anirudherabelly 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); }}
ivdekov Could you explain your approach to the problem? What does the line
cArr[s1.charAt(i)-97]++do?
chunktri 97 is the ASCII code of character a. Since the string only consist of lower case letters, there's an array of size 26. To store the count of letter a means subtract 97 from character and add one to it's array value
arjun033 nice!
jeffreyeffendy_1 You can just use one array here to store. Here is a better version of your code. I am halving your space requirement by 26 * 4 bytes.
int cArr[]=new int[26]; for(int i=0;i<s1.length();i++)cArr[s1.charAt(i)-97]++; for(int i=0;i<s2.length();i++)cArr[s2.charAt(i)-97]--; int count=0; for(int i=0;i<26;i++)count+=Math.abs(cArr[i]); System.out.println(count);Cheers!
jfuji6 Nice optimization!
onlysumon We can swap the string and reduce the count loop from O(len(s1)) + O(len(s2)) to O(max(len(s1), len(s2)).
sicter This solution with some syntactic sugar:
int[] chars = new int[26]; for (char c1 : s1.toCharArray()) { chars[c1-97]++; } for (char c2 : s2.toCharArray()) { chars[c2-97]--; } int count = 0; for(int i: chars) {count += Math.abs(i);} System.out.println(count);Note on for-each in Java- http://stackoverflow.com/a/256861/3527434
davidrieper61 Loved this man.
ajanacs [deleted]ajanacs How did this work?
uchiha111 nice one buddy!!!!
zakiralig184 Cool bro!!!! ur code optimization!!!!!!
adityamalhotra11 Instead of using an array of size 26, why not use a hashMap, which will have a size 26 in the worst case and in other cases, always less than that. This reduces space complexity and time pretty much remains O(n).
bingo_22 This approach has been immensely helpful. Thanks a lot.
anirudherabelly thankU!! @anmol_1995
lonewolff my java logic goes like this
public static void main(String[] args) { Scanner sc=new Scanner(System.in); String s1=sc.next(); String s2=sc.next(); int d=0; for(int i=0,k;i<s1.length();i++) if((k=s2.indexOf(s1.charAt(i)))!=-1) { s2=s2.substring(0,k)+s2.substring(k+1); d++; } System.out.println(s1.length()+s2.length()-d); }
voiceofall I don't understand the solution. What about the test case: "cde", "adc". The result should be 4 not 2.
Iancarson01 Check on your "d" and "c",delete e and a =2
15131A05K8_ nice logic
cse_115 nice..
ayushjain_creat1 There is no need of posting code for such easy problems.
fernando_luz I made a similar answer in C++, using unordered_map.
int makingAnagrams(const string &s1, const string &s2){ unordered_map<char, int> letters; for(char c: s1){ ++letters[c]; } for(char c: s2){ --letters[c]; } int countDeletions = 0; for(pair<char, int> pr: letters){ if (pr.second != 0){ countDeletions += abs(pr.second); } } return countDeletions; }
ayushjain_creat1 You just traslated his code too c++..
fernando_luz I use only one container to store the letters instead 2 containers.
And I reach the solution without check the discussion forum.
More differences:
- I use unordered_map providing a fastest lookup to an element;
- I check another improvement in the code, I dont need the if conditional inside the last loop to counter the values.
- I use prefix increment instead pos fix increment.
int makingAnagrams(const string &s1, const string &s2){ unordered_map<char, int> letters; for(char c: s1){ ++letters[c]; } for(char c: s2){ --letters[c]; } int countDeletions = 0; for(pair<char, int> pr: letters){ countDeletions += abs(pr.second); } return countDeletions; }
This solution is very elegant. :-)
manoj_banik_cs Your solution is good. But you can avoid map, just a single int array can serve the purpose. I did it, even i didn't use the "abs" function also.
sachetwasti61 I had done something similar but only two testcases pass why?
import java.util.*; import java.io.*; public class Hack45{ static PrintStream p = System.out; static int [] arr = new int[26]; public static void main(String[] args){ StringBuilder sb = new StringBuilder(); String s,s1; Scanner in = new Scanner(System.in); s = in.next(); s1 = in.next(); int i; for(i=0; i<s.length(); i++){ arr[s.charAt(i)-'a']++; } for(i=0; i<s1.length(); i++){ if(arr[s1.charAt(i)-'a']!=0) arr[s1.charAt(i)-'a']--; else arr[s1.charAt(i)-'a']++; } int count = 0; for(i=0; i<26; i++){ count+=arr[i]; } p.println(count); } }
sarathy_v_krish1 C++ solution :
int makingAnagrams(string s1, string s2) { map<char, int> m1, m2; int deletions=0; for (char ch:s1) m1[ch]++; for (char ch:s2) m2[ch]++; for (int i=0;i<26;i++) deletions+=abs(m1['a'+i]-m2['a'+i]); return deletions; }
pawankushwah850 string s1,s2; cin>>s1>>s2; int co=0; for(char i='a'; i<='z'; i++) co+=abs(count(s1.begin(), s1.end(),i)-count(s2.begin(), s2.end(),i)); cout<<co; return 0;sarathy_v_krish1 Looks smaller, yes, but you are iterating over each string 26 times...less efficient.
amit_raj_cse20 Python Oneliner Code
def makingAnagrams(s1, s2): return len(s1)+len(s2)-2*(len(s1)-sum((Counter(s1)-Counter(s2)).values()))
DomNomNom 4 clean lines - python3
from collections import Counter counts = Counter(input()) counts.subtract(input()) print(sum(abs(x) for x in counts.values()))
glegoux [deleted]abhishekscript from collections import Counter a=Counter(raw_input()) b=Counter(raw_input()) print sum( ((a-b).values()) + ((b-a).values()) )
Done without loop
tasdemir_nuri What about sum(). Isn't that a loop?
abhishekscript Obviously it is. What I meant is
print(sum(abs(x) for x in counts.values()))
and this
print sum( ((a-b).values()) + ((b-a).values()) )
without for. Thanks
kevindu [deleted]Norman99 Nice. I need to learn how to use Counter!
My solution with defaultdict:
from collections import defaultdict def count_chars(word): counts = defaultdict(lambda: 0) for ch in word: counts[ch] += 1 return counts word1, word2 = count_chars(input()), count_chars(input()) allchars = set(list(word1.keys()) + list(word2.keys())) print(sum(abs(word1[ch] - word2[ch]) for ch in allchars))
lucinda04zhang create dicts in one line and without the library
all_keys = set(s1 + s2) k1 = {i:s1.count(i) for i in all_keys} k2 = {i:s2.count(i) for i in all_keys}
meacreatio Thanks for giving a great example of some of the cool stuff
Countercan do.
manoharsingh2110 Need to learn python after seeing this :)
DivJain def makingAnagrams(s1, s2): s3=list(s2) for i in s1: if i in s3: s3.remove(i) no_common=len(s2)-len(s3) return len(s1)+len(s2)-2*no_common
My Solution
akhileshjoshi123 pure set concept illustration :) thanks !
alex_zehet cool solution!
mike_buttery Thanks your code helped. Version without import.
def makingAnagrams(s1, s2): d = {c:s1.count(c) for c in set(s1)} for c in s2: d[c] = d.get(c, 0) - 1 return sum(map(abs, d.values()))
snkt_pat2 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;
bazuka_ what is the prolem with my code it passes all the test casses except three cases.
#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int max (int a,int b){return a>b?a:b;} int main() { string s2,s1;int a[123]={0},b[123]={0},cnt=0; cin>>s1>>s2; for(int j=0;j<max(s1.size(),s2.size());j++) { a[s1[j]]++;b[s2[j]]++;} for(int j=97;j<123;j++) if(a[j]!=b[j]) cnt+=abs(a[j]-b[j]); cout<<cnt; /* Enter your code here. Read input from STDIN. Print output to STDOUT */ return 0; }
mohitthapark include
include
include
include
include
using namespace std;
int main() { string s2,s1; int a[123]={0},b[123]={0},cnt=0; cin>>s1>>s2; for(int j=0;j<(s1.size());j++) { a[s1[j]]++; } for(int i=0;i<(s2.size());i++) { b[s2[i]]++; } /for(int j=97;j<123;j++) {cout</ for(int j=97;j<123;j++) if(a[j]!=b[j]) { cnt+=abs(a[j]-b[j]); //cout<
cout<<cnt; /* Enter your code here. Read input from STDIN. Print output to STDOUT */ return 0;}
check out your code for test case: aaaaaaaaaaaaaaaaaaaaaaaaaaaaa bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb It is overwriting the value of array a 2 times
sunitswn91 j<max(s1.size(),s2.size())
You are accessing an array (either a or b depending on whose size is lesser) beyond its range, the behaviour might be undefined.
zeyad1998 can you explain to me the logic behind your code?
Dhruv_94 [deleted]Dhruv_94 Hey, do you mind explaining what the bold part does exactly? Thanks!
for(int i=0;i<in1.length();i++) **array[in1[i]-'a']++;** for(int i=0;i<in2.length();i++) **array[in2[i]-'a']--;**
zeyad1998 he puts the no. of letter in its correct no. in the alphabet array he made. and ++ means he increment this place by 1
dineshnadimpalli Awesome logic! Loved it!
khoangcachxalam1 thanks for your good idea !although i do not still know all images your code !
ASHWIN_18 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))
by001 What i love in python is the straightforward & clean approach:
sum([abs(s1.count(c) - s2.count(c)) for c in set(s1 + s2)])
and that's it! pure developer's pleasure! Of course it runs in O(N^2), so it's shortness over performance. will be a 3-liner to run in O(N)
rushil231100 Amazing
advaithsurya I'd be happy to write the extra lines to avoid O(N^2). hahaha
RUPESH1439530 def makingAnagrams(s1, s2): c1 = Counter(s1) c2 = Counter(s2) return sum((c1-c2).values())+sum((c2-c1).values())
Prashant_agarwal java solution satisfying all test cases
static int makingAnagrams(String s1, String s2){ int l1,l2,i,j,c=0,e=0; l1=s1.length(); l2=s2.length(); char[] c1= s1.toCharArray(); char[] c2 = s2.toCharArray(); for(i=0;i<l1;i++) { for(j=0;j<l2;j++) { if(c1[i]==c2[j]){ c++; e++; c2[j]='@'; break; } } c=0; } **Strong Text** return (l1-e)+(l2-e); } public static void main(String[] args) { Scanner in = new Scanner(System.in); String s1 = in.next(); String s2 = in.next(); int result = makingAnagrams(s1, s2); System.out.println(result); }}
bobby_teja Simple and easy logic!!
rajhere i used single array container..check my java solution, hope it will be helpful :) https://www.hackerrank.com/challenges/making-anagrams/forum/comments/432111
ankushrasgon07 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
man198407 def Making_Anagrams(s1, s2):
same = 0 s1 = list(s1) s2 = list(s2) for i in s1: if i in s2: s2.remove(i) same += 1 return (len(s1) - same) + len(s2)
rajhere working java solution
static int makingAnagrams(String s1, String s2){ // Complete this function int[] temp = new int[26]; for(int i=0,j=0;i<s1.length() || j<s2.length();i++,j++) { if(i<s1.length()) temp[s1.charAt(i)-'a']++; if(j<s2.length()) temp[s2.charAt(j)-'a']--; } int count=0; for(int i=0;i<26;i++) count+=Math.abs(temp[i]); return count;}
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