Brute-force approach
Here I present a few approaches to deduce "minimum insertions" required to convert a string into a palindrome.
The basic brute force approach is quite simple, given a string with length L, start comparing, the first character from left and the last character while scanning inwards.
Here is a basic test for a palindrome.
def ispalindrome(s):
L = len(s)
for i in range(L):
if s[i] != s[L - i - 1]:
return False,i,L-i -1
return True,0,0
The above code returns True if the string is a palindrome or returns False with mismatching indices.
result,left,right = ispalindrome("aba") will return True,0,0 while result,left,right = ispalindrome("abc") will return False,0,2 indicating the characters at 0th position, a, does not match with character at 2nd position, c
Once the indices are found a basic recursive solution can be written to reflect the missing strings from the mismatching indices.
def to_palendrome_left_reflect(s):
p,l,r = ispalindrome(s)
if not p:
if l == 0:
s = s[r] + s
else:
s = s[0:l]+ s[r] + s[l:]
return to_palendrome_left_reflect(s)
else:
return s
def to_Palendrome_rigth_reflect(s):
p,l,r = ispalindrome(s)
if not p:
if l == 0:
s = s + s[l]
else:
s = s[0:r+1] + s[l] + s[r+1:]
return to_Palendrome_rigth_reflect(s)
else:
return s
Some tests are as follows.
print to_palendrome_left_reflect("ab") #bab
print to_palendrome_left_reflect("abc")#cbabc
print to_Palendrome_rigth_reflect("ab") #aba
print to_Palendrome_rigth_reflect("abc")#abcba
To find which is smaller we write a quick function as shown below.
def find_Minimum_Insertions(s):
L = len(s)
left_insertion = to_palendrome_left_reflect(s)
right_insertion = to_Palendrome_rigth_reflect(s)
delta_l = len(left_insertion) - L
delta_r = len(right_insertion) - L
print "LEFT:",left_insertion
print "RIGHT:",right_insertion
if delta_l < delta_r:
print "MIN-LEFT",delta_l
else:
print "MIN-RIGHT",delta_r
Its important to consider that above mentioned techniques ignores deletion or susbtitution.
Another approach which only calculates count, while considering all possible operations.
def findMinInsertions(s,start,end):
if start == end:
return 0
if start == end -1:
if s[start] == s[end]:
return 0
return 1
if s[start] == s[end]:
return findMinInsertions(s,start+1,end - 1)
else:
a = findMinInsertions(s,start,end - 1)
b = findMinInsertions(s,start+1,end)
return min(a,b)+1
Dynamic Programming approach (fastest)
def findMin_Insertions_count_Dynamicly(s):
n = len(s)
table = [[0 for i in range(n)] for j in range(n)]
start = 0
end = 0
for y in range(0,n-1):
a = 0
for x in range(n):
if x>y-1:
table[y][x] = a
a+=1
return table[0][n-1]
This, by far is the fastest of all methods, but only works for string which are not already apalindrome , the trick is to have a n x n matrices and then fill in a particual order from left to right, increasing the gaps per row as shown below, for string abcde
a b c d e ---------- 0 1 2 3 4 0 0 1 2 3 0 0 0 1 2 0 0 0 0 1 0 0 0 0 0
The M[0][n-1] is 3
Hope this helps.
