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- Prepare
- Algorithms
- Strings
- Making Anagrams
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Making Anagrams
Making Anagrams
+ 19 comments Java implementation of the problem!!
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner sc=new Scanner(System.in); String s1=sc.nextLine(); String s2=sc.nextLine(); int cArr[]=new int[26]; int cArr1[]=new int[26]; for(int i=0;i<s1.length();i++)cArr[s1.charAt(i)-97]++; for(int i=0;i<s2.length();i++)cArr1[s2.charAt(i)-97]++; int count=0; for(int i=0;i<26;i++)count+=Math.abs(cArr[i]-cArr1[i]); System.out.println(count); }
}
+ 12 comments 4 clean lines - python3
from collections import Counter counts = Counter(input()) counts.subtract(input()) print(sum(abs(x) for x in counts.values()))
+ 5 comments passed all test cases in 0s
string in1,in2; cin>>in1>>in2; int array[26] = {0}; for(int i=0;i<in1.length();i++) array[in1[i]-'a']++; for(int i=0;i<in2.length();i++) array[in2[i]-'a']--; int sum = 0; for(int i=0;i<26;i++) sum+=abs(array[i]); cout<<sum; return 0;
+ 1 comment def makingAnagrams(s1, s2): for i in s1: if i in s2: s1=s1.replace(i,"",1) s2=s2.replace(i,"",1) return(len(s1)+len(s2))
Here is my simple python 3 solution
+ 1 comment simple
def makingAnagrams(s1, s2): # Complete this function k=[c for c in s1] l=list() for m in s2: if m in k: k.remove(m) else: l.append(m) return (len(k)+len(l))
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