Latest Java Programmin Aptitude Question SOLUTION: In the Byteland country a string “S” is said to super ASCII string if and only ifcount of each character in the string is equal to its ASCII value. In the Byteland country ASCI

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(#M40039498) JAVA PROGRAMMIN Programming question Super ASCII String checker Keep an EYE

In the Byteland country a string “S” is said to super ASCII string if and only ifcount
of each character in the string is equal to its ASCII value.
In the Byteland country ASCII code of ‘a’ is 1, ‘b’ is 2...’z’ is 26.
Your task is to find out whether the given string is a super ASCII string or not.

Asked In Java Programmin pankti (10 years ago)

Unsolved Read Solution (7)

Is this Puzzle helpful?   (17)   (10) Submit Your Solution Technical

Read Solution (7) :

sandipan  (9 years ago)

#include
int number(char st[],char a,int i,int c);

int main( void ){


char st[400];
int v1=0,test,l=0;
int a[30];
char c1;
int pf=0;
int i;
scanf("%d",&test);
while(l!=test){
scanf("%s",&st);
for(i=0;i

Is this solution Helpfull? Yes (3) | No   

sandipan  (9 years ago)

#include
int number(char st[],char a,int i,int c);

int main( void ){


char st[400];
int v1=0,test,l=0;
int a[30];
char c1;
int pf=0;
int i;
scanf("%d",&test);
while(l!=test){
scanf("%s",&st);
for(i=0;i

Is this solution Helpfull? Yes (1) | No   

venkatachalam  (9 years ago)

hfasdfhliasdhf pearhgfipraohgflijsanf tj wqapoewhntoiqew; 9rqhepftraewij abtfiuewdbvoewar ewgfhijewrbv

Is this solution Helpfull? Yes | No (2)  

Kumar Jishu  (9 years ago)

Sorry dude. Help yourself.

Is this solution Helpfull? Yes | No   

pratik  (9 years ago)

import java.util.*;

class AsciiChecker
{
public static void main(String a[])
{
int i,count=0,j;
Scanner sc=new Scanner(System.in);
ArrayList result=new ArrayList();
int t=sc.nextInt();sc.nextLine();
if(t>=1 && t0)
{
String s=sc.nextLine();
char c[]=s.toCharArray();
if(s.length()>0 && s.length()

Is this solution Helpfull? Yes | No (1)  

Shivam Mishra  (9 years ago)

#include
#include
#include
int main()
{
char s[50]={''};
int i=0,b,d=0,l;
gets(s);
l=strlen(s);
for(i=0;s[i]!='';i++)
{
s[i]-=96;

}
for(i=0;s[i]!='';i++)
{
for(b=0;s[b]!='';b++)
{
if(s[i]==s[b])
d++;
}
if(s[i]==d)
d=0;
else
{
printf("nNO");
break;
}
}
if((i)==l)
printf("Yes");
}
getch();

Is this solution Helpfull? Yes | No (2)  

Vikas P  (7 years ago)

This code You can use In python :

from collections import OrderedDict
n=input("Enter no. of Strings :")
#print n
li=[]
for j in range(1,n+1) :
li.append(raw_input("Enter String : ").lower())
#print li
#st="bbaccc".lower()
for st in li :
dic = {}
for i in set(st):
b = st.count(i, 0, len(st))
dic[i] = b
#print dic

od ="".join(OrderedDict.fromkeys(st).keys())
#print(od)
#print len(od)
c=0

for i in od :
#print "ascii of ",i," is ",ord(i)-96,"count : ",dic[i]
if (ord(i)-96)==dic[i] :
c +=1
if c==len(od) :
print st,"Yes"
else :
print st,"No"

Is this solution Helpfull? Yes | No (1)  

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