Sherlock and the Valid String
Problem Statement :
Problem Statement : Sherlock considers a string to be valid if all characters of the string appear the same number of times. It is also valid if he can remove just 1 character at 1 index in the string, and the remaining characters will occur the same number of times. Given a string s, determine if it is valid. If so, return YES, otherwise return NO. Function Description Complete the isValid function in the editor below. isValid has the following parameter(s): string s: a string Returns string: either YES or NO Input Format A single string s. Sample Input 0 aabbcd Sample Output 0 NO
Solution :
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;
}
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");
}
}
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;
}
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"
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