Home Strings [Hackerrank] – Two Strings Solution

# [Hackerrank] – Two Strings Solution

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Question: Given two strings, determine if they share a common sub-string.

Input: “a” and “art”

Output: YES

#### Problem Statement

Let me try to simplify this problem statement first.

You are given two strings, and you need to find out if they share a common sub-string or not. If you can find a common sub-string, output “YES”, else output “NO”. Note that you don’t have to find out what is the common sub-string.

Another important aspect of this problem is that the sub-string can be as small as 1 character. A sub-string is basically any string that can be formed using one or more characters of the original string.

Let us look at a sample test case:

So, if you are given with strings “CAT” and “RAT”, you can see that they share a common sub-string “AT”, hence in this case your answer should be “YES”

#### Brute Force Method:

One easy way to solve this problem would be:

1. Find out all the sub-strings of first string.
2. Find out all the sub-strings of second string.
3. Once you have all the sub-strings, see if you can find any common sub-strings.
4. Output “YES” or “NO” based upon your findings.

Note that, this method works fine, but you will waste a lot of time in finding out all the valid sub-strings. So, we need to find a way to optimize it somehow.

#### Efficient Method:

There is a very big hint hidden in the problem statement itself. The problem says that the sub-string could be as small as one character. So, you do not need even need to find out all the sub-strings. If you can find a common character, that should be all.

So let us assume you are given with two strings:

str1 = “barbell”
str2 = “trapper”

An efficient way to solve this problem could be:

1. Add all the characters of str1 to setA. So, setA would look like
setA = < b, a, r, e, l >
2. Add all the characters of str2 to setB. So, setB would become
setB = < t, r, a, p, e >
3. Take an intersection of both the sets. An intersection should give you the common characters.
4. After the intersection, you will get the common characters as:
< a, r, e >
5. If you can find some characters after intersection, the result is “YES”, else “NO”.

#### Code:

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.hljs.shcb-line-numbers .shcb-loc::before {
border-right: 1px solid #ddd;
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.hljs > mark.shcb-loc { background-color: #ddf6ff; }  public String twoStrings(String s1, String s2) {

Set<Character> s1Set = new HashSet<>();
Set<Character> s2Set = new HashSet<>();

//creating the set of string1
for(char c : s1.toCharArray()) {
}
//creating the set of string2
for(char c : s2.toCharArray()) {
}

// store the set intersection in s1Set
s1Set.retainAll(s2Set);

if (!s1Set.isEmpty())
return "YES";

return "NO";
}


Time Complexity: O(n)
Space Complexity: O(1)
The full solution and test cases are also available on Github.

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