximum time of X?'). We can then binary search for the smallest X for which the answer is 'yes'.

The possible values for the minimum maximum working time are within a specific range. The lower bound is the largest single job time (as one worker must take it), and the upper bound is the sum of all job times (the case with one worker).

We perform a binary search on this range of possible answers. For each mid value in the binary search, we check if it's possible to assign all jobs such that no worker's time exceeds mid. This check is done by a helper function, isPossible, which itself uses a backtracking search. The key difference from the first approach is that the pruning condition is now a fixed limit (mid), which can make the search faster.

1class Solution {2    public int minimumTimeRequired(int[] jobs, int k) {3        Arrays.sort(jobs);4        int[] reversedJobs = new int[jobs.length];5        int sum = 0;6        int maxJob = 0;7        for (int i = 0; i < jobs.length; i++) {8            reversedJobs[i] = jobs[jobs.length - 1 - i];9            sum += jobs[i];10            maxJob = Math.max(maxJob, jobs[i]);11        }12 13        int low = maxJob, high = sum;14        int ans = high;15 16        while (low <= high) {17            int mid = low + (high - low) / 2;18            if (isPossible(reversedJobs, k, mid)) {19                ans = mid;20                high = mid - 1;21            } else {22                low = mid + 1;23            }24        }25        return ans;26    }27 28    private boolean isPossible(int[] jobs, int k, int limit) {29        return canAssign(jobs, 0, new int[k], limit);30    }31 32    private boolean canAssign(int[] jobs, int jobIndex, int[] workerTimes, int limit) {33        if (jobIndex == jobs.length) {34            return true;35        }36 37        for (int i = 0; i < workerTimes.length; i++) {38            if (workerTimes[i] + jobs[jobIndex] <= limit) {39                workerTimes[i] += jobs[jobIndex];40                if (canAssign(jobs, jobIndex + 1, workerTimes, limit)) {41                    return true;42                }43                workerTimes[i] -= jobs[jobIndex];44            }45            // Symmetry breaking46            if (workerTimes[i] == 0) {47                break;48            }49        }50        return false;51    }52}

• DP State: dp[i][mask] represents the minimum possible maximum working time if we use i workers to complete the set of jobs represented by mask. A 1 at the j-th position in mask means job j is included.
• Precomputation: First, we precompute an array sum[mask] that stores the total time for each subset of jobs. This can be done in O(n * 2^n) or O(2^n).
• Algorithm Steps:
  1. Precompute sum[mask] for all 2^n masks.
  2. Initialize a DP table dp[k+1][1<<n].
  3. Base Case: For one worker (i=1), the max time is just the sum of jobs in the subset: dp[1][mask] = sum[mask].
  4. Transition: We iterate from i = 2 to k workers. To compute dp[i][mask], we consider all possible ways to form the i-th worker's assignment. We can give a subset of jobs sub (a submask of mask) to the i-th worker. The remaining i-1 workers must handle the jobs in mask \ sub. The maximum time for this specific split is max(sum[sub], dp[i-1][mask ^ sub]). We want to find the sub that minimizes this value. dp[i][mask] = min( max(dp[i-1][mask ^ sub], sum[sub]) ) for all non-empty proper submasks sub of mask.
  5. The final answer is dp[k][(1<<n) - 1].

Given the small constraint on the number of jobs (n <= 12), we can use dynamic programming with bitmasking to represent subsets of jobs. This approach directly calculates the minimum maximum time without binary search.

We build the solution worker by worker. The state of our DP will involve the number of workers used and the subset of jobs completed.

DP State: dp[i][mask] represents the minimum possible maximum working time if we use i workers to complete the set of jobs represented by mask. A 1 at the j-th position in mask means job j is included.

First, we precompute an array sum[mask] that stores the total time for each subset of jobs. Then, we build our dp table. The base case is for i=1 worker, where dp[1][mask] is simply sum[mask]. For i > 1, we compute dp[i][mask] by iterating through all submasks sub of mask. We assign the jobs in sub to the i-th worker and the jobs in mask ^ sub to the previous i-1 workers. The maximum time for this split is max(sum[sub], dp[i-1][mask ^ sub]). We take the minimum over all possible submasks sub.

1class Solution {2    public int minimumTimeRequired(int[] jobs, int k) {3        int n = jobs.length;4        int[] sum = new int[1 << n];5        for (int mask = 1; mask < (1 << n); mask++) {6            // Find the lowest set bit (a job in the mask)7            int j = Integer.numberOfTrailingZeros(mask);8            int prevMask = mask ^ (1 << j);9            sum[mask] = sum[prevMask] + jobs[j];10        }11 12        int[][] dp = new int[k + 1][1 << n];13        14        // Base case: 1 worker15        for (int mask = 0; mask < (1 << n); mask++) {16            dp[1][mask] = sum[mask];17        }18 19        // Fill DP table for i = 2 to k workers20        for (int i = 2; i <= k; i++) {21            for (int mask = 1; mask < (1 << n); mask++) {22                dp[i][mask] = Integer.MAX_VALUE;23                // Iterate through all submasks of mask24                for (int sub = mask; sub > 0; sub = (sub - 1) & mask) {25                    int remainingMask = mask ^ sub;26                    int currentMax = Math.max(dp[i - 1][remainingMask], sum[sub]);27                    dp[i][mask] = Math.min(dp[i][mask], currentMax);28                }29            }30        }31 32        return dp[k][(1 << n) - 1];33    }34}

• Similar to the second approach, we binary search for the minimum possible maximum time, limit.
• Decision Problem: 'Is it possible to partition the jobs array into k or fewer subsets, such that the sum of each subset does not exceed limit?'
• DP State for isPossible: We use a DP table dp[mask] to store information about partitioning the jobs in mask. A simple and effective state is dp[mask] = (workers_needed, time_of_last_worker). We aim to minimize workers_needed first, and then time_of_last_worker as a tie-breaker.
• Algorithm Steps:
  1. Perform binary search on the answer limit in the range [max(jobs), sum(jobs)].
  2. For each limit, call isPossible(jobs, k, limit).
• isPossible(jobs, k, limit) function:
  1. Define dp[mask] as a pair {num_workers, last_worker_time}.
  2. Initialize dp[0] = {1, 0}. All other dp[mask] are initialized to an invalid state like {k+1, 0}.
  3. Iterate mask from 1 to (1<<n) - 1.
  4. For each mask, iterate through each job j that is part of the mask.
  5. Let prev_mask = mask ^ (1 << j). Get the state dp[prev_mask].
  6. Consider adding job j to the partition for prev_mask. We can either add it to the last worker if time permits, or start a new worker.
  7. Update dp[mask] with the lexicographically smaller of the new state and its current value.
  8. After filling the DP table, if dp[(1<<n) - 1].num_workers <= k, return true.

This is the most efficient approach, combining the binary search strategy with a fast DP-based check for the decision problem.

Similar to the second approach, we binary search for the minimum possible maximum time, limit. The core of this method is a highly efficient isPossible(limit) check using dynamic programming.

The decision problem is: 'Is it possible to partition the jobs array into k or fewer subsets, such that the sum of each subset does not exceed limit?'

We can solve this with DP. Let dp[mask] store the time taken by the last worker for an optimal partition of jobs in mask, and workers[mask] store the number of workers needed. We want to minimize workers, and then dp as a tie-breaker.

We iterate through each mask from 1 to (1<<n)-1. For each job j in the mask, we look at the state for mask ^ (1<<j). We try to add job j to this previous state. If we can add it to the last worker without exceeding limit, we do so. Otherwise, we must start a new worker. We update the state for mask if we find a better partition (fewer workers, or same workers with less time on the last one).

After computing the DP table, we check if workers[(1<<n)-1] is at most k. If so, the limit is achievable.

1class Solution {2    public int minimumTimeRequired(int[] jobs, int k) {3        int n = jobs.length;4        int sum = 0;5        int maxJob = 0;6        for (int job : jobs) {7            sum += job;8            maxJob = Math.max(maxJob, job);9        }10 11        int low = maxJob, high = sum;12        int ans = high;13 14        while (low <= high) {15            int mid = low + (high - low) / 2;16            if (isPossible(jobs, k, mid)) {17                ans = mid;18                high = mid - 1;19            } else {20                low = mid + 1;21            }22        }23        return ans;24    }25 26    private boolean isPossible(int[] jobs, int k, int limit) {27        int n = jobs.length;28        int[] dp = new int[1 << n]; // Stores time of last worker29        int[] workers = new int[1 << n]; // Stores num of workers30        Arrays.fill(dp, Integer.MAX_VALUE);31        Arrays.fill(workers, Integer.MAX_VALUE);32        dp[0] = 0;33        workers[0] = 1;34 35        for (int mask = 1; mask < (1 << n); mask++) {36            for (int j = 0; j < n; j++) {37                if ((mask & (1 << j)) != 0) {38                    int prevMask = mask ^ (1 << j);39                    if (workers[prevMask] == Integer.MAX_VALUE) continue;40 41                    // Option 1: Add to the last worker of the previous state42                    if (dp[prevMask] + jobs[j] <= limit) {43                        if (workers[prevMask] < workers[mask] || 44                           (workers[prevMask] == workers[mask] && dp[prevMask] + jobs[j] < dp[mask])) {45                            workers[mask] = workers[prevMask];46                            dp[mask] = dp[prevMask] + jobs[j];47                        }48                    } 49                    // Option 2: Start a new worker50                    else {51                        if (workers[prevMask] + 1 < workers[mask] ||52                           (workers[prevMask] + 1 == workers[mask] && jobs[j] < dp[mask])) {53                            workers[mask] = workers[prevMask] + 1;54                            dp[mask] = jobs[j];55                        }56                    }57                }58            }59        }60        return workers[(1 << n) - 1] <= k;61    }62}