![]() ![]() if the count ≤ 1 ⇒ palindrome is possible. In case of odd occurrences, the count will get incremented by 1. If all letters are appearing even times, the count will remain 0. Instead of using isCountOdd pointer, we can use another variable count which will store values of map%2. 266 - Palindrome Permutation Leetcode 266 - Palindrome Permutation Posted on Aug2 minute read Welcome to Subscribe On Youtube Question Formatted question description: Given a string s, return true if a permutation of the string could form a palindrome and false otherwise. if it is odd- palindrome is not possible.if isCountOdd is 1- this means we already have a letter that has occurred as odd so we look at the count value of the char.if the count is odd-this means the palindrome is possible if no other letters have odd count values, keep a variable isCountOdd to keep a track when we encounter an odd count value.if the count is even- this means it will help us to form palindrome, do nothing.After sorting all the counts of each letter, we look at the values. Iterate through each letter in the string and increase the value of each char to count occurrences. We will count the number of occurrences of each letter using a hashmap. Adam Arold at 19:59 3 Do you mean: is there a permutation of the string that is a palindrome The title and the question have mistakes, I'm afraid. This should give us the intuition to solve this problem. 2,349 12 21 asked at 19:55 user5080124 You simply have to permutate all characters and check each resulting sequence whether its reverse is equal to the sequence itself. “aaxaa”, ‘x’ does not has any partner hence does not form a pair. In case of an odd length of string, only one letter having no pair is allowed. There can exist more than 1 pair of the same letter, eg. If a string s can become a palindrome, there is exactly one character. Let’s call this left and right occurrence of the letter “pair”. The Key Idea for Solving This Coding Question. To make a string palindrome, we need an even number of occurrences of letters so that the string reads the same from left and right. If this count happens to exceed 1 at any step, we conclude that a palindromic permutation isnt possible for the string s. Examples & ExplanationsĮxplanation: we can not arrange letters of "code" to form a palindromeĮxplanation: permutation "aba" of given string is palindrome, hence trueĮxplanation: the given string is itself palindrome, there exists multiple permuations that form palindrome Approach Palindrome Permutation LeetCode Solution – We are given a string and asked if a permutation of the given string could form a palindrome. Java code for Palindrome Permutation LeetCode Solution.C++ code for Palindrome Permutation LeetCode Solution. ![]()
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