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308 lines
6.6 KiB
Markdown
308 lines
6.6 KiB
Markdown
Recursion and Backtracking 2
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----------------------------
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```
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https://www.interviewbit.com/problems/kth-permutation-sequence/
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https://www.interviewbit.com/problems/number-of-squareful-arrays/
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https://www.interviewbit.com/problems/gray-code/
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https://www.interviewbit.com/problems/combination-sum-ii/
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https://www.interviewbit.com/problems/nqueens/
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https://www.interviewbit.com/problems/all-unique-permutations/
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https://www.interviewbit.com/problems/sorted-permutation-rank/
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https://www.interviewbit.com/problems/word-break-ii/
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https://www.interviewbit.com/problems/palindrome-partitioning/
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https://www.interviewbit.com/problems/unique-paths-iii/
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```
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Permutations
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------------
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> Given String containing distinct characters. Print all permutations of the string
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>
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Approach:
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Fix the first char
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Can essentially be achieved by swapping
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```python
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def permute(S, i):
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if i == len(S):
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print(S)
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for j in range(i, len(S)):
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S[i], S[j] = S[j], S[i]
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permute(S, i+1)
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S[i], S[j] = S[j], S[i] # backtrack
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```
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-- --
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Lexicographic permutations
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--------------------------
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asked in Amazon
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- start with sorted elements
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- instead of swap, do right rotation(i to j). Undo = left rotate
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- swap was O(1), whereas rotation is O(n)
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-- --
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Unique Permutations, when string has duplicates
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-----------------------------------------------
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Basically, if we're swapping S[i] with S[j], but S[j] already occured earlier from S[i] .. S[j-1], then swapping will result in repetition
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```python
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def permute_distinct(S, i):
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if i == len(S):
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print(S)
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for j in range(i, len(S)):
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if S[j] in S[i:j]:
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continue
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S[i], S[j] = S[j], S[i]
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permute_distinct(S, i+1)
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S[i], S[j] = S[j], S[i] # backtrack
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```
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-- --
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Kth Permutation Sequence
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------------------------
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> Given A[N] and k,
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> find the kth permutation
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>
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[Kth Permutation Sequence - InterviewBit](https://www.interviewbit.com/problems/kth-permutation-sequence/)
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$\dfrac{k}{(n-1)!}$ will give us the index of the first digit. Remove that digit, and continue
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```python
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def get_perm(A, k):
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perm = []
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while A:
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# get the index of current digit
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div = factorial(len(A)-1)
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i, k = divmod(k, div)
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perm.append(A[i])
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# remove handled number
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del A[index]
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return perm
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```
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-- --
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Sorted Permutation Rank
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-----------------------
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> Given S, find the rank of the string amongst its permutations sorted lexicographically.
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> Assume that no characters are repeated.
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>
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```
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Input : 'acb'
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Output : 2
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The order permutations with letters 'a', 'c', and 'b' :
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abc
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acb
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bac
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bca
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cab
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cba
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```
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**Hint:**
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If the first character is X, all permutations which had the first character less than X would come before this permutation when sorted lexicographically.
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Number of permutation with a character C as the first character = number of permutation possible with remaining $N-1$ character = $(N-1)!$
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**Approach:**
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rank = number of characters less than first character * (N-1)! + rank of permutation of string with the first character removed
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```
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Lets say out string is “VIEW”.
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Character 1 : 'V'
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All permutations which start with 'I', 'E' would come before 'VIEW'.
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Number of such permutations = 3! * 2 = 12
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Lets now remove ‘V’ and look at the rank of the permutation ‘IEW’.
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Character 2 : ‘I’
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All permutations which start with ‘E’ will come before ‘IEW’
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Number of such permutation = 2! = 2.
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Now, we will limit ourself to the rank of ‘EW’.
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Character 3:
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‘EW’ is the first permutation when the 2 permutations are arranged.
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So, we see that there are 12 + 2 = 14 permutations that would come before "VIEW".
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Hence, rank of permutation = 15.
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```
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[Sorted Permutation Rank - InterviewBit](https://www.interviewbit.com/problems/sorted-permutation-rank/)
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-- --
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Number of Squareful Arrays
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--------------------------
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> Given A[N]
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> array is squareful if for every pair of adjacent elements, their sum is a perfect square
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>
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> Find and return the number of permutations of A that are squareful
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>
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> Example:
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> A = [2, 2, 2]
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> output: 1
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>
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> A = [1, 17, 8]
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> output: 2
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> [1, 8, 17], [17, 8, 1]
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>
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```python
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def check(a, b):
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sq = int((a + b) ** 0.5)
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return (sq * sq) == (a + b)
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if len(A) == 1: # corner case
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return int(check(A[0], 0))
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count = 0
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def permute_distinct(S, i):
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global count
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if i == len(S):
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count += 1
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for j in range(i, len(S)):
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if S[j] in S[i:j]: # prevent duplicates
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continue
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if i > 0 and (not check(S[j], S[i-1])): # invalid solution - branch and bound
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continue
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S[i], S[j] = S[j], S[i]
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permute_distinct(S, i+1)
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S[i], S[j] = S[j], S[i] # backtrack
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permute_distinct(A, 0)
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return count
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```
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-- --
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Gray Code
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---------
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> Given a non-negative integer N representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.
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> The gray code is a binary numeral system where two successive values differ in only one bit.
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G(n+1) can be constructed as:
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0 G(n)
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1 R(n)
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```
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Example G(2) to G(3):
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0 00
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0 01
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0 11
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0 10
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----
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1 10
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1 11
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1 01
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1 00
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```
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```python
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def gray(self, n):
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codes = [0, 1] # length 1
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for i in range(1, n):
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new_codes = [s | (1 << i) for s in reversed(codes)]
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codes += new_codes
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return codes
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```
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-- --
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Combination Sum II
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------------------
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[Combination Sum II - InterviewBit](https://www.interviewbit.com/problems/combination-sum-ii/)
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We already did this. Why couldn't you solve it!?
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-- --
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N Queens
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--------
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[NQueens - InterviewBit](https://www.interviewbit.com/problems/nqueens/)
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Backtracking
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- Place one queen per row
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- backtrack if failed
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-- --
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Word Break II
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-------------
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> Given a string A and a dictionary of words B, add spaces in A to construct a sentence where each word is a valid dictionary word.
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>
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```
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Input 1:
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A = "catsanddog",
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B = ["cat", "cats", "and", "sand", "dog"]
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Output 1:
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["cat sand dog", "cats and dog"]
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```
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```python
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def wordBreak(A, B):
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B = set(B)
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sents = []
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def foo(i, start, sent):
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word = A[start:i+1]
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if i == len(A):
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if word in B:
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sents.append((sent + ' ' + word).strip())
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return
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if word in B:
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foo(i+1, i+1, sent + ' ' + word)
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foo(i+1, start, sent)
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foo(0, 0, '')
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``` |