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| 1 | +package com.ctci.arraysandstrings; |
| 2 | + |
| 3 | +import java.util.HashMap; |
| 4 | +import java.util.Map; |
| 5 | + |
| 6 | +/** |
| 7 | + * @author rampatra |
| 8 | + * @since 21/11/2018 |
| 9 | + */ |
| 10 | +public class PalindromePermutation { |
| 11 | + |
| 12 | + /** |
| 13 | + * This method exploits the fact that a palindrome will contain at most |
| 14 | + * one character with odd counts. All other characters should be of even |
| 15 | + * counts. |
| 16 | + * |
| 17 | + * @param str input string |
| 18 | + * @return {@code true} if {@code str} is a permutation of a palindrome |
| 19 | + */ |
| 20 | + private static boolean isPermutationOfPalindrome(String str) { |
| 21 | + Map<Character, Integer> charCounts = new HashMap<>(); |
| 22 | + Integer freq; |
| 23 | + int oddCounts = 0; // keep count of odds so that we don't have to loop through the hashmap the second time |
| 24 | + for (int i = 0; i < str.length(); i++) { |
| 25 | + char c = str.charAt(i); |
| 26 | + if (c != 32) { |
| 27 | + freq = charCounts.get(c) == null ? 0 : charCounts.get(c); |
| 28 | + if ((freq + 1) % 2 == 1) { |
| 29 | + oddCounts++; |
| 30 | + } else { |
| 31 | + oddCounts--; |
| 32 | + } |
| 33 | + charCounts.put(c, freq + 1); |
| 34 | + } |
| 35 | + } |
| 36 | + return oddCounts <= 1; |
| 37 | + } |
| 38 | + |
| 39 | + |
| 40 | + /** |
| 41 | + * This approach sets a bit in a number based on the character in the string and |
| 42 | + * then un-sets the bit if it sees the character again. Finally, checks if the |
| 43 | + * bitVector has at most one bit set only. |
| 44 | + * |
| 45 | + * @param str input string |
| 46 | + * @return {@code true} if {@code str} is a permutation of a palindrome |
| 47 | + */ |
| 48 | + private static boolean isPermutationOfPalindromeViaBits(String str) { |
| 49 | + int bitVector = 0; |
| 50 | + int index; |
| 51 | + |
| 52 | + for (int i = 0; i < str.length(); i++) { |
| 53 | + index = getIndex(str.charAt(i)); |
| 54 | + if (index != -1) { |
| 55 | + bitVector = toggleBitAt(bitVector, index); |
| 56 | + } |
| 57 | + } |
| 58 | + return (bitVector & (bitVector - 1)) == 0; |
| 59 | + } |
| 60 | + |
| 61 | + /** |
| 62 | + * Calculates the index to set the bit according to the character {@code c}. |
| 63 | + * |
| 64 | + * @param c |
| 65 | + * @return the index to set the bit as per the character {@code c} |
| 66 | + */ |
| 67 | + private static int getIndex(char c) { |
| 68 | + char a = 'a'; |
| 69 | + char z = 'z'; |
| 70 | + |
| 71 | + // assuming string contains only lowercase characters |
| 72 | + if (c < a || c > z) { |
| 73 | + return -1; |
| 74 | + } |
| 75 | + |
| 76 | + return c - a; |
| 77 | + } |
| 78 | + |
| 79 | + /** |
| 80 | + * Toggles the bit at index {@code index} in {@code bitVector}. |
| 81 | + * |
| 82 | + * @param bitVector |
| 83 | + * @param index |
| 84 | + * @return the resulting {@code bitVector} after toggling the bit |
| 85 | + */ |
| 86 | + private static int toggleBitAt(int bitVector, int index) { |
| 87 | + return bitVector ^ (1 << index); |
| 88 | + } |
| 89 | + |
| 90 | + public static void main(String[] args) { |
| 91 | + System.out.println(isPermutationOfPalindrome("tactc oapapa")); |
| 92 | + System.out.println(isPermutationOfPalindrome("maam")); |
| 93 | + System.out.println(isPermutationOfPalindrome("maa m")); |
| 94 | + System.out.println(isPermutationOfPalindrome("rammmar")); |
| 95 | + System.out.println(isPermutationOfPalindrome("rammmara")); |
| 96 | + System.out.println("---------"); |
| 97 | + System.out.println(isPermutationOfPalindromeViaBits("tactc oapapa")); |
| 98 | + System.out.println(isPermutationOfPalindromeViaBits("maam")); |
| 99 | + System.out.println(isPermutationOfPalindromeViaBits("maa m")); |
| 100 | + System.out.println(isPermutationOfPalindromeViaBits("rammmar")); |
| 101 | + System.out.println(isPermutationOfPalindromeViaBits("rammmara")); |
| 102 | + } |
| 103 | +} |
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