HackerRank Two Strings Explained - Java

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[maheshnaik]

We can declare one HashSet, in second for loop check for contains, this way we can reduce some lines.

[sahilbisht]

solution which will reduce some lines of code :-)
public static String twoStrings(String s1, String s2) {
HashSet<Character>s1_chars = new HashSet();
for (int i = 0; i < s1.length() ; i++) {
s1_chars.add(s1.charAt(i));
}
for( int i = 0; i < s2.length(); i++) {
if {
return "YES";
}
}
return "NO";
}

[Y.Albasel]

finally, a proper English tutorial Thank you !

[arunkumarvaithiyanathan]

public static String twoStrings(String s1, String s2) {
HashSet<Character>hs=new HashSet();
for(char c:s1.toCharArray()){
hs.add(c);
}
for(char c:s2.toCharArray()){
if(hs.contains(c)){
return "YES";
}
}
return "NO";
}

[divyashreemh]

public static String twoStrings(String s1, String s2) {


Set<Character> set = new HashSet<>();

for(int i=0;i<s1.length();i++){
set.add(s1.charAt(i));
}

for(int i=0;i<s2.length();i++){

return "YES";
} }

return "NO";
}

[travelingDeveloper]

For ppl who are commenting that we could just implement hashmap and iterate over the shorter string it won't work if strings have repeating characters

[MithleshSingh]

static String twoStrings(String s1, String s2) {
String ans = "NO";
for (int i = 97; i <= 122; i++) {
String me = String.valueOf((char) i);
if (s1.contains(me) && s2.contains(me)) {
return "YES";
}
}
return ans;
}

[krokill]

Awesome video! Just wanted to share my solution with a single int instead of a HashSet.
static String twoStrings(String s1, String s2) {
int checker = 0;
for (char c : s1.toCharArray()) {
int val = c - 'a';
if ((checker & (1 << val)) <= 0)
checker |= (1 << val);
}
for (char c : s2.toCharArray()) {
int val = c - 'a';
if ((checker & (1 << val)) > 0)
return "YES";
}
return "NO";
}

[m.houdeib]

This is my JAVA solution:

public static String twoStrings(String s1, String s2) {
HashSet<Character> set = new HashSet<Character>();
for(int i = 0 ; i < s1.length() ; i++){
set.add(s1.charAt(i)) ;
if(set.size() == 26) break ;
}
for(int i = 0 ; i < s2.length() ; i++){
return "YES" ;
}
return "NO" ;
}

[genli]

Great video. But I have a question, On hacker rank, how to debug using the hidden test case. For example, if I failed hidden test cases 3, and 4. And I don't know how to change the code to fix it. How can I debug the code with the test cases

[AaamrK]

We can just use 1 HashSet and loop the second string and return true for the first match.

[Marc-sfxr]

I think this solution is simpler and uses constant space and linear complexity:

static String twoStrings(String s1, String s2) {
boolean[] letters = new boolean[26];
for(char c : s1.toCharArray()) letters[c - 'a'] = true;
for(char c : s2.toCharArray()) if(letters[c - 'a']) return "YES";
return "NO";
}

[sakaradhikari]

public static String twoStrings(String s1, String s2) {

return s1.chars()
.mapToObj(x -> (char) x)
.distinct()
.anyMatch(y -> ? "YES" : "NO";
}