HackerRank Anagram Solution

Hello Programmers, In this post, you will learn how to solve HackerRank Anagram Solution. This problem is a part of the HackerRank Algorithms Series.

HackerRank Anagram Solution
HackerRank Anagram Solution

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

HackerRank Anagram Solution

Task

Two words are anagrams of one another if their letters can be rearranged to form the other word.

Given a string, split it into two contiguous substrings of equal length. Determine the minimum number of characters to change to make the two substrings into anagrams of one another.

Example

s = abccde

Break s into two parts: abc’ and ‘cde’. Note that all letters have been used, the substrings are contiguous and their lengths are equal. Now you can change ‘aand ‘b’ in the first substring to ‘d’ and ‘e’ to have ‘dec’ and ‘cde’ which are anagrams. Two changes were necessary.

Function Description

Complete the anagram function in the editor below.

anagram has the following parameter(s):

  • string s: a string

Returns

  • int: the minimum number of characters to change or -1.

Input Format

The first line will contain an integer, q, the number of test cases.
Each test case will contain a string s.

Constraints

  • 1 <= q <= 100
  • 1 <= |s| <= 104
  • s consists only of characters in the range ascii[a-z].

Sample Input

6
aaabbb
ab
abc
mnop
xyyx
xaxbbbxx

Sample Output

3
1
-1
2
0
1

Explanation

Test Case #01: We split s into two strings S1 =’aaaand S2 =’bbb’. We have to replace all three characters from the first string with ‘b’ to make the strings anagrams.

Test Case #02: You have to replace ‘a’ with ‘b, which will generate “bb”.

Test Case #03: It is not possible for two strings of unequal length to be anagrams of one another.

Test Case #04: We have to replace both the characters of first string (“mn”) to make it an anagram of the other one.

Test Case #05: S1 and S2 are already anagrams of one another.

Test Case #06: Here S1 = “xaxb” and S2 = “bbxx”. You must replace ‘a’ from S1 with ‘b’ so that S1 = “xbxb”.

HackerRank Anagram Solution

Anagram Solution in C

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<malloc.h>
#define lld long long int
#define llu long long unsigned int
int compare(const void * a, const void * b){return *(lld *)a-*(lld *)b;}
long long int readint() {long long int n=0,count=0,counti=0; char c;while(1){c=getchar_unlocked();if(c=='-')count=1;else if((c==' '||c=='\n'||c==EOF) && counti==1)break;else if(c>='0' && c<='9'){counti=1;n=(n<<3)+(n<<1)+c-'0';}}if(count==0)return n;else return -n;}
#define min(a,b)(a>b?b:a)
#define max(a,b)(a<b?b:a)
#define sort(arr,n) qsort(arr,n,sizeof(arr[0]),compare)
#define sd(n) scanf("%d",&n)
#define sl(n) scanf("%lld",&n)
#define su(n) scanf("%llu",&n)
#define rep(i,start,end) for(i=start; i<end; i++)
#define pdn(n) printf("%d\n",n)
#define pln(n) printf("%lld\n",n)
#define pun(n) printf("%llu\n",n)
#define pd(n) printf("%d",n)
#define pl(n) printf("%lld",n)
#define pu(n) printf("%llu",n)
#define pn printf("\n")
#define ps printf(" ")
int mod(int a)
{
    if(a>0)return a;
    else return -a;
}
int main()
{
    int t;
    char str[10009];
    sd(t);
    getchar();
    while(t--)
    {
        int ar[26]={},arr[26]={};
        scanf("%s",str);
        getchar();
        int len=strlen(str),i;
        if(len%2==0)
        {
            for(i=0; i<(len/2); i++)
                ar[str[i]-'a']++;
            for(i=(len/2); i<len; i++)
                arr[str[i]-'a']++;
            int ans=0;
            for(i=0; i<26; i++)
                ans+=mod(ar[i]-arr[i]);
            pdn(ans/2);
        }
        else
            printf("-1\n");
    }
    return 0;
}

Anagram Solution in Cpp

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for ( __typeof((c).begin()) it=(c).begin(); it!=(c).end(); it++ )
using namespace std;
#define N 10010
char s[N],sa[N],sb[N];
int main()
{
    int t;
    scanf("%d",&t);
    while ( t-- ) {
        scanf("%s",s);
        int n=strlen(s);
        if ( n%2!=0 ) {
            puts("-1");
            continue;
        }
        int cnt[26]={};
        for ( int i=0; i<n/2; i++ ) cnt[s[i]-'a']++;
        for ( int i=n/2; i<n; i++ ) cnt[s[i]-'a']--;
        int ans=0;
        for ( int i=0; i<26; i++ ) ans+=abs(cnt[i]);
        printf("%d\n",ans/2);
    }
    return 0;
}

Anagram Solution in Java

import java.io.*;
import java.util.*;
public class Solution {
	static void solve() throws IOException {
		int tests = nextInt();
		while (tests-- > 0) {
			String s = nextToken();
			int answer = solve(s);
			out.println(answer);
		}
	}
	private static int solve(String s) {
		if ((s.length() & 1) != 0) {
			return -1;
		}
		int k = s.length() >> 1;
		char[] c1 = s.substring(0, k).toCharArray();
		char[] c2 = s.substring(k, 2 * k).toCharArray();
		int[] cnt1 = get(c1);
		int[] cnt2 = get(c2);
		int result = 0;
		for (int i = 0; i < 256; i++) {
			result += Math.abs(cnt1[i] - cnt2[i]);
		}
		return result >> 1;
	}
	private static int[] get(char[] c1) {
		int[] ret = new int[256];
		for (char cc : c1) {
			++ret[cc];
		}
		return ret;
	}
	static BufferedReader br;
	static StringTokenizer st;
	static PrintWriter out;
	public static void main(String[] args) throws IOException {
		InputStream input = System.in;
		PrintStream output = System.out;
		File file = new File("a.in");
		if (file.exists() && file.canRead()) {
			input = new FileInputStream(file);
		}
		br = new BufferedReader(new InputStreamReader(input));
		out = new PrintWriter(output);
		solve();
		out.close();
	}
	static int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	static String nextToken() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null) {
				return null;
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}
}
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Anagram Solution in Python

t=input()
for _ in range(t):
    s=raw_input()
    l=len(s)
    if(len(s) %2 != 0):print -1
    else:
        s1=s[:l/2]
        s2=s[l/2:]
        cnt1=[0 for _ in xrange(26)]
        cnt2=[0 for _ in xrange(26)]
        for c in s1:
            cnt1[ord(c)-ord('a')]+=1
        for c in s2:
            cnt2[ord(c)-ord('a')]+=1
        ans=0
        for i in xrange(26):
            ans+=abs(cnt1[i]-cnt2[i])
        print ans/2

Anagram Solution using JavaScript

var dup_fun = function (dup) {
    return function (c) {
        if (dup[c] === undefined) { dup[c] = 0; }
        dup[c]++;
    };
};
function processData(input) {
    var lines = input.split('\n');
    var T = parseInt(lines.shift(), 10);
    
    var data = lines.splice(0, T);
    var res = [];
    data.forEach(function (s) {
        if (s.length % 2 != 0) {
            res.push(-1);
            return;
        }
        var len = Math.floor(s.length / 2);
        var s1 = s.substr(0, len).split('');
        var s2 = s.substr(len).split('');
        var dup1 = {}; var dup_fun1 = dup_fun(dup1);
        var dup2 = {}; var dup_fun2 = dup_fun(dup2);
        s1.forEach(dup_fun1);
        s2.forEach(dup_fun2);
        var dist = 0;
        for (var i in dup1) {
            var d1 = dup1[i];
            var d2 = (dup2[i] === undefined) ? 0 : dup2[i];
            dist += Math.max(0, d1 - d2);
        }
        res.push(dist);
    });
    res.forEach(function (n) {
        process.stdout.write(n + '\n');
    });
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});
process.stdin.on("end", function () {
   processData(_input);
});

Anagram Solution in Scala

var dup_fun = function (dup) {
    return function (c) {
        if (dup[c] === undefined) { dup[c] = 0; }
        dup[c]++;
    };
};
function processData(input) {
    var lines = input.split('\n');
    var T = parseInt(lines.shift(), 10);
    
    var data = lines.splice(0, T);
    var res = [];
    data.forEach(function (s) {
        if (s.length % 2 != 0) {
            res.push(-1);
            return;
        }
        var len = Math.floor(s.length / 2);
        var s1 = s.substr(0, len).split('');
        var s2 = s.substr(len).split('');
        var dup1 = {}; var dup_fun1 = dup_fun(dup1);
        var dup2 = {}; var dup_fun2 = dup_fun(dup2);
        s1.forEach(dup_fun1);
        s2.forEach(dup_fun2);
        var dist = 0;
        for (var i in dup1) {
            var d1 = dup1[i];
            var d2 = (dup2[i] === undefined) ? 0 : dup2[i];
            dist += Math.max(0, d1 - d2);
        }
        res.push(dist);
    });
    res.forEach(function (n) {
        process.stdout.write(n + '\n');
    });
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});
process.stdin.on("end", function () {
   processData(_input);
});

Ezoicreport this adAnagram Solution in Pascal

type 
  arr=array[0..26]of longint;
  procedure count(var ch:char;var A:arr);
  begin
  inc(A[ord(ch)-ord('a')+1]);
  end;
var
  i,ans,t,id:longint;
  Sa:ansistring;
  A,B:arr;
begin
  readln(t);
  for id:=1 to t do begin
    ans:=0;
    fillchar(A,sizeof(A),0);
    fillchar(B,sizeof(B),0);
    readln(Sa);
    if length(Sa) mod 2 =1 then begin
      writeln(-1);
      continue;
    end;
    for i:=1 to length(Sa) div 2 do count(Sa[i],A);
    for i:=(length(Sa) div 2 +1) to (length(Sa)) do count(Sa[i],B);
    for i:=1 to 26 do inc(ans,abs(A[i]-B[i]) );
    writeln(ans div 2);
  
  end;
end.

Disclaimer: This problem (Anagram) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

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