mujain Very poor description of a simple problem..I am not sure what are we achieving by complicating the problem description.. what was the significance of length 'a' and 'b'when in example it already calls out it should be equal length in order to be anagram. This is merely making it verbose and adding complexities like |a-b| < 1 .. not contributing to the gut of the problem. My suggestion is to remove this problem or modify signifincantly. only reason I can see to keep it as is if we intent to confuse reader.
nixomose Agreed. It took me longer to understand the question than it did to write a solution.
[DELETED] [deleted]
Maher_Saleem agreed
meacreatio Agreed - and this question is not unique in this. Many questions on this site are harder to understand than they are to solve.
ealeksandro I was about to say the same thing. Surprising that after two years, nobody corrected the problem.
|S1.len-S2.len| must be 0 or S1,S2 can't be an anagram.
hiteshmadhyan191 same again after two years.
nguyenquan2707 Agreed.
hiteshmadhyan191 [deleted]
richard_luo06951 This is how real-world problems look like. You have to analyze it
andadeacu I agree to delete this problem.
edrouwendaal I agree on this also leaving a wrong test case, blocks one's logic. In real life of course this could happen. Or during job interview questions. But for just training skills it's not the best question.
deeteecee Why not just split the input by a space to make it more understandable rather than telling them they have a S1+S2 string...
HarryDC The core of the problem statement says 'Your challenge is to help him find the minimum number of characters of the first string he needs to change to make it an anagram of the second string'. This does not imply the require reshuffling of the string. For example changing this to: Your challenge is to help him find the minimum number of characters of the first string he needs to change to enable him to make it an anagram of the second string. Would make things clearer without giving away too much.
hydra_smith Agreed, the original one is kind of confusing.
dheeraj I've made this change! Thanks for notifying us!
particle Doesn't it follow from the definition of an anagram?
bramyers [deleted]
tao_zhang My java solution:
import java.util.*; public class Solution { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while (t-->0){ String str = sc.next(); int len = str.length(), count = 0; if (len%2!=0){ System.out.println(-1); continue; } String s1 = str.substring(0,len/2); String s2 = str.substring(len/2,len); char[] s1Chars = s1.toCharArray(); for (char c : s1Chars){ int index = s2.indexOf(c); if (index == -1){ count++; } else { s2 = s2.substring(0,index)+s2.substring(index+1); } } System.out.println(count); } } }
mithildeeva awesome! such simple logic. Thankyou!
adikiit123 Will it work with the string bbxxxaxb ?
humoyun_ahmad core of this solution is indexOf() method
tvucic1 Beautiful!!
Gurparshad amazing , very helpful
varlyanijatin88 Solution is Amazing. But can you tell me why did you added
s2.substring(index+1);Thanks in advance.
shweta78 to remove duplicacy... e.g. in let s1='abbc' and s2='fabc'. Since count of b isn't same in s1 and s2, not using s2 = s2.substring(0,index)+s2.substring(index+1); would consider b to be present ignoring the count. Expl: s2.substring(0,index) would calc the substring from beginning to pos index-1(including) and s2.substring(index+1) means starting from index+1 position thereby omitting the character at pos index.
ritikagarwal what does while (t-->0) means in above programs
prageeth_maha Super late to the game but, it means while( t-- > 0) {...} It does decrement. Same can be like while(t++
kris_tobiasson Here's my JS solution:
if(s.length%2 !== 0){ return -1; } let sub = s.substring(0, s.length * 0.5).split(``); for(let i = sub.length; i < s.length; i++){ let id = sub.indexOf(s.charAt(i)); if(id !== -1){ sub.splice(id,1); } } return sub.length;
saxenarajat26 "this is working but time limit exceeds can u help me with this!!"
function anagram(s) { var l=s.length,c=0,c1,c2 if(l%2==0){ c1=s.slice(0,l/2).split("") c2=s.slice(l/2).split("") }else{ return "-1" } var l2=c1.length for(var i=0;i
kris_tobiasson i would love to help you, but your code is incomplete and it isnt formatted correctly. If you supply the whole solution properly formatted, I would be happy to help.
By the way, with comments there is no code block button, but you can put the code within three backticks at the beginning and three backticks at the end of the code block to have it properly formatted.
Cheers
saxenarajat26 ` 'use strict';
const fs = require('fs');
process.stdin.resume(); process.stdin.setEncoding('utf-8');
let inputString = ''; let currentLine = 0;
process.stdin.on('data', inputStdin => { inputString += inputStdin; });
process.stdin.on('end', _ => { inputString = inputString.replace(/\s*/, ''));
main();});
function readLine() { return inputString[currentLine++]; }
// Complete the anagram function below. function anagram(s) { var l=s.length,c=0,c1,c2 if(l%2==0){ c1=s.slice(0,l/2).split("") c2=s.slice(l/2).split("") var l2=c1.length for(var i=0;i
return c }
function main() { const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
const q = parseInt(readLine(), 10); for (let qItr = 0; qItr < q; qItr++) { const s = readLine(); let result = anagram(s); ws.write(result + "\n"); } ``` ws.end();}
kris_tobiasson All I need is the full anagram() function.. Anyhow, if you take a look in the comment you just sent, you will see that in the middle of the anagram() function, it cuts off at a for loop statment
for(var i=0;iat the i and then goes straight toreturn c;, so you are missing some code. Just supply the whole anagram function, no need for the other code.saxenarajat26 [deleted]kris_tobiasson I saw the code in your PM. The performance hit was in your for loop. You were checking to see if one substring had the other substring's character twice (once with includes() and once with indexOf()). Both those functions will search for the given character in a string/array, but includes() will return true or false and indexOf() will return the index of the character or -1 if it doesn't contain the character. So, if you only use indexOf() and check if the return value is === -1 instead of searching through the string twice. Below is an edited version of the code you sent me and you will see what i mean:
let l=s.length,c=0,c1,c2; if(l%2==0){ c1=s.slice(0,l/2).split(""); c2=s.slice(l/2).split(""); let l2=c1.length; for(let i=0;i<l2;i++){ let id = c2.indexOf(c1[i]); if(id !== -1){ c2[id]=" "; } else{ c+=1; } } } else{ return "-1"; } return c;
saxenarajat26 thanks i got it !! that includes was unnecessary taking so much time!!
palashmax Fun question: How do you fail TestCase #0 when you have already passed it? Isn't it the same as in the question or the one you use to check your solution(Run code)?
kumarjivi Same issue with me, When I submit the code it says almost all cases fail, including test Case#1. You got solution for this?
tilper Test case #1 is the same as the sample for this problem.
hugo_ratiney I pass only #1 and #14... You can get almost the same score with :
print("3\n1\n-1\n2\n0\n1\n")
yashgandhi900 I m sure that you are drunk.. Alcoholic...
sushilll [deleted]motwanirushil wrong
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