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