What is the brute force solution for Minimum Number of Work Sessions to Finish the Tasks?

The brute force approach for Minimum Number of Work Sessions to Finish the Tasks is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying all possible combinations of tasks to see how many work sessions are needed. This is straightforward but inefficient, as it explores every possible arrangement of tasks. The optimal Optimal Solution approach improves this to O(n * 2^n).

What are the interview tips for Minimum Number of Work Sessions to Finish the Tasks?

Explain your thought process clearly when discussing brute-force vs. optimal solutions. Be prepared to discuss the trade-offs between time and space complexity. Practice writing clean and efficient code under time constraints.

What are common mistakes when solving Minimum Number of Work Sessions to Finish the Tasks?

Not considering all combinations of tasks in the brute-force approach. Overlooking edge cases where tasks fit perfectly into sessions.

#1986

Minimum Number of Work Sessions to Finish the Tasks

Medium

ArrayDynamic ProgrammingBacktrackingBit ManipulationBitmaskDynamic ProgrammingBit Manipulation

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n)
Space	O(1)	O(2^n)

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Intuition

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

The optimal solution uses dynamic programming with bit masking to efficiently track which tasks have been completed. This reduces the number of states we need to explore and allows us to reuse previously computed results.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array to store the minimum sessions needed for each state of completed tasks.
2Step 2: Iterate through all possible states of completed tasks using bit masking.
3Step 3: For each state, try adding tasks to the current session and update the DP array accordingly.

solution.py15 lines

1# Full working Python code
2def minSessions(tasks, sessionTime):
3    n = len(tasks)
4    dp = [float('inf')] * (1 << n)
5    dp[0] = 0
6    for mask in range(1 << n):
7        total = 0
8        for i in range(n):
9            if mask & (1 << i) == 0:
10                if total + tasks[i] <= sessionTime:
11                    total += tasks[i]
12                else:
13                    dp[mask | (1 << i)] = min(dp[mask | (1 << i)], dp[mask] + 1)
14    return dp[(1 << n) - 1]
15

ℹ

Complexity note: The time complexity is O(n * 2^n) because we explore all possible states of tasks using bit masking, and for each state, we may check all tasks. The space complexity is O(2^n) due to the DP array storing results for each state.

1Dynamic programming can significantly reduce the number of states to explore.
2Bit masking is a powerful technique for problems involving subsets.

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