Valid permutations for di sequence¶
Time: O(N^2); Space: O(N); hard
We are given S, a length n string of characters from the set {‘D’, ‘I’}. (These letters stand for “decreasing” and “increasing”.)
A valid permutation is a permutation P[0], P[1], …, P[n] of integers {0, 1, …, n}, such that for all i:
If S[i] == ‘D’, then P[i] > P[i+1], and;
If S[i] == ‘I’, then P[i] < P[i+1].
How many valid permutations are there? Since the answer may be large, return your answer modulo 10^9 + 7.
Example 1:
Input: “DID”
Output: 5
Explanation:
The 5 valid permutations of (0, 1, 2, 3) are:
(1, 0, 3, 2)
(2, 0, 3, 1)
(2, 1, 3, 0)
(3, 0, 2, 1)
(3, 1, 2, 0)
Example 2:
Input: “DDID”
Output: 9
Constraints:
1 <= len(S) <= 200
S consists only of characters from the set {‘D’, ‘I’}
1. Dynamic Programming [O(N^3), O(N^2)]¶
Intuition
When writing the permutation P = P_0, P_1, …, P_N from left to right, we only care about the relative rank of the last element placed. For example, if N = 5 (so that we have elements {0, 1, 2, 3, 4, 5}), and our permutation starts 2, 3, 4, then it is similar to a situation where we have placed ?, ?, 2 and the remaining elements are {0, 1, 3}, in terms of how many possibilities there are to place the remaining elements in a valid way.
To this end, let dp(i, j) be the number of ways to place every number up to and inlcuding P_i, such that P_i when placed had relative rank j. (Namely, there are j remaining numbers less than P_i.)
Algorithm
When placing P_i following a decreasing instruction S[i-1] == ‘D’, we want P_{i-1} to have a higher value. When placing P_i following an increasing instruction, we want P_{i-1} to have a lower value. It is relatively easy to deduce the recursion from this fact.
[16]:
from functools import lru_cache
class Solution1(object):
"""
Time: O(N^3), where N is the length of S
Space: O(N^2)
"""
def numPermsDISequence(self, S):
"""
:type S: str
:rtype: int
"""
MOD = 10**9 + 7
N = len(S)
@lru_cache(None)
def dp(i, j):
# How many ways to place P_i with relative rank j?
if i == 0:
return 1
elif S[i-1] == 'D':
return sum(dp(i-1, k) for k in range(j, i)) % MOD
else:
return sum(dp(i-1, k) for k in range(j)) % MOD
return sum(dp(N, j) for j in range(N+1)) % MOD
[17]:
s = Solution1()
S = "DID"
assert s.numPermsDISequence(S) == 5
S = "DDID"
assert s.numPermsDISequence(S) == 9
2. Dynamic Programming with Optimization [O(N^2), O(N^2)]¶
Optimization
Actually, we can do better than this. For any given i, let’s look at how the sum of D_k = dp(i-1, k) is queried. Assuming S[i-1] == ‘I’, we query D_0, D_0 + D_1, D_0 + D_1 + D_2, … etc. The case for S[i-1] == ‘D’ is similar.
Thus, we don’t need to query the sum every time. Instead, we could use (for S[i-1] == ‘I’) the fact that dp(i, j) = dp(i, j-1) + dp(i-1, j-1). For S[i-1] == ‘D’, we have the similar fact that dp(i, j) = dp(i, j+1) + dp(i-1, j).
These two facts make the work done for each state of dp have O(1) (amortized) complexity, leading to a total time complexity of O(N^2) for this solution.
[18]:
from functools import lru_cache
class Solution2(object):
def numPermsDISequence(self, S):
MOD = 10**9 + 7
N = len(S)
@lru_cache(None)
def dp(i, j):
# How many ways to place P_i with relative rank j?
if not(0 <= j <= i):
return 0
if i == 0:
return 1
elif S[i-1] == 'D':
return (dp(i, j+1) + dp(i-1, j)) % MOD
else:
return (dp(i, j-1) + dp(i-1, j-1)) % MOD
return sum(dp(N, j) for j in range(N+1)) % MOD
[19]:
s = Solution2()
S = "DID"
assert s.numPermsDISequence(S) == 5
S = "DDID"
assert s.numPermsDISequence(S) == 9
3. Dynamic Programming [O(N^2), O(N)]¶
[20]:
class Solution3(object):
"""
Time: O(N^2)
Space: O(N)
"""
def numPermsDISequence(self, S):
"""
:type S: str
:rtype: int
"""
MOD = 10**9 + 7
dp = [1]*(len(S)+1)
for c in S:
if c == "I":
dp = dp[:-1]
for i in range(1, len(dp)):
dp[i] += dp[i-1]
else:
dp = dp[1:]
for i in reversed(range(len(dp)-1)):
dp[i] += dp[i+1]
return dp[0] % MOD
[21]:
s = Solution3()
S = "DID"
assert s.numPermsDISequence(S) == 5
S = "DDID"
assert s.numPermsDISequence(S) == 9
4: Divide and Conquer [O(N^2), O(N^2)]¶
Intuition
Let’s place the zero of the permutation first. It either goes between a ‘DI’ part of the sequence, or it could go on the ends (the left end if it starts with ‘I’, and the right end if it ends in ‘D’.) Afterwards, this splits the problem into two disjoint subproblems that we can solve with similar logic.
Algorithm
Let dp(i, j) be the number of valid permutations (of n = j-i+2 total integers from 0 to n-1) corresponding to the DI sequence S[i], S[i+1], …, S[j]. If we can successfully place a zero between S[k-1] and S[k], then there are two disjoint problems S[i], …, S[k-2] and S[k+1], …, S[j].
To count the number of valid permutations in this case, we should choose k-i elements from n-1 (n total integers, minus the zero) to put in the left group; then the answer is this, times the number of ways to arrange the left group [dp(i, k-2)], times the number of ways to arrange the right group [dp(k+1, j)].
[22]:
from functools import lru_cache
class Solution4(object):
"""
Time: O(N^2), where NN is the length of S
Space: O(N^2)
"""
def numPermsDISequence(self, S):
"""
:type S: str
:rtype: int
"""
MOD = 10**9 + 7
fac = [1, 1]
for x in range(2, 201):
fac.append(fac[-1] * x % MOD)
facinv = [pow(f, MOD-2, MOD) for f in fac]
def binom(n, k):
return fac[n] * facinv[n-k] % MOD * facinv[k] % MOD
@lru_cache(None)
def dp(i, j):
if i >= j: return 1
ans = 0
n = j - i + 2
if S[i] == 'I': ans += dp(i+1, j)
if S[j] == 'D': ans += dp(i, j-1)
for k in range(i+1, j+1):
if S[k-1:k+1] == 'DI':
ans += binom(n-1, k-i) * dp(i, k-2) % MOD * dp(k+1, j) % MOD
ans %= MOD
return ans
return dp(0, len(S) - 1)
[23]:
s = Solution4()
S = "DID"
assert s.numPermsDISequence(S) == 5
S = "DDID"
assert s.numPermsDISequence(S) == 9