Tuesday, 4 August 2015

TCS codevite 2014 problem: Super ASCII String Checker

Problem : Super ASCII String Checker

In the Byteland country a string "S" is said to super ascii string if and only if count 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.

Input Format:

First line contains number of test cases T, followed by T lines, each containing a string "S".

Output Format:

For each test case print "Yes" if the String "S" is super ascii, else print "No"
Constraints:

1<=T<=100
1<=|S|<=400, S will contains only lower case alphabets ('a'-'z').

Sample Input and Output

SNo.InputOutput
1
2
bba
scca
Yes
No





Solution:



#include<iostream>
using namespace std;
int c[26];
int main(){
    int t,len,flag=1;;
    string s;
    cin>>t;
    while(t--){
        cin>>s;
        len=s.length();
        for(int i=0;i<len;i++)
            c[int(s[i])-int('a')]++;
        for(int i=0;i<26;i++){
            if(!(c[i]==0 || c[i]==i+1))
                flag=0;
        }
        if(flag==1)
            cout<<"Yes\n";
        else
            cout<<"No\n";
        for(int i=0;i<26;i++)
            c[i]=0;
        flag=1;
    }
    return 0;
}

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5 comments:

  1. Unknown5 November 2015 at 04:30

    can please give solution in C

    ReplyDelete
  2. Unknown2 July 2016 at 04:49

    #include
    using namespace std;

    int main() {

    int t;
    cin>> t;
    START:
    while(t--){
    string s;

    cin>> s;
    int ar[27]={0};

    for(int i=0;i<s.length();i++)ar[s.at(i)-'a'+1]++;

    for(int i=0;i<s.length();i++){
    if(ar[s.at(i)-'a'+1]!=s.at(i)-'a'+1){
    cout<< "No"<<endl;
    goto START;
    }
    }
    cout<<"Yes"<< endl;

    }

    return 0;
    }

    ReplyDelete
  3. Unknown12 January 2017 at 09:00

    c[int(s[i])-int('a')]++;
    for(int i=0;i<26;i++){
    if(!(c[i]==0 || c[i]==i+1))
    flag=0;
    PLEASE TELL ME THE WORKING OF THIS CODE PRESENT HERE

    ReplyDelete
    Replies
    1. Unknown14 January 2017 at 08:28

      int(s[i])-int('a') gives 1 less than ascii value of a character. For example if s[i] is equal to b then it evaluates to int('b')-int('a') = 1

      so using this property we are storing count of a character with ascii value x at index x-1

      The condition !(c[i]==0 || c[i]==i+1) can be simplified to (c[i]!=0 && c[i]!=i+1) using de morgan's law. c[i] stores count of a character with ascii value i+1. S0 we check whether it is equal to i+1.If not we set the flag to zero

      Delete
  4. Unknown28 June 2020 at 07:17

    #include
    int main() {
    char s[30];
    int i,num[30]={0},isascii,n;
    scanf("%d",&n);
    while(n--)
    {
    scanf("%s",s);
    i=0;
    isascii=1;
    while(s[i]!='\0')
    {
    if((s[i]>='a')&&(s[i]<='z'))
    num[s[i]-'a']++;
    s[i]='\0';
    i++;
    }
    for(i=0;i<26;i++)
    {
    if((num[i]>0)&&(num[i]!=(i+1)))
    isascii=0;
    num[i]=0;
    }
    if(isascii)
    printf("yes\n");
    else
    printf("no");
    }
    return 0;
    }

    ReplyDelete

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