Answer:
Check the explanation
Explanation:
a) This done can be executed through the use of brute force by the given algorithm: Consider any two indices i and j then we can check if string from s[i....j] is palindrome or not in O(L) where L is length of string.
Algorithm to check if string is palindrome :
Takes String s as argument and the two indices i,j between which the string is to be considered
bool palindromeCheck(s,i,j) :
y = j
for x in range(i,j) :
if(s[x]==s[y]) :
y--
continue
else :
return false
return true
Now taking a look at the part a) we will be calling the function to confirm and execute palindrome for all i<=j that is a total of points of the order of O(n^2) and the string checking will take O(n) in the worst case making it O(n^3).
b) To advance over we became aware of the fact that repeated calculation is being done i.e when we confirmed the string s[i...j] we also tested in between strings like s[i+1,j-1] etc in the same time. Consequently we used memorization to store the outcome that we calculate and reuse them we need them later.
We know that P[i][i] will always be one since a single character string will always be a palindrome.
Using this as our base case to check for P[i][j] we need to check if the ith and the jth character are equal and also if P[i+1][j-1] is true i.e is a palindrome.
We come up with the following recurrence
P[i][j] = 1 if i=j
P[i][j] = P[i+1][j-1] ; if i>=j and s[i]==s[j]
0; otherwise
There are order of O(n^2) states and each one of the state necessitates only constant and regular time character checking for its evaluation and then assigning the value of 1, P[i+1][j-1],0 as a result of that which is again a constant time operation. Therefore all the P[][] matrix values can be found in O(n^2) time.
Pseudo code :
Let n be the length of the string
for x in range(1,n) :
P[x][x]=1
for sz in range(2,n) : # Since we have already computed answer for string of size 1
for x in range (1,n-sz+1):
y=x+sz - 1
if x == y :
if sz==2 :
P[x][y] = 1
else :
P[x][y] = P[x+1][y-1]
else :
P[x][y] = 0
From the implementation also it is clear that the running time of the algorithm is O(n^2).
