Walter Teng.2147 Number of Ways to Divide a Long Corridor - Hard
Problem:
Along a long library corridor, there is a line of seats and decorative plants. You are given a 0-indexed string corridor of length n consisting of letters 'S' and 'P' where each 'S' represents a seat and each 'P' represents a plant.
One room divider has already been installed to the left of index 0, and another to the right of index n - 1. Additional room dividers can be installed. For each position between indices i - 1 and i (1 <= i <= n - 1), at most one divider can be installed.
Divide the corridor into non-overlapping sections, where each section has exactly two seats with any number of plants. There may be multiple ways to perform the division. Two ways are different if there is a position with a room divider installed in the first way but not in the second way.
Return the number of ways to divide the corridor. Since the answer may be very large, return it modulo 109 + 7. If there is no way, return 0.
Example 1:
Input: corridor = "SSPPSPS" Output: 3 Explanation: There are 3 different ways to divide the corridor. The black bars in the above image indicate the two room dividers already installed. Note that in each of the ways, each section has exactly two seats.
Example 2:
Input: corridor = "PPSPSP" Output: 1 Explanation: There is only 1 way to divide the corridor, by not installing any additional dividers. Installing any would create some section that does not have exactly two seats.
Example 3:
Input: corridor = "S" Output: 0 Explanation: There is no way to divide the corridor because there will always be a section that does not have exactly two seats.
Constraints:
n == corridor.length1 <= n <= 105corridor[i]is either'S'or'P'.
Problem Analysis:
Method 1 (DP):
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form a 2d array to cache (i, seats) -> count
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recusively perform
dfs- base case: if we reach the end of the corridor return 1 if seats are 2 else 0
- if current seats count is 2 and the current position is a seat, reset count to 0 and increment by 1 for the current seat and perfom dfs
- else if current seats count is 2 and current seat is plant, there is option to put divider or not. reset count if divider is placed. else maintain seat count and perform recursion
- else if current seats count is not 2, increment seat count if current position is seat and perform recursion
-
Time Complexity: O(n)
-
Space Complexity: O(n)
Method 2 (Combinatorics):
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count all the seats available
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for every 2 seats count the distance between each pair of seats.
-
multiply these values to get the combinations
-
Time Complexity: O(n)
-
Space Complexity: O(n)
Solutions:
Dynamic Programming
class Solution:
def numberOfWays(self, corridor: str) -> int:
MOD = 10**9 + 7
# memorization
cache = [[-1] * 3 for i in range(len(corridor))] # (i, seats) -> count
def dfs(i, seats):
if i == len(corridor):
return 1 if seats == 2 else 0
if cache[i][seats] != -1:
return cache[i][seats]
res = 0
if seats == 2:
if corridor[i] == "S":
# reset to 0 + 1 seat after counted 2 seat
res = dfs(i+1, 1)
else:
# else if current seat is plant
# reset to 0 after putting divider or
# choose not to put divider
res = (dfs(i+1, 0) + dfs(i+1, 2)) % MOD
else:
if corridor[i] == "S":
# add since current is seat
res = dfs(i+1, seats + 1)
else:
# dont add since current is plant
res = dfs(i+1, seats)
cache[i][seats] = res
return res
return dfs(0,0)```
### Combinatorics
```python
class Solution:
def numberOfWays(self, corridor: str) -> int:
MOD = 10**9 + 7
seats = []
for i, c in enumerate(corridor):
if c == "S":
seats.append(i)
length = len(seats)
# base case
if length <2 or length %2 == 1:
return 0
res = 1
i = 1
# count valid combinations by counting seat pairs
while i < length - 1:
res = (res * (seats[i+1] - seats[i])) % MOD
i += 2
return res