Given a string str consist of only lowercase alphabets and an integer K, the task is to count the number of substrings of size K such that any permutation of the substring is a palindrome. you just need to do some auditing: if the string length is odd, there should be one and only one char whose total number over the string is odd; if the string length is even, all chars' total numbers over the string should be even. For example , “race car”, “anna”, “abcba” are palindromes. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation. The Problem: Write a function that checks whether any permutation of a string is a palindrome. A better solution is suggested from the above hint. A native solution is to generate the permutation of the string, then check whether it is a palindrome. And we have to print all possible palindromic permutation that can be generated using the characters of the string in alphabetical order. if that's the case, you are really complicating this problem. share | improve this question | follow | edited Nov 30 '16 at 7:26. Examples: Input: str = “abbaca”, K = 3 Output: 3 Explanation: The substrings of size 3 whose any permutation is palindrome are {“abb”, “bba”, “aca”}. java bit-manipulation permutation palindrome bitvector. 1,101 3 3 gold badges 8 8 silver badges 18 18 bronze badges. Decimal Decimal. Now, the String “car race” is a permutation of “race car” which is a palindrome. Pramod Gharu. In this article, we will see Java algorithm to find out if a String is a permutation of a palindrome. Permutation An ordering of a set of items; in this case the letters of the string. You can certainly check all permutations, but there is a much more efficient approach. Here is how you can implement this: This Java program asks the user to provide a string input and checks it for the Palindrome String. The problem “Minimum insertions to form a palindrome with permutations allowed” states that you are given a String with all letters in lowercase. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview … Note that in order for a string to be a palindrome, then every letter is mirrored around the center of the string. Scanner class is a part of java.util package, so we required to import this package in our Java … If a palindromic permutation exists, we just need to generate the first half of the string. asked Nov 30 '16 at 6:45. That means a collection of letters can form a palindrome if there is at most one letter that has an odd count. Java代码 Hide Tags Backtracking: Hide Similar Problems (M) Next Permutation (M) Permutations II (E) Palindrome Permutation */ // Validate if can build palindromic, add half of the char, and record the odd char. Palindrome is a word or phrase that is the same when you traverse from beginning or end. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation. 1. ... // Java program to find number of // palindromic permutations of a // given string . like aba, abbba. Scanner class and its function nextLine() is used to obtain the input, and println() function is used to print on the screen. Print all palindrome permutations of a string; Program to reverse a string (Iterative and Recursive) ... A string can permute to a palindrome if number of odd occurring characters are at most one. If palindrome is not created using the string print ‘-1’. 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