What is the time complexity of Minimum Number of Work Sessions to Finish the Tasks?

The optimal solution has a time complexity of O(n * 2^n).

What is the space complexity of Minimum Number of Work Sessions to Finish the Tasks?

The space complexity is O(2^n).

What data structures are used to solve Minimum Number of Work Sessions to Finish the Tasks?

This problem uses Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Number of Work Sessions to Finish the Tasks?

Insight 1: Since the number of tasks is small (<=14), we can use bitmasking to represent the state of completed tasks. Each bit in the mask corresponds to a task, with a '1' indicating that the task has been completed and a '0' indicating it hasn't. Insight 2: Dynamic programming can be used to efficiently determine the minimum number of sessions needed for each possible state (combination of completed tasks).

Which companies ask LeetCode 1986. Minimum Number of Work Sessions to Finish the Tasks in interviews?

This problem is frequently asked at Swiggy.

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Minimum Number of Work Sessions to Finish the Tasks - LeetCode 1986 Solution

Q: How to solve Minimum Number of Work Sessions to Finish the Tasks?

The solution uses dynamic programming with bitmasking. A `dp` array is created where `dp[mask]` stores a tuple (or pair) representing the minimum number of sessions needed to complete the tasks represented by `mask`, and the remaining time in the last session. The algorithm iterates through all possible subsets of tasks (represented by bitmasks). For each subset, it considers each task within that subset. It tries to either add the task to the current session (if there is enough time) or start a new session with that task. The `dp` array is updated to store the best (minimum number of sessions) solution for each state. The algorithm leverages the constraints on the size of the input to create an efficient approach.

Q: Which companies ask LeetCode 1986. Minimum Number of Work Sessions to Finish the Tasks in interviews?

This problem is frequently asked at Swiggy.

Minimum Number of Work Sessions to Finish the Tasks - Complete Solution Guide

Minimum Number of Work Sessions to Finish the Tasks is LeetCode problem 1986, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n tasks assigned to you. The task times are represented as an integer array tasks of length n , where the i th task takes tasks[i] hours to finish. A work session is when you work for at most sessionTime consecutive hours and then take a break. You should finish the given tasks in a way that satisfies the following conditions: If you start a task in a work session, you must complete it in the same work session. You can start a new task immediately after finishing the previous one. You

Detailed Explanation

The problem asks us to find the minimum number of work sessions required to complete a set of tasks, given a maximum time limit for each session. Each task must be completed within a single session, and tasks can be completed in any order. The input consists of an array `tasks` representing the time each task takes, and `sessionTime` which is the maximum time available in a work session. The constraints limit the number of tasks to a maximum of 14, implying that a bitmasking approach is feasible.

Solution Approach

The solution uses dynamic programming with bitmasking. A `dp` array is created where `dp[mask]` stores a tuple (or pair) representing the minimum number of sessions needed to complete the tasks represented by `mask`, and the remaining time in the last session. The algorithm iterates through all possible subsets of tasks (represented by bitmasks). For each subset, it considers each task within that subset. It tries to either add the task to the current session (if there is enough time) or start a new session with that task. The `dp` array is updated to store the best (minimum number of sessions) solution for each state. The algorithm leverages the constraints on the size of the input to create an efficient approach.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `2^n`, where n is the number of tasks. Each element `dp[mask]` is a tuple (sessions, time_left), initialized to (n+1, sessionTime+1), representing an impossibly high number of sessions and time left. `dp[0]` is initialized to (0, 0), because no tasks done requires 0 sessions and 0 time spent.
Step 2: Iterate through all possible masks from 1 to `2^n - 1`. Each mask represents a subset of tasks.
Step 3: For each mask, iterate through all tasks from 0 to `n - 1`.
Step 4: Check if the `i`-th task is present in the current mask (i.e., if `(mask >> i) & 1` is true).
Step 5: If the `i`-th task is present, consider the previous state, `prev_mask = mask ^ (1 << i)`, which is the mask without the `i`-th task.
Step 6: Retrieve the number of sessions and time left from `dp[prev_mask]`. If `dp[prev_mask].sessions == 0`, this is the initial state and thus start first session, new_sessions = 1, new_time = tasks[i]
Step 7: If adding the current task to the previous session is possible (`dp[prev_mask].time_left + tasks[i] <= sessionTime`), then update the new session with the previous sessions and reduce the remaining time. new_sessions = prev_sessions, new_time = prev_time + tasks[i]
Step 8: Otherwise, a new session needs to be started, so the current task will take the full amount of the current session. new_sessions = prev_sessions + 1, new_time = tasks[i]
Step 9: Update `dp[mask]` with the minimum number of sessions and minimum remaining time.
Step 10: After iterating through all masks, the minimum number of sessions required to complete all tasks is stored in `dp[(1 << n) - 1].sessions`.

Key Insights

Insight 1: Since the number of tasks is small (<=14), we can use bitmasking to represent the state of completed tasks. Each bit in the mask corresponds to a task, with a '1' indicating that the task has been completed and a '0' indicating it hasn't.
Insight 2: Dynamic programming can be used to efficiently determine the minimum number of sessions needed for each possible state (combination of completed tasks).
Insight 3: The DP state represents the minimum number of sessions required to complete the tasks represented by the bitmask, along with the remaining time left in the current session.

Complexity Analysis

Time Complexity: O(n * 2^n)

Space Complexity: O(2^n)

Topics

This problem involves: Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask.

Companies

Asked at: Swiggy.

Minimum Number of Work Sessions to Finish the Tasks | LeetCode 1986

Medium

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Time: O(n * 2^n)Space: O(2^n)

Solution Code

Available in:

class Solution {
    public int minSessions(int[] tasks, int sessionTime) {
        int n = tasks.length;
        
        // dp[mask] = (min_sessions, min_time_in_last_session)
        // Initialize with a value larger than any possible outcome
        int[][] dp = new int[1 << n][2];
        for (int i = 0; i < (1 << n); i++) {
            dp[i][0] = n + 1;
            dp[i][1] = sessionTime + 1;
        }
        
        // Base case: 0 tasks require 0 sessions and 0 time.
        dp[0][0] = 0;
        dp[0][1] = 0;
        
        for (int mask = 1; mask < (1 << n); mask++) {
            for (int i = 0; i < n; i++) {
                // If task i is in the current subset (mask)
                if ((mask >> i) % 2 == 1) {
                    // Consider the state before adding task i
                    int prev_mask = mask ^ (1 << i);
                    int prev_sessions = dp[prev_mask][0];
                    int prev_time = dp[prev_mask][1];
                    
                    int new_sessions;
                    int new_time;
                    
                    if (prev_sessions == 0) { // This means prev_mask was 0
                        // This is the first task being added to an empty set.
                        // It will always start the first session.
                        new_sessions = 1;
                        new_time = tasks[i];
                    } else if (prev_time + tasks[i] <= sessionTime) {
                        // Add task i to the last session of the previous configuration.
                        new_sessions = prev_sessions;
                        new_time = prev_time + tasks[i];
                    } else {
                        // Start a new session for task i.
                        new_sessions = prev_sessions + 1;
                        new_time = tasks[i];
                    }

                    // Update dp[mask] if we found a better state.
                    // Python tuples are compared lexicographically by default.
                    if (new_sessions < dp[mask][0] || (new_sessions == dp[mask][0] && new_time < dp[mask][1])) {
                        dp[mask][0] = new_sessions;
                        dp[mask][1] = new_time;
                    }
                }
            }
        }
        
        // The result for all tasks is the number of sessions for the mask with all bits set.
        return dp[(1 << n) - 1][0];
    }
}

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Problem Info

Description

Problem 1986: Minimum Number of Work Sessions to Finish the Tasks - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Companies

Swiggy

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1986

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a DP array `dp` of size `2^n`, where n is the number of tasks. Each element `dp[mask]` is a tuple (sessions, time_left), initialized to (n+1, sessionTime+1), representing an impossibly high number of sessions and time left. `dp[0]` is initialized to (0, 0), because no tasks done requires 0 sessions and 0 time spent.
2
Step 2: Iterate through all possible masks from 1 to `2^n - 1`. Each mask represents a subset of tasks.
3
Step 3: For each mask, iterate through all tasks from 0 to `n - 1`.
4
Step 4: Check if the `i`-th task is present in the current mask (i.e., if `(mask >> i) & 1` is true).
5
Step 5: If the `i`-th task is present, consider the previous state, `prev_mask = mask ^ (1 << i)`, which is the mask without the `i`-th task.
6
Step 6: Retrieve the number of sessions and time left from `dp[prev_mask]`. If `dp[prev_mask].sessions == 0`, this is the initial state and thus start first session, new_sessions = 1, new_time = tasks[i]
7
Step 7: If adding the current task to the previous session is possible (`dp[prev_mask].time_left + tasks[i] <= sessionTime`), then update the new session with the previous sessions and reduce the remaining time. new_sessions = prev_sessions, new_time = prev_time + tasks[i]
8
Step 8: Otherwise, a new session needs to be started, so the current task will take the full amount of the current session. new_sessions = prev_sessions + 1, new_time = tasks[i]
9
Step 9: Update `dp[mask]` with the minimum number of sessions and minimum remaining time.
10
Step 10: After iterating through all masks, the minimum number of sessions required to complete all tasks is stored in `dp[(1 << n) - 1].sessions`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n * 2^n)

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity is determined by the nested loops that iterate through all possible combinations of tasks (represented by bitmasks). The space complexity arises from storing the dynamic programming table, which needs to hold information for each possible subset of tasks.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Backtracking

A backtracking approach could be used to explore all possible task orderings and session assignments. However, it would likely be significantly slower than the dynamic programming approach due to the exponential number of possible orderings.

Time: O(n!), where n is the number of tasks in the worst case.Space: O(n) for the recursion stack.

Trade-offs: Backtracking is easier to understand conceptually, but it's much less efficient than the DP approach, especially for larger values of n. DP avoids redundant computations by storing and reusing previously computed results.

✨

Key Insights

1Insight 1: Since the number of tasks is small (<=14), we can use bitmasking to represent the state of completed tasks. Each bit in the mask corresponds to a task, with a '1' indicating that the task has been completed and a '0' indicating it hasn't.
2Insight 2: Dynamic programming can be used to efficiently determine the minimum number of sessions needed for each possible state (combination of completed tasks).
3Insight 3: The DP state represents the minimum number of sessions required to complete the tasks represented by the bitmask, along with the remaining time left in the current session.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `tasks` is empty. The code implicitly handles this as `1 << 0 == 1` so the outermost loop will not execute and the initial dp[0] value which is 0 will be returned.
Edge case 2: All tasks can be completed within one session (`sum(tasks) <= sessionTime`). The dynamic programming algorithm correctly finds this optimal solution.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding bitmasking. Bitmasking is crucial for efficiently representing the state of which tasks have been completed.

✗Common mistake 2: Incorrectly initializing the `dp` array. The initial values must be large enough so that any valid solution will be smaller, allowing the algorithm to find the optimal minimum number of sessions.

✗Common mistake 3: Failing to consider all possible combinations of tasks. The algorithm must systematically explore all subsets of tasks to ensure the optimal solution is found.

✗Common mistake 4: Incorrectly using the prev_mask. Must ensure that the previous mask is actually the mask without the current task.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling with resource constraints. This algorithm can be applied to optimize the allocation of tasks to workers, considering limitations on worker availability or resource capacity.

🏦Real-world scenario 2: Optimizing resource usage in cloud computing. This type of problem arises in cloud resource management where you need to allocate compute instances (sessions) to complete a set of jobs (tasks) under a certain cost constraint (sessionTime).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Number of Work Sessions to Finish the Tasks"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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