Holes in a string

general
1

Count the holes
if D,O,P,Q,A have 1 hole
and B has 2 hole

input word: BABOY

output:6 holes

2

@cris18 traverse the string and do follows:

int main()
    {
    char st[100];
    scanf("%s",st);
    int l = strlen(st);
    for(int i=0; i<l; i++) {
    if(st[i] == 'A'||st[i] == 'D'||st[i] == 'P'||st[i] == 'O'||st[i] == 'Q')
      count += 1;
    if(st[i] == 'B')
       count += 2;
    }
    printf("%d Holes\n",count);
    return 0;
    }

Happy Coding!!!

1 Like
3

Just timepass :wink:

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int holes[26] = {1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0};

int main() {
    char s[100];
    cin>>s;
    int len = strlen(s);
    int hcount = 0;
    for(int i=0; i<len; i++) {
		hcount += holes[s[i] - 'A'];
    }
    cout<<hcount<<endl;
    return 0;
}
4

R has no holes? :confused: R should have 1 hole as well. My code includes that.

5

can you help with this in java language?

6

can you help me with this in java language?

7

@cris18

Here is the Java solution you seem to need.

public static void main(String[] args) throws IOException {

    BufferedReader in = new BufferedReader( new InputStreamReader( System.in ) );
    BufferedWriter out = new BufferedWriter( new OutputStreamWriter( System.out ) ):

    String s = in.readLine(); // Accepts the word as input.

    int count = 0;
    for( int i = 0; i < s.length(); i++ ) {

        // Checks if the character has one hole. R also has a hole.
        if( s.charAt( i ) == 'D' || s.charAt( i ) == 'O' || s.charAt( i ) == 'P' || s.charAt( i ) == 'Q' || s.charAt( i ) == 'A' |} s.charAt( i ) == 'R' ))
              count++;
        // Checks if the character has two holes.
        else if ( s.charAt( i ) == 'B' )
              count += 2;

    }

    out.println( count );
    out.flush();
    out.close();
}

This solution assumes that the word given as input will be one line. If it isn’t, a slight modification to the input acceptance will have to be made. This runs O(N) time where N is the length of the word.

8

import java.util.*;

class Hello

{

public static void main(String args[]){

System.out.print("Enter String : ");

Scanner sc=new Scanner(System.in);

String input=sc.next();

int holes=0;

for(int i=0;i<input.length();i++)

{

char ch=input.charAt(i);

if(ch==‘D’||ch==‘O’||ch==‘P’||ch==‘Q’||ch==‘A’)

holes++;

else if(ch==‘B’)

holes=holes+2;

}

System.out.println(" Holes in string “+input+” are = "+holes);

}

}

DEMO -

Enter String : BABA

Holes in string BABA are = 6

9

check the answers below, fellow users have done it! :slight_smile:

10

/**
Title: HOLES IN THE TEXT
By: Chester C. Naguit
**/

import java.util.;
import java.text.;
import java.io.*;

class HoleInTheText{
static void sop(Object o){System.out.print(o);}
static void sopln(Object o){System.out.println(o);}
static void sopf(String s, Object o){System.out.printf(s,o);}
static Scanner sc = new Scanner(System.in);

static int[] count = new int[40];
static String str[] = new String[100];
static int t=0;
public static void main(String args[]){ 
	boolean b = true;
	while(b){
	t = sc.nextInt();
	sc.nextLine();
		if(t>40){
			sopln("Out of Range! 40 no. of test only!");
		}else{
			b = false;
			Count(t);
		}
	}
}

static void Count(int t){
	boolean b = false;
	for(int s=0; s<t; s++){
		str[s] = sc.nextLine();
		char[] c = str[s].toCharArray(); // convert string into array of char
		for(int i=0; i<c.length; i++){
			if(c[i]=='P'||c[i]=='O'||c[i]=='A'||c[i]=='D'||c[i]=='R') count[s]++;
			else if(c[i]=='B') count[s]+=2;
		}
	}
	for(int i=0; i<t; i++){
		sopln(count[i]);
	}
}

}

11

import java.util.*;
public class Holes {

public static void main(String[] args) {
   Scanner sc=new Scanner(System.in);
   String s;
   int holes=0;
   char c[]=new char[10];
   System.out.println("enter the given string to find holes");
   s=sc.next();
   c=s.toCharArray();
   for(int i=0;i<c.length;i++)
   {
	   if(c[i]=='A' || c[i]=='D' || c[i]=='P' || c[i]=='Q' || c[i]=='O')
		   holes++;
	   else if(c[i]=='B')
		   holes=holes+2;
	   else
		   holes=holes+0;
   }
   System.out.println("total no hoels is in your string is:"+holes);
}

}