Java Anagrams import java.util.Scanner; public class Solution { static boolean isAnagram(String a, String b) { boolean x= true; if (a.length()!=b.length()){ x=false; } a= a.toLowerCase(); b=b.toLowerCase(); int al[]=new int [256]; for(char c : a.toCharArray()){ int ...
hackerrank solutions java GitHub | hackerrank tutorial in java | hackerrank 30 days of code solutions | hackerrank algorithms solution | hackerrank cracking the coding interview solutions | hackerrank general programming solutions | hackerrank implementation solutions | hackerrank data structures solutions in ...
Java Hard 50 Palindrome Index Java O(n) O(1) Easy 25 Anagram Java O(n) O(1) Easy 25 Making Anagrams Java O(n) O(n) Easy 30 Game of Thrones - I Java C# O(n) O(1) Easy 30 Two Strings Java C# O(|a| + |b|) O(1) Easy 25 a and b are lengths of the inp...
public static boolean isAnagram(String a,String b) { char a1[]=a.toCharArray(); char b1[]=b.toCharArray(); java.util.Arrays.sort(a1); java.util.Arrays.sort(b1);
import java.util.*; public class Solution { public static void main(String args[] ) throws Exception { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ Scanner in = new Scanner(System.in); String str = in.nextLine(); ...
题解:一个字符串能够通过变换变成一个回文串的充要条件是它里面最多有一种字母,在字符串里面出现的次数是奇数,其他种的字符在字符串里面出现的次数都是偶数。 1importjava.io.*;2importjava.util.*;3importjava.text.*;4importjava.math.*;5importjava.util.regex.*;67publicclassSolution {89staticString Gam...
#TitleSolutionTimeSpaceDifficultyPointsNote Solve Me First Java C# O(1) O(1) Easy 1 Simple Array Sum Java C# O(n) O(1) Easy 10 Compare the Triplets Java C# O(1) O(1) Easy 10 A Very Big Sum Java C# O(n) O(1) Easy 10 Diagonal Difference Java C# O(n) O(1)...
Java Hard 50 Palindrome Index Java O(n) O(1) Easy 25 Anagram Java O(n) O(1) Easy 25 Making Anagrams Java O(n) O(n) Easy 30 Game of Thrones - I Java C# O(n) O(1) Easy 30 Two Strings Java C# O(|a| + |b|) O(1) Easy 25 a and b are lengths of the inp...
Java Anagrams My solution, from scratch, of course. My biggest problem was that I could not import the classes I needed. Then, "oh yeah, lets just access it directly", QED! static boolean isAnagram(String a, String b) { // Complete the function if (a.length() != b.length()) ...
Java Anagrams In this case, I have to search each character in the first string in the second string. I was thinking to use HashMaps at first, but, I realized, I can't do that here. So, I came up with this solution: static boolean isAnagram(String a, String b) {...