# Hackerrank: Cracking the Coding Interview – Strings: Making Anagrams

The solution to this problem involves figuring out that if we just take the differences in the counts of the number of distinct characters in each string then that is the optimal amount of deletions we need to make. It should also be noted that while doing the calculations we need to ignore negative values and make them positive instead.

Solution:

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static int numberNeeded(String first, String second) {
char[] a = first.toCharArray();
char[] b = second.toCharArray();
int[] ac = new int[26];
int[] bc = new int[26];
for (int i=0; i<a.length; i++) {
ac[a[i]-'a']++;
}
for (int i=0; i<b.length; i++) {
bc[b[i]-'a']++;
}
int ans = 0;
for (int i=0; i<ac.length; i++) {
ans += Math.abs(ac[i] - bc[i]);
}
return ans;
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String a = in.next();
String b = in.next();
System.out.println(numberNeeded(a, b));
}
}
```

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