Codeforces Round #429 (Div. 2) A. Generous Kefa【hash/判断字符串是否有一种字符个数大于m】

A. Generous Kefa

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Kefa found n baloons. For convenience, we denote color of i-th baloon as si — lowercase letter of the Latin alphabet. Also Kefa has k friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 100) — the number of baloons and friends.

Next line contains string s — colors of baloons.

Output

Answer to the task — «YES» or «NO» in a single line.

You can choose the case (lower or upper) for each letter arbitrary.

Examples

input

4 2
aabb

output

YES

input

6 3
aacaab

output

NO

Note

In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.

In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

 【题意】：判断字符串是否有一种字符个数大于m。

【分析】：把每个字符映射为一个0~26的数值，判断数值出现的次数是否超过m。

【代码】：

#include<bits/stdc++.h>
using namespace std;
char a;
int mp[50];
int m,n;
int f;
int main()
{
    while(cin>>n>>m)
    {
        getchar();//
        for(int i=0;i<n;i++)
        {
            scanf("%c",&a);
            mp[a-'a']++;
        }
        f=1;
        for(int i=0;i<26;i++)
        {
            if(mp[i]!=0)
            {
                if(mp[i]>m)
                {
                    f=0;
                    break;
                }
            }
        }
        if(f) puts("YES");
        else puts("NO");
    }
    return 0;
}

　　

int n, k; cin >> n >> k;  
    string s; cin >> s;  
    int cnt[30] = {0};  
    for (int i = 0; i < s.size(); i++) {  
        cnt[s[i] - 'a']++;  
    }  

char str[110];  
int a[26];  
        scanf("%s",str);  
        int len=strlen(str);  
        memset(a,0,sizeof(a));  
        for(int i=0;i<len;i++)  
        {  
            a[str[i]-'a']++;  
        }