Convert string to palindrome string with minimum insertions
Solution 1
Let S[i, j]
represents a sub-string of string S
starting from index i
and ending at index j
(both inclusive) and c[i, j]
be the optimal solution for S[i, j]
.
Obviously, c[i, j] = 0 if i >= j
.
In general, we have the recurrence:
Solution 2
To elaborate on VenomFangs answer, there is a simple dynamic programming solution to this one. Note that I'm assuming the only operation allowed here is insertion of characters (no deletion, updates). Let S be a string of n characters. The simple recursion function P for this is:
= P [i+1 .. j-1], if S[i] = S[j]
P[i..j]
= min (P[i..j-1], P[i+1..j]) + 1,
If you'd like more explanation on why this is true, post a comment and i'd be happy to explain (though its pretty easy to see with a little thought). This, by the way, is the exact opposite of the LCS function you use, hence validating that your solution is in fact optimal. Of course its wholly possible I bungled, if so, someone do let me know!
Edit: To account for the palindrome itself, this can be easily done as follows: As stated above, P[1..n] would give you the number of insertions required to make this string a palindrome. Once the above two-dimensional array is built up, here's how you find the palindrome:
Start with i=1, j=n. Now, string output = "";
while(i < j)
{
if (P[i][j] == P[i+1][j-1]) //this happens if no insertions were made at this point
{
output = output + S[i];
i++;
j--;
}
else
if (P[i][j] == P[i+1][j]) //
{
output = output + S[i];
i++;
}
else
{
output = S[j] + output;
j--;
}
}
cout<<output<<reverse(output);
//You may have to be careful about odd sized palindromes here,
// I haven't accounted for that, it just needs one simple check
Does that make better reading?
Comments
-
whitepearl over 3 years
In order to find the minimal number of insertions required to convert a given string(s) to palindrome I find the longest common subsequence of the string(lcs_string) and its reverse. Therefore the number of insertions to be made is length(s) - length(lcs_string)
What method should be employed to find the equivalent palindrome string on knowing the number of insertions to be made?
For example :
1) azbzczdzez
Number of insertions required : 5 Palindrome string : azbzcezdzeczbza
Although multiple palindrome strings may exist for the same string but I want to find only one palindrome?
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whitepearl almost 12 yearsThank you @kyun. But I successfully found out the number of insertions to be made. I have specified that I need to find the palindrome string formed after insertions? Can you give me a optimal solution? Thanking you in advance.
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foobar over 7 yearsthis solution won't give right answer in most of the cases. Try OROP , only 1 character is required, ie P in the beginning, but your solution will give answer 2.
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Abhijit Sarkar about 5 yearsExplain the recurrence.
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Abhijit Sarkar about 5 years"If you'd like more explanation on why this is true, post a comment"? Why don't you explain anyway, if you can, instead of baiting?