1038A - Equality - CodeForces Solution

implementation strings *800

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Python Code: 

from cmath import *
from math import *
from decimal import *
from collections import *
import numbers
from operator import index
import sys
def _input(): return map(int, input().split())
def _list(): return list(map(int, input().split()))
def _int(): return int(input())


def solves():
    n,k=_input()
    s=input()
    mn=inf
    char=''
    alpha=['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
    for i in range (k):
        if alpha[i] not in s:
            print(0)
            return
        if (s.count(alpha[i])<=mn):
            mn=min(mn,s.count(alpha[i]))
    print(mn*k)
         
        

t=1
for _ in range(0,t):
    solves()

C++ Code: 

#include<bits/stdc++.h>

typedef long long int ll;
typedef unsigned long long int ull;

#define dbg printf("in\n")
#define nl printf("\n");
#define N 105
#define inf 10000000000000000

#define sf(n) scanf("%d", &n)
#define sff(n,m) scanf("%d%d",&n,&m)
#define sfl(n) scanf("%I64d", &n)
#define sffl(n,m) scanf("%I64d%I64d",&n,&m)

#define pf(n) printf("%d ",n)
#define pff(n,m) printf("%d %d\n",n,m)
#define pffl(n,m) printf("%I64d %I64d\n",n,m)
#define pfl(n) printf("%I64d\n",n)
#define pfs(s) printf("%s",s)

#define pb push_back
#define pp pair<int,int>

using namespace std;

int main()
{
    //freopen("in.txt","r",stdin);

    int i,j,k;
    int n,m;

    string s;

    int a[26];
    memset(a,0,sizeof(a));

    sff(n,k);
    cin>>s;

    for(i=0;i<n;i++)
    {
        a[s[i]-'A']++;
    }

    int mn=n;
    for(i=0;i<k;i++)
        mn=min(mn,a[i]);

    pf(mn*k);


    return 0;
}


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