How do you solve Number of Ways to Divide a Long Corridor on LeetCode?

Number of Ways to Divide a Long Corridor (LeetCode #2147) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The number of 'S' must be even to form pairs.

What is the time complexity of Number of Ways to Divide a Long Corridor?

The optimal solution for Number of Ways to Divide a Long Corridor runs in O(n) time and uses O(n) space. This complexity is linear because we only traverse the string a couple of times and store the indices of 'S'.

What is the brute force solution for Number of Ways to Divide a Long Corridor?

The brute force approach for Number of Ways to Divide a Long Corridor is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible ways to divide the corridor by checking every possible segment of two seats. This method is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Ways to Divide a Long Corridor?

Number of Ways to Divide a Long Corridor uses the following concepts: Math, String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Combinatorial Counting.

What are the interview tips for Number of Ways to Divide a Long Corridor?

Always start with edge cases, such as empty strings or strings with no 'S'. Think about how to break the problem down into smaller parts, like counting pairs. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Number of Ways to Divide a Long Corridor?

Not checking if the count of 'S' is even before proceeding. Overcomplicating the problem by trying to visualize every possible configuration instead of focusing on pairs.

#2147

Number of Ways to Divide a Long Corridor

Hard

MathStringDynamic ProgrammingDynamic ProgrammingCombinatorial Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

By counting the number of 'S' and ensuring they can be grouped into pairs, we can calculate the number of ways to place dividers more efficiently.

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Algorithm

4 steps

1Step 1: Count the total number of 'S' in the corridor.
2Step 2: If the count of 'S' is odd or less than 2, return 0 as we can't form pairs.
3Step 3: Calculate the number of pairs of 'S' and the number of plants in between them.
4Step 4: Use combinatorial mathematics to calculate the number of ways to place dividers.

solution.py9 lines

1def countWays(corridor):
2    mod = 10**9 + 7
3    seats = [i for i, c in enumerate(corridor) if c == 'S']
4    if len(seats) % 2 != 0:
5        return 0
6    ways = 1
7    for i in range(1, len(seats) // 2):
8        ways = (ways * (seats[2 * i] - seats[2 * i - 1])) % mod
9    return ways

ℹ

Complexity note: This complexity is linear because we only traverse the string a couple of times and store the indices of 'S'.

1The number of 'S' must be even to form pairs.
2The spaces between pairs of 'S' determine the number of valid configurations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.