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# emitL CodeChef Solution

## emitL CodeChef Solution in C++14

``````#include <iostream>
#include <vector>
using namespace std;

int main() {
long long t;
cin>>t;
while(t--){
string s;
cin>>s;

vector<long long>v(5,0);
bool f=1;
long long n=s.size();
if(n<9){
cout<<"NO"<<endl;
}

else if(n==9){

for(int i=0; i<n; i++){
if(s[i]=='L'){
v[0]++;
}
if(s[i]=='T'){
v[1]++;
}
if(s[i]=='I'){
v[2]++;
}
if(s[i]=='M'){
v[3]++;
}
if(s[i]=='E'){
v[4]++;
}

}
for(int i=0; i<4; i++){
if(v[i]!=2){
f=0;
}
}
if(v[4]!=1){
f=0;
}

if(f==0){
cout<<"NO"<<endl;
}
else{
cout<<"YES"<<endl;
}

}
else{
for(int i=0; i<n; i++){
if(s[i]=='L'){
v[0]++;
}
if(s[i]=='T'){
v[1]++;
}
if(s[i]=='I'){
v[2]++;
}
if(s[i]=='M'){
v[3]++;
}
if(s[i]=='E'){
v[4]++;
}

}

for(int i=0; i<5; i++){
if(v[i]<2){
f=0;
}
}
if(f==0){
cout<<"NO"<<endl;
}
else{
cout<<"YES"<<endl;
}

}

}
return 0;
}``````

## emitL CodeChef Solution in PYTH 3

``````# cook your dish here
def solve(s):

if len(s) < 5:
return 'NO'
mp = {
'L': 0,
'T':0,
'I':0,
'M':0,
'E':0
}
for i in range(len(s)):
if s[i] in mp:
mp[s[i]] += 1

# return mp.values()
for key,val in mp.items():

if val < 2:
if key == 'E' and len(s) == 9:
continue
else:
return 'NO'

return 'YES'

for _ in range(int(input())):

s = input()

print(solve(s))

``````

## emitL CodeChef Solution in C

``````#include <stdio.h>
#include <string.h>

int main ()
{
int y;

scanf( "%d", &y);

while (y--)

{

char string[101];

scanf( "%s", string );

int length = strlen(string);

int count[5] = {0};

for ( int i = 0 ; i <= length - 1 ; i++ )

{
switch ( string[i] )

{
case 'L':

count[0]++;

break;
case 'T':
count[1]++;
break;
case 'I':
count[2]++;
break;
case 'M':
count[3]++;
break;
case 'E':
count[4]++;
break;
}
}

if ( length <= 8 )
{
printf( "NO\n" );
}
else if ( length == 9 )
{
if ( count[0] == 2 && count[1] == 2 && count[2] == 2 && count[3] == 2 && count[4] == 1 )
{
printf( "YES\n" );
}
else
{
printf( "NO\n" );
}
}
else
{
if ( count[0] >= 2 && count[1] >= 2 && count[2] >= 2 && count[3] >= 2 && count[4] >= 2 )
{
printf( "YES\n" );
}
else
{
printf( "NO\n" );
}
}

}
} ``````

## emitL CodeChef Solution in JAVA

``````/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
String s = sc.next();
long counte =0,countm =0,counti =0,countt =0,countl =0;
for(int i=0; i<s.length();i++){
char ch = s.charAt(i);
if(ch=='E')
counte++;
else if(ch=='M')
countm++;
else if(ch=='I')
counti++;
else if(ch=='T')
countt++;
else if(ch=='L')
countl++;

}if(s.length()>9){
if(counte>1 &&countm>1 &&counti>1 &&countt>1 &&countl>1)
System.out.println("YES");
else
System.out.println("NO");
}else if(s.length()==9){
if(counte>=1 &&countm>1 &&counti>1 &&countt>1 &&countl>1)
System.out.println("YES");
else
System.out.println("NO");
}else{
System.out.println("NO");
}
}
}
}``````

## emitL CodeChef Solution in PYPY 3

``````t=int(input())
while t>0:
t-=1
s=str(input())
arr=[0,0,0,0,0]
for i in s:
if i.lower()=='l':
arr[0]+=1
elif i.lower()=='t':
arr[1]+=1
elif i.lower()=='i':
arr[2]+=1
elif i.lower()=='m':
arr[3]+=1
elif i.lower()=='e':
arr[4]+=1
if arr[0]>1 and arr[1]>1 and arr[2]>1 and arr[3]>1 and arr[4]>1:
print("YES")
elif arr[0]>1 and arr[1]>1 and arr[2]>1 and arr[3]>1 and arr[4]==1:
if len(s)==9:
print("YES")
else:
print("NO")
else:
print("NO")``````

## emitL CodeChef Solution in PYTH

``````t = int(raw_input())
for i in range(t):
st = raw_input().strip()
A = [0 for x in range(91)]
for x in st:
n = ord(x)
A[n] += 1
# endfor x
if (A[76] > 1) and (A[84] > 1) and (A[73] > 1) and (A[77] > 1) and (A[69] > 0):
res = 'YES'
else:
res = 'NO'
# endif
if (len(st) > 9) and (A[69] == 1):
res = 'NO'
# endif
print res
# endfor i
``````

## emitL CodeChef Solution in C#

``````    using System;

class Salem
{
public static bool checkLetter(string strin, char[] letters, int times)
{
int index;
for (int i = 0; i < times; i++ )
{
index = strin.IndexOf(letters[i]);
if (index != -1)
{
if (strin.IndexOf(letters[i], ++index) == -1)
return false;
}
else
return false;
}
if (strin.Contains("E"))
return true;
else
return false;
}

public static void Main()
{
for (int t = 0; t < amount; t++)
{
char[] search = { 'L', 'T', 'I', 'M', 'E'};

if (str.IndexOfAny(search) == -1 || str.Length < 9)
{
Console.WriteLine("NO");
}
else if (str.Length == 9)
{
bool flag = checkLetter(str, search, 4);
if (flag == false)
Console.WriteLine("NO");
else
Console.WriteLine("YES");
}
else if (str.Length > 9)
{
bool flag = checkLetter(str, search, 5);
if (flag == false)
Console.WriteLine("NO");
else
Console.WriteLine("YES");
}
}
return;
}
}``````

## emitL CodeChef Solution in GO

``````package main

import (
"bufio"
"os"
"strconv"
"fmt"
)

func nextInt(scanner *bufio.Scanner) int {
scanner.Scan()
num, _ := strconv.Atoi(scanner.Text())
return num
}

func nextString(scanner *bufio.Scanner) string{
scanner.Scan()
str := scanner.Text()
return str
}

func main() {
scanner.Split(bufio.ScanWords)
cases := nextInt(scanner)
for i:= 0; i < cases; i++ {
arr := [26]int{0}
str := nextString(scanner)
for _, val := range str {
if val == 'L' || val == 'T' || val == 'I' || val == 'M' || val == 'E' {
arr[val-65] += 1
}
}

if len(str) == 9 && arr['L' - 65] == 2 && arr['T' - 65] == 2 && arr['I' - 65] == 2 && arr['M' - 65] == 2 && arr['E' -65] == 1 {
fmt.Println("YES")
} else if arr['L' - 65] > 1 && arr['T' - 65] > 1 && arr['I' - 65] > 1 && arr['M' - 65] > 1 && arr['E' -65] > 1 {
fmt.Println("YES")
} else {
fmt.Println("NO")
}
}
}``````
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