How can I remove the repetitive characters in Histogram [JAVA]
class Mclass {
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
System.out.println(a[i]+":"+count);
count = 0;
Output:
a:2
b:2
c:2
d:1
a:2
b:2
c:2
Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?
java loops histogram
add a comment |
class Mclass {
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
System.out.println(a[i]+":"+count);
count = 0;
Output:
a:2
b:2
c:2
d:1
a:2
b:2
c:2
Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?
java loops histogram
add a comment |
class Mclass {
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
System.out.println(a[i]+":"+count);
count = 0;
Output:
a:2
b:2
c:2
d:1
a:2
b:2
c:2
Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?
java loops histogram
class Mclass {
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
System.out.println(a[i]+":"+count);
count = 0;
Output:
a:2
b:2
c:2
d:1
a:2
b:2
c:2
Here I want to stop the loop until it counts d = 1. but it again prints the same variable? How can i do that?
java loops histogram
java loops histogram
asked Nov 13 '18 at 19:49
Haseeb AhmedHaseeb Ahmed
175
175
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Starting from what you already did, first sort the array then try to count
import java.util.*;
class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Arrays.sort(a); // sort the array
for (int i=0; i<a.length; i++)
for(int j=i; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
continue;
i=j-1;
break;
System.out.println(a[i]+":"+count);
count = 0;
output
a:2
b:2
c:2
d:1
add a comment |
If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set
and print only when Set
doesn't contain the character and after printing, add it to Set
so next time on wards it doesn't get printed.
Change your code to this,
class Mclass
public static void main(String args)
Set<String> doneSet = new HashSet<String>();
char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
int count = 0;
for (int i = 0; i < a.length; i++)
for (int j = 0; j < a.length; j++)
if (a[j] == a[i])
count += 1;
if (!doneSet.contains(String.valueOf(a[i])))
System.out.println(a[i] + ":" + count);
doneSet.add(String.valueOf(a[i]));
count = 0;
This gives following output as you intend,
a:2
b:2
c:2
d:1
add a comment |
Do not print inside loop
Save your count and print outside the loop.
Do something like this:
public class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Map<String,Integer> output = new HashMap<>();
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
output.put(Character.toString(a[i]), count);
//System.out.println(a[i]+":"+count);
count = 0;
System.out.println(output);
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Starting from what you already did, first sort the array then try to count
import java.util.*;
class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Arrays.sort(a); // sort the array
for (int i=0; i<a.length; i++)
for(int j=i; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
continue;
i=j-1;
break;
System.out.println(a[i]+":"+count);
count = 0;
output
a:2
b:2
c:2
d:1
add a comment |
Starting from what you already did, first sort the array then try to count
import java.util.*;
class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Arrays.sort(a); // sort the array
for (int i=0; i<a.length; i++)
for(int j=i; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
continue;
i=j-1;
break;
System.out.println(a[i]+":"+count);
count = 0;
output
a:2
b:2
c:2
d:1
add a comment |
Starting from what you already did, first sort the array then try to count
import java.util.*;
class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Arrays.sort(a); // sort the array
for (int i=0; i<a.length; i++)
for(int j=i; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
continue;
i=j-1;
break;
System.out.println(a[i]+":"+count);
count = 0;
output
a:2
b:2
c:2
d:1
Starting from what you already did, first sort the array then try to count
import java.util.*;
class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Arrays.sort(a); // sort the array
for (int i=0; i<a.length; i++)
for(int j=i; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
continue;
i=j-1;
break;
System.out.println(a[i]+":"+count);
count = 0;
output
a:2
b:2
c:2
d:1
edited Nov 13 '18 at 21:34
answered Nov 13 '18 at 21:23
The Scientific MethodThe Scientific Method
1,7462515
1,7462515
add a comment |
add a comment |
If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set
and print only when Set
doesn't contain the character and after printing, add it to Set
so next time on wards it doesn't get printed.
Change your code to this,
class Mclass
public static void main(String args)
Set<String> doneSet = new HashSet<String>();
char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
int count = 0;
for (int i = 0; i < a.length; i++)
for (int j = 0; j < a.length; j++)
if (a[j] == a[i])
count += 1;
if (!doneSet.contains(String.valueOf(a[i])))
System.out.println(a[i] + ":" + count);
doneSet.add(String.valueOf(a[i]));
count = 0;
This gives following output as you intend,
a:2
b:2
c:2
d:1
add a comment |
If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set
and print only when Set
doesn't contain the character and after printing, add it to Set
so next time on wards it doesn't get printed.
Change your code to this,
class Mclass
public static void main(String args)
Set<String> doneSet = new HashSet<String>();
char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
int count = 0;
for (int i = 0; i < a.length; i++)
for (int j = 0; j < a.length; j++)
if (a[j] == a[i])
count += 1;
if (!doneSet.contains(String.valueOf(a[i])))
System.out.println(a[i] + ":" + count);
doneSet.add(String.valueOf(a[i]));
count = 0;
This gives following output as you intend,
a:2
b:2
c:2
d:1
add a comment |
If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set
and print only when Set
doesn't contain the character and after printing, add it to Set
so next time on wards it doesn't get printed.
Change your code to this,
class Mclass
public static void main(String args)
Set<String> doneSet = new HashSet<String>();
char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
int count = 0;
for (int i = 0; i < a.length; i++)
for (int j = 0; j < a.length; j++)
if (a[j] == a[i])
count += 1;
if (!doneSet.contains(String.valueOf(a[i])))
System.out.println(a[i] + ":" + count);
doneSet.add(String.valueOf(a[i]));
count = 0;
This gives following output as you intend,
a:2
b:2
c:2
d:1
If you don't want to print the character that has already been printed, you need to maintain it somewhere like in a Set
and print only when Set
doesn't contain the character and after printing, add it to Set
so next time on wards it doesn't get printed.
Change your code to this,
class Mclass
public static void main(String args)
Set<String> doneSet = new HashSet<String>();
char a = 'a', 'b', 'c', 'd', 'a', 'b', 'c' ;
int count = 0;
for (int i = 0; i < a.length; i++)
for (int j = 0; j < a.length; j++)
if (a[j] == a[i])
count += 1;
if (!doneSet.contains(String.valueOf(a[i])))
System.out.println(a[i] + ":" + count);
doneSet.add(String.valueOf(a[i]));
count = 0;
This gives following output as you intend,
a:2
b:2
c:2
d:1
answered Nov 13 '18 at 20:06
Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi
8,0252927
8,0252927
add a comment |
add a comment |
Do not print inside loop
Save your count and print outside the loop.
Do something like this:
public class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Map<String,Integer> output = new HashMap<>();
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
output.put(Character.toString(a[i]), count);
//System.out.println(a[i]+":"+count);
count = 0;
System.out.println(output);
add a comment |
Do not print inside loop
Save your count and print outside the loop.
Do something like this:
public class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Map<String,Integer> output = new HashMap<>();
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
output.put(Character.toString(a[i]), count);
//System.out.println(a[i]+":"+count);
count = 0;
System.out.println(output);
add a comment |
Do not print inside loop
Save your count and print outside the loop.
Do something like this:
public class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Map<String,Integer> output = new HashMap<>();
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
output.put(Character.toString(a[i]), count);
//System.out.println(a[i]+":"+count);
count = 0;
System.out.println(output);
Do not print inside loop
Save your count and print outside the loop.
Do something like this:
public class Mclass
public static void main(String args)
char a= 'a','b','c','d','a','b','c';
int count = 0;
Map<String,Integer> output = new HashMap<>();
for (int i=0; i<a.length; i++)
for(int j=0; j<a.length; j++)
if ( a[j] == a[i] )
count += 1;
output.put(Character.toString(a[i]), count);
//System.out.println(a[i]+":"+count);
count = 0;
System.out.println(output);
answered Nov 13 '18 at 20:02
Sayantan MandalSayantan Mandal
273213
273213
add a comment |
add a comment |
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