solve the hackerrank substring challenge (Java solution)
In this article we will solve the HackerRank Substring challenge (Java solution). Let’s take a look at the question
A substring is a group of contiguous characters in a string. For instance, all substrings of abc are [a, b, c, ab, bc, abc].
In this challenge, you will be given a binary representation of a number. You must determine the total number of substrings present that match the following conditions:
As an example, consider the string 001101. The 4 substrings matching the two conditions include [0011, 01, 10, 01]. Note that 01 appears twice, from indexes 1-2 and 4-5. There are other substrings, e.g. 001 and 011 that match the first condition but not the second.
Let’s solve the hackerrank substring challenge using Java
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
/*
* Complete the function below.
*/ static int counting(String s) {
int curr = 1, prev = 0, ans = 0;
for(int i = 1; i< s.length(); i++)
if(s.charAt(i) == s.charAt(i-1)) curr++;
else{
ans += Math.min(curr, prev);
prev = curr;
curr = 1;
}
return ans + Math.min(curr, prev);
}
public static void main(String[] args) throws IOException {
Scanner in = new Scanner(System.in);
final String fileName = System.getenv("OUTPUT_PATH");
BufferedWriter bw = null;
if (fileName != null) {
bw = new BufferedWriter(new FileWriter(fileName));
}
else {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
int res;
String s;
try {
s = in.nextLine();
} catch (Exception e) {
s = null;
}
res = counting(s);
bw.write(String.valueOf(res));
bw.newLine();
bw.close();
}
}
Here, we declared three variables cur, prev and ans. We’re also looping through to get the characters at the given index.
This is how we solve the hackerrank substring challenge using Java
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public static int CountUniqueSubstrings(string s)
{
int count = 0;
int n = s.Length;
HashSet uniqueChars = new HashSet();
for (int i = 0; i < n; i++)
{
uniqueChars.Clear();
for (int j = i; j < n; j++)
{
if (uniqueChars.Contains(s[j]))
{
break;
}
uniqueChars.Add(s[j]);
count++;
}
}
return count;
}