HackerEarth Bob and String problem solution

In this HackerEarth Bob and String problem solution, Bob and Khatu both love the string. Bob has a string S and Khatu has a string T. They want to make both string S and T to anagrams of each other. Khatu can apply two operations to convert string T to an anagram of string S which are given below:

1. Delete one character from the string T.
2. Add one character from the string S.

Khatu can apply above both operations as many times as he wants. Find the minimum number of operations required to convert string T so that both T and S will become anagram of each other.

HackerEarth Bob and String problem solution.

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int test;
    cin>>test;
    while(test--)
    {
        string S,T;
        cin>>S>>T;
        int hashS[26]={0};
        int hashT[26]={0};
        for(int i=0;i<S.size();i++)
            hashS[S[i]-'a']++;

        for(int i=0;i<T.size();i++)
            hashT[T[i]-'a']++;

        int count=0;
        for(int i=0;i<26;i++)
        {
            if(hashS[i]-hashT[i]>0)
                count+=hashS[i]-hashT[i];
            if(hashS[i]-hashT[i]<0)
                count+=hashT[i]-hashS[i];
        }   
        cout<<count<<endl;
    }
    return 0;
}

Second solution

#include<bits/stdc++.h>

using namespace std;

#define vi vector < int >
#define pii pair < int , int >
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define foreach(it,v) for( __typeof((v).begin())it = (v).begin() ; it != (v).end() ; it++ )
#define ll long long
#define llu unsigned long long
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define dbg(x) { cout<< #x << ": " << (x) << endl; }
#define dbg2(x,y) { cout<< #x << ": " << (x) << " , " << #y << ": " << (y) << endl; }
#define all(x) x.begin(),x.end()
#define mset(x,v) memset(x, v, sizeof(x))
#define sz(x) (int)x.size()

int main()
{
    int t;
    cin >> t;
    assert(1 <= t && t <= 10);
    while(t--)
    {
        string a , b;
        cin >> a >> b;
        int n = sz(a) , m = sz(b) , i;
        assert(1 <= n && n <= 100000);
        assert(1 <= m && m <= 100000);
        int c[256] = {0};
        for(i=0;i<n;i++)
        {
            c[a[i]]++;
        }
        for(i=0;i<m;i++)
        {
            c[b[i]]--;
        }
        int ans = 0;
        for(i=0;i<256;i++)
        {
            ans += abs(c[i]);
        }
        cout << ans << endl;
    }
    return 0;
}