 # PERMPAL - Editorial

Author: Praveen Dhinwa
Tester: Hanlin Ren
Editorialist: Hanlin Ren

SIMPLE

None

### PROBLEM:

Given a string s[1\sim n], does there exist a permutation p of 1,2,\dots,n such that, if we let t*=s[p*], then t is palindrome?

### EXPLANATION:

This problem is equivalent to: can we rearrange the characters of s into a permutation?

Let’s count the number of occurrences of every characters in a-z. For a char c, let cnt[c] be the number of times c occur in s.

If there are two chars c_1,c_2(c_1 e c_2) such that both cnt[c_1] and cnt[c_2] is odd, then the answer is no.

• To see this, consider any palindromic string t and any char c. Suppose t*=c.
• If the length n of t is even, then t[n-i+1]=c. Since n is even, n-i+1 and i has different parity and hence i e n-i+1. Also n-(n-i+1)+1=i, so we say i and n-i+1 is a pair. Any pair must use the same character, so any character must occur an even number of times.
• If the length n of t is odd, the above argument holds, too, except that the central char(t[\frac{n+1}{2}]) is matched with itself. In this case, this char appears an odd number of times, and is the only char that appears an odd number of times.
• See figure below. n=7, pairs are (1,7),(2,6),(3,5), and t is matched with itself, and hence appears an odd number of times. • In conclusion, if C is the number of chars c that cnt[c] is odd, and s can be arranged into a palindrome, then C\le 1.

The converse is also true: if there are at most 1 characters c with cnt[c] odd, then the answer is yes. Here is one of the constructions.

• Let’s construct the permutation char-by-char. Let c iterate from a to z.
• If cnt[c] is odd, we ignore c(that is, we consider c later). At most one such c exists.
• We maintain two pointers l and r. Initially l=0,r=n+1. The permutation p[1\sim l] and p[r\sim n] is already filled.
• We scan s. Each time we meet a c, say t*=c,
1. if this is the 1,3,5,\dots,(odd) time we meet c, we assign it to the left: l\gets l+1 and p[l]\gets i;
2. otherwise (this is the 2,4,6,\dots time we meet c), we assign it to the right: r\gets r-1 and p[r]\gets i.
3. The picture below shows how we fill p when s=abdacbabdccba, i=b. • At last, in case that n is odd, we need to deal with the c where cnt[c] is odd. That’s easy: p[l+1\sim r-1] is unfilled, and we fill them by char c.
• This is a valid answer. Time complexity: O(\sigma\cdot n). (\sigma=26)

### ALTERNATIVE SOLUTIONS

This is a constructive problem. I think there’ll be many solutions. Please feel free to share your solution.

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.
Tester’s solution can be found here.

Here is my solution https://www.codechef.com/viewsolution/17423799. please help me finding my mistake i am unable to find it.

https://www.codechef.com/viewsolution/17340080 Can anyone tell me what was wrong with my solution??

This is my solution https://www.codechef.com/viewsolution/17360282. It works only for subtask #1. can anyone please tell me what went wrong with my solution?

this is my solution https://www.codechef.com/viewsolution/19769109 .

You are not outputting a correct permutation. The string generated by your permutation is not a palindrome.

Check your solution on the following test cases.

10
ggxggvvxpp
szzsbbhkhk
sjkxkjxsoo
oppdooopoo
ialilpijpp
efefxecxe
ppxuouxpp
vyuvvvyuh
ggwgzzzeg
ggywwwgwg

thank you so much. found the mistake 