Sherlock and the Valid String

                            Solution in C :

In  C:





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int fre[26];
int main() {

    char S[100001];
    scanf("%s",&S);
    int len=strlen(S);
    int i=0;
    for(i=0;i<len;i++)
        {
        fre[S[i]-'a']++;
        }
    int flag=0;
    int same=0;
    int count=0;
    for(i=0;i<26;i++)
        {
        if(flag==0&&fre[i]!=0)
            {
            flag=1;
            same=fre[i];
        }
        if(flag==1&&fre[i]!=0&&fre[i]!=same)
            {
            count++;
        }
        
    }
    if(count<=1)
    printf("YES");
    else
        printf("NO");
    return 0;
}
                        

                        Solution in C++ :

In  C++ :





#define _USE_MATH_DEFINES
#include <algorithm>
#include <cstdio>
#include <functional>
#include <iostream>
#include <cfloat>
#include <climits>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <vector>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> i_i;
typedef pair<ll, int> ll_i;
typedef pair<double, int> d_i;
typedef pair<ll, ll> ll_ll;
typedef pair<double, double> d_d;
struct edge { int u, v; ll w; };

ll MOD = 1000000007;
ll _MOD = 1000000009;
double EPS = 1e-10;

bool check(vector<int>& a) {
	set<int> s;
	for (int k = 0; k < 26; k++)
		s.insert(a[k]);
	return s.size() <= 2;
}

int main() {
	string S; cin >> S;
	vector<int> a(26);
	for (int i = 0; i < S.length(); i++)
		a[S[i] - 'a']++;
	bool ok = check(a);
	for (int k = 0; k < 26; k++)
		if (a[k]) {
			a[k]--;
			if (check(a)) ok = true;
			a[k]++;
		}
	cout << (ok ? "YES" : "NO") << endl;
}
                    

                        Solution in Java :

In Java :





import java.io.*;
import java.util.Arrays;
import java.util.Locale;


public class Main {

	public static void main(String[] args) throws Exception {
		new Main().run();
	}


	private void run() throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		char[] s = br.readLine().trim().toCharArray();
		int[] cnt = new int[26];
		for (char c : s) {
			cnt[c - 'a']++;
		}
		
		Arrays.sort(cnt);
		
		int cntMin = 0;
		int cntMax = 0;
		
		int min = Integer.MAX_VALUE;
		int max = 0;
		
		for (int i = 0; i < 26; i++) {
			if (cnt[i] > 0) {
				min = Math.min(min, cnt[i]);
				max = Math.max(max, cnt[i]);
			}
		}
		
		//System.out.println(Arrays.toString(cnt));
		
		//System.out.println(min);
		//System.out.println(max);
		
		for (int i = 0; i < 26; i++) {
			if (cnt[i] > 0) {
				cntMin += Math.abs(min - cnt[i]);
			}
			if (cnt[i] < max) {
				cntMax += cnt[i];
			}
		}
		
		//System.out.println(cntMin);
		//System.out.println(cntMax);
		
		System.out.println(cntMin <= 1 || cntMax <= 1 ? "YES" : "NO");
	}


}
                    

                        Solution in Python : 
                            
In  Python3 :





rem = []
s = input().strip()
ch = []
for i in range(26):
    ch.append(0)
for i in range(len(s)):
    ch[ord(s[i])-97] += 1
#print(ch)
set = set()
for i in range(26):
    set.update(ch)
set.discard(0)
#print(set)
for x in set:
    remove = 0
    for c in ch:
        if c > 0:
            if c > x:
                remove += c - x
            elif c < x:
                remove += c
    rem.append(remove)
#print(min(rem))
if min(rem) <= 1:
    print("YES")
else:
    print("NO")