Programming Forums > Application Development > Java

xavier · Nov 13th, 2004, 3:02 AM

Hi, my problem is as folows: I have 2 strings : exapmple: s1="aabbccdef" and s2="abcddefd" . Now i have to find out wich letter is found more times in the 2 strings and is common to both. I turned them into char .. with toCharArray() but i'm having problems with comparing the 2. In this case the aswere should be "d".

The first thing would be to take all the commone letters and put them in ....
char c3=new char[ ? ] . I'm not sure how long it should be. from then it should be simple.
Thanks for the help i hope u'll give me .

xavier · Nov 14th, 2004, 5:48 AM

And another thing : in this quick sort I get:

(Toggle Plain Text)

Exception in thread "main" java.lang.StackOverflowError

i'm not sure y!! -- actualy i don't know y!

(Toggle Plain Text)

import java.util.*;
class quicksort{
	public static void main(String args[]){
 final int N=1000;
        final int L=1000;
          int a[]=new int[N];
 Random r=new Random();
  for(int i=0;i<N;i++)
 	a[i]=r.nextInt(L);
 
 Date d=new Date();
 long t0=d.getTime();
 
 quickSort(a,0,N-1);
 
 d=new Date();
 long t1=d.getTime();
 System.out.println("Sorting time="+((t1-t0)/1000)+"sec");
  
	}
	
	public static void quickSort(int a[],int left, int right){
 int p=Pivot(a,left,right);
 quickSort(a,left,p-1);
 quickSort(a,p+1,right);
	}
	
	public static int Pivot(int a[],int left,int right){
 int i,j;
 i=left;
 j=right;
 int di=0;
 int dj=1;
 while(i<j){
 	if(a[i]>a[j]){
  int aux=a[i];
  	a[i]=a[j];
  	a[j]=aux;
  	
  	//aux=di;
  
  	aux=di;
  	di=dj;
  	dj=aux;
 	}
 i=i+di;
 j=j-dj;
 }
	return i;
	}
}

HELP please!!

arial · Nov 14th, 2004, 12:30 PM

Couldnt you just compare the strings somehow? instead of using arrays? (Not sure if that's possible, I'm new to Java too)

T3FLoN · Nov 14th, 2004, 12:46 PM

just an idea ur comparing letters right...well u can put the alphabet in an array of 26 and using a for loop u can make it take each letter in ur string and compare it to the next using a nested second for loop.

Then create urself a c3 empty string and make some ifs and elses that will replace the char stored everytime it finds a char with a higher index.

Set your for loop to do this 26 times.

That method should work fine since it covers any range of Strings u might input.

xavier · Nov 14th, 2004, 2:42 PM

the problem is that i can't get all the common ellements into another char[]
Example:
c1:abbc
=========>c3:aaabbbbcc this is what i'm having problems getting.
c2:bbbcdaa

even if i sort first the 2 strings :
c1:abbc
c2:aabbbcd

I can't just comare the 2 and put them in c3. For example the first 'a' would be put in twice !!!!
I thaught about reducing the strings:
s1:abc
s2:abcd

Then compare c1 with s2 and c2 with s1 --> then put what's commone from each in c3 ... after that it would be simple.

(Toggle Plain Text)

private static int removeExtra(char a[]){
 int i;
        int d=a.length;
 for(i=0;i<a.length-1;i++){
 	if(a[i]==a[i+1]){
  a[i]=a[i+1];
  d--;
 	}
         }
      }

that's how i'm thinking of doing it. but i have to return th letters that remain .. and not sure how to do it...... and what to write in main.

Hmmmmmm.......... <_<

tempest · Nov 14th, 2004, 11:22 PM

Have you considered bubble sort?

(Toggle Plain Text)

public static void bubbleSort(int a[]){
  int len = a.length();
  for(int i=0;i<len - 1;i++)
    for(int j=0;j<len - 1 - i;j++)
      if(a[j] > a[j+1]) {
        int tmp = a[j];
        a[j] = a[j+1];
        a[j+1] = tmp;
      }
}

xavier · Nov 15th, 2004, 9:25 AM

First ... i think your sorting doesn't work... the second for should be :

(Toggle Plain Text)

 for(int j=0;j<len;j++)

But the SORTING is not the problem.

Quote:

CODE

private static int removeExtra(char a[]){
int i;
int d=a.length;
for(i=0;i<a.length-1;i++){
if(a[i]==a[i+1]){
a[i]=a[i+1];
d--;
}
}
}

that's how i'm thinking of doing it. but i have to return th letters that remain .. and not sure how to do it...... and what to write in main.

That's what i don't know:: getting from : aabbccccddeeeffff to -> abcdef

eccles · Nov 21st, 2004, 9:17 PM

I realise its a bit ugly - but perhaps something like this might help??
then you can just call it with your two character arrays and it will return an array representing the (unique) intersection.

then you can iterate over that, and you get the bonus that you know how many chars are actually common to each, so you might use that to initialize an array to store the number of occurences in or whatever...

I haven't tried to compile this but if it does have problems you should be able to fix them.

(Toggle Plain Text)

public char[] intersection( char[] array1, char[] array2 )
{
	Vector str1 = array1.toVector();
	Vector str2 = array2.toVector();
	Vector intersection = new Vector();


	for ( Iterator i = new str1.iterator(); i.hasNext(); )
	{
 if ( str2.contains( i.next() ) && !( intersection.contains(i.next()) ) )
 {
 	intersection.add( i.next() );
 }
	}
	
	char[] intersect = new char[intersection.size()];
	for ( int i=0; i < intersection.size(); i++ ) 
	{
 intersect[i] = (char)intersection.get(i);
	}
	return intersect;
}

P.S. I didn't use the toArray() method because we want a char array, not an object array.
which reminds me - the use of the toVector() method may not work for the same reason.

it's all very inelegant really.

xavier · Nov 22nd, 2004, 7:45 AM

I did it in the end ... lilke this :

(Toggle Plain Text)

class comun{

	public static void main(String args[]){
 	final int N=100;
 String s1="abccd";
 String s2="abbdccga";
 char c1[];
 char c2[];
 c1=s1.toCharArray();
 c2=s2.toCharArray();	
 int i;int j;
 char test[]=new char[N];
 char test1[]=new char[N];
 int k=0;
 int p=0;
 for(i=0;i<c1.length;i++){
 	boolean esteComun=false;
 	for(j=0;j<c2.length;j++){
  if(c1[i]==c2[j]){
  	esteComun=true;
  	break;
  }
 	}
 if(esteComun==true){

 	test[k]=c1[i];
 	k++;

 }
 }

	char w[] = new char[k];
 System.arraycopy(test, 0, w, 0, k);



 for(i=0;i<c2.length;i++){
 	boolean esteComun=false;
 	for(j=0;j<c1.length;j++){
  if(c2[i]==c1[j]){
  	esteComun=true;
  	break;
  }
 	}
 if(esteComun==true){

 	test1[p]=c2[i];
 	p++;

 }
	
 }
 
	char q[] = new char[p];
	System.arraycopy(test1, 0, q, 0, p);

	char x[] = new char[k+p];
	for(i=0;i<w.length;i++)x[i]=w[i];
 for(i=0;i<q.length;i++)x[i+w.length]=q[i];

	

	sort (x);
//	for(i=0;i<x.length;i++)System.out.print(x[i]);
	
	int cont=1;
	int max=0;
	char val=0;

	for (i=0;i<x.length-1;i++){
 if(x[i]==x[i+1]){
 	cont++;
 	if(cont>max){max=cont;
 	val=x[i];}
 }
 else cont=1;
	}
	System.out.print(val);

}
	private static void sort(char a[]){
 int i;
 for(i=0;i<a.length-1;i++)
 	for(int j=i;j<a.length;j++)
  if(a[i]>a[j]){char aux=a[i];
    a[i]=a[j];
    a[j]=aux;

    }
 }
}

I think i did the same thig twice in the midle there ... but u get the picture... it works

:banana:
10x for your help

Nov 13th, 2004, 3:02 AM	#1
xavier Professional Programmer Join Date: Oct 2004 Location: .ro Posts: 406 Rep Power: 5	Hi, my problem is as folows: I have 2 strings : exapmple: s1="aabbccdef" and s2="abcddefd" . Now i have to find out wich letter is found more times in the 2 strings and is common to both. I turned them into char .. with toCharArray() but i'm having problems with comparing the 2. In this case the aswere should be "d". The first thing would be to take all the commone letters and put them in .... char c3=new char[ ? ] . I'm not sure how long it should be. from then it should be simple. Thanks for the help i hope u'll give me . __________________ Don't take life too seriously, it's not permanent !

Nov 14th, 2004, 5:48 AM	#2
xavier Professional Programmer Join Date: Oct 2004 Location: .ro Posts: 406 Rep Power: 5	And another thing : in this quick sort I get: (Toggle Plain Text) Exception in thread "main" java.lang.StackOverflowError Exception in thread "main" java.lang.StackOverflowError i'm not sure y!! -- actualy i don't know y! (Toggle Plain Text) import java.util.; class quicksort{ public static void main(String args[]){ final int N=1000; final int L=1000; int a[]=new int[N]; Random r=new Random(); for(int i=0;i<N;i++) a[i]=r.nextInt(L); Date d=new Date(); long t0=d.getTime(); quickSort(a,0,N-1); d=new Date(); long t1=d.getTime(); System.out.println("Sorting time="+((t1-t0)/1000)+"sec"); } public static void quickSort(int a[],int left, int right){ int p=Pivot(a,left,right); quickSort(a,left,p-1); quickSort(a,p+1,right); } public static int Pivot(int a[],int left,int right){ int i,j; i=left; j=right; int di=0; int dj=1; while(i<j){ if(a[i]>a[j]){ int aux=a[i]; a[i]=a[j]; a[j]=aux; //aux=di; aux=di; di=dj; dj=aux; } i=i+di; j=j-dj; } return i; } } import java.util.; class quicksort{ public static void main(String args[]){ final int N=1000; final int L=1000; int a[]=new int[N]; Random r=new Random(); for(int i=0;i<N;i++) a[i]=r.nextInt(L); Date d=new Date(); long t0=d.getTime(); quickSort(a,0,N-1); d=new Date(); long t1=d.getTime(); System.out.println("Sorting time="+((t1-t0)/1000)+"sec"); } public static void quickSort(int a[],int left, int right){ int p=Pivot(a,left,right); quickSort(a,left,p-1); quickSort(a,p+1,right); } public static int Pivot(int a[],int left,int right){ int i,j; i=left; j=right; int di=0; int dj=1; while(i<j){ if(a[i]>a[j]){ int aux=a[i]; a[i]=a[j]; a[j]=aux; //aux=di; aux=di; di=dj; dj=aux; } i=i+di; j=j-dj; } return i; } } HELP please!! __________________ Don't take life too seriously, it's not permanent !

Nov 14th, 2004, 12:30 PM	#3
arial Newbie Join Date: Nov 2004 Location: UK Posts: 3 Rep Power: 0	Couldnt you just compare the strings somehow? instead of using arrays? (Not sure if that's possible, I'm new to Java too) __________________ It's nice to be nobody.

Nov 14th, 2004, 12:46 PM	#4
T3FLoN Newbie Join Date: Nov 2004 Posts: 21 Rep Power: 0	just an idea ur comparing letters right...well u can put the alphabet in an array of 26 and using a for loop u can make it take each letter in ur string and compare it to the next using a nested second for loop. Then create urself a c3 empty string and make some ifs and elses that will replace the char stored everytime it finds a char with a higher index. Set your for loop to do this 26 times. That method should work fine since it covers any range of Strings u might input. __________________ <span style='font-family:Courier'><span style='font-size:12pt;line-height:100%'><span style='color:orange'>☼☼☼ Just kill em all, and let God sort em out. ☼☼☼</span></span></span>

Nov 14th, 2004, 2:42 PM	#5
xavier Professional Programmer Join Date: Oct 2004 Location: .ro Posts: 406 Rep Power: 5	the problem is that i can't get all the common ellements into another char[] Example: c1:abbc =========>c3:aaabbbbcc this is what i'm having problems getting. c2:bbbcdaa even if i sort first the 2 strings : c1:abbc c2:aabbbcd I can't just comare the 2 and put them in c3. For example the first 'a' would be put in twice !!!! I thaught about reducing the strings: s1:abc s2:abcd Then compare c1 with s2 and c2 with s1 --> then put what's commone from each in c3 ... after that it would be simple. (Toggle Plain Text) private static int removeExtra(char a[]){ int i; int d=a.length; for(i=0;i<a.length-1;i++){ if(a[i]==a[i+1]){ a[i]=a[i+1]; d--; } } } private static int removeExtra(char a[]){ int i; int d=a.length; for(i=0;i<a.length-1;i++){ if(a[i]==a[i+1]){ a[i]=a[i+1]; d--; } } } that's how i'm thinking of doing it. but i have to return th letters that remain .. and not sure how to do it...... and what to write in main. Hmmmmmm.......... <_< __________________ Don't take life too seriously, it's not permanent !

Nov 14th, 2004, 11:22 PM	#6
tempest Programming Guru Join Date: Oct 2004 Posts: 1,041 Rep Power: 6	Have you considered bubble sort? (Toggle Plain Text) public static void bubbleSort(int a[]){ int len = a.length(); for(int i=0;i<len - 1;i++) for(int j=0;j<len - 1 - i;j++) if(a[j] > a[j+1]) { int tmp = a[j]; a[j] = a[j+1]; a[j+1] = tmp; } } public static void bubbleSort(int a[]){ int len = a.length(); for(int i=0;i<len - 1;i++) for(int j=0;j<len - 1 - i;j++) if(a[j] > a[j+1]) { int tmp = a[j]; a[j] = a[j+1]; a[j+1] = tmp; } } __________________

Nov 21st, 2004, 9:17 PM	#8
eccles Newbie Join Date: Nov 2004 Posts: 16 Rep Power: 0	I realise its a bit ugly - but perhaps something like this might help?? then you can just call it with your two character arrays and it will return an array representing the (unique) intersection. then you can iterate over that, and you get the bonus that you know how many chars are actually common to each, so you might use that to initialize an array to store the number of occurences in or whatever... I haven't tried to compile this but if it does have problems you should be able to fix them. (Toggle Plain Text) public char[] intersection( char[] array1, char[] array2 ) { Vector str1 = array1.toVector(); Vector str2 = array2.toVector(); Vector intersection = new Vector(); for ( Iterator i = new str1.iterator(); i.hasNext(); ) { if ( str2.contains( i.next() ) && !( intersection.contains(i.next()) ) ) { intersection.add( i.next() ); } } char[] intersect = new char[intersection.size()]; for ( int i=0; i < intersection.size(); i++ ) { intersect[i] = (char)intersection.get(i); } return intersect; } public char[] intersection( char[] array1, char[] array2 ) { Vector str1 = array1.toVector(); Vector str2 = array2.toVector(); Vector intersection = new Vector(); for ( Iterator i = new str1.iterator(); i.hasNext(); ) { if ( str2.contains( i.next() ) && !( intersection.contains(i.next()) ) ) { intersection.add( i.next() ); } } char[] intersect = new char[intersection.size()]; for ( int i=0; i < intersection.size(); i++ ) { intersect[i] = (char)intersection.get(i); } return intersect; } P.S. I didn't use the toArray() method because we want a char array, not an object array. which reminds me - the use of the toVector() method may not work for the same reason. it's all very inelegant really.

Nov 22nd, 2004, 7:45 AM	#9
xavier Professional Programmer Join Date: Oct 2004 Location: .ro Posts: 406 Rep Power: 5	I did it in the end ... lilke this : (Toggle Plain Text) class comun{ public static void main(String args[]){ final int N=100; String s1="abccd"; String s2="abbdccga"; char c1[]; char c2[]; c1=s1.toCharArray(); c2=s2.toCharArray(); int i;int j; char test[]=new char[N]; char test1[]=new char[N]; int k=0; int p=0; for(i=0;i<c1.length;i++){ boolean esteComun=false; for(j=0;j<c2.length;j++){ if(c1[i]==c2[j]){ esteComun=true; break; } } if(esteComun==true){ test[k]=c1[i]; k++; } } char w[] = new char[k]; System.arraycopy(test, 0, w, 0, k); for(i=0;i<c2.length;i++){ boolean esteComun=false; for(j=0;j<c1.length;j++){ if(c2[i]==c1[j]){ esteComun=true; break; } } if(esteComun==true){ test1[p]=c2[i]; p++; } } char q[] = new char[p]; System.arraycopy(test1, 0, q, 0, p); char x[] = new char[k+p]; for(i=0;i<w.length;i++)x[i]=w[i]; for(i=0;i<q.length;i++)x[i+w.length]=q[i]; sort (x); // for(i=0;i<x.length;i++)System.out.print(x[i]); int cont=1; int max=0; char val=0; for (i=0;i<x.length-1;i++){ if(x[i]==x[i+1]){ cont++; if(cont>max){max=cont; val=x[i];} } else cont=1; } System.out.print(val); } private static void sort(char a[]){ int i; for(i=0;i<a.length-1;i++) for(int j=i;j<a.length;j++) if(a[i]>a[j]){char aux=a[i]; a[i]=a[j]; a[j]=aux; } } } class comun{ public static void main(String args[]){ final int N=100; String s1="abccd"; String s2="abbdccga"; char c1[]; char c2[]; c1=s1.toCharArray(); c2=s2.toCharArray(); int i;int j; char test[]=new char[N]; char test1[]=new char[N]; int k=0; int p=0; for(i=0;i<c1.length;i++){ boolean esteComun=false; for(j=0;j<c2.length;j++){ if(c1[i]==c2[j]){ esteComun=true; break; } } if(esteComun==true){ test[k]=c1[i]; k++; } } char w[] = new char[k]; System.arraycopy(test, 0, w, 0, k); for(i=0;i<c2.length;i++){ boolean esteComun=false; for(j=0;j<c1.length;j++){ if(c2[i]==c1[j]){ esteComun=true; break; } } if(esteComun==true){ test1[p]=c2[i]; p++; } } char q[] = new char[p]; System.arraycopy(test1, 0, q, 0, p); char x[] = new char[k+p]; for(i=0;i<w.length;i++)x[i]=w[i]; for(i=0;i<q.length;i++)x[i+w.length]=q[i]; sort (x); // for(i=0;i<x.length;i++)System.out.print(x[i]); int cont=1; int max=0; char val=0; for (i=0;i<x.length-1;i++){ if(x[i]==x[i+1]){ cont++; if(cont>max){max=cont; val=x[i];} } else cont=1; } System.out.print(val); } private static void sort(char a[]){ int i; for(i=0;i<a.length-1;i++) for(int j=i;j<a.length;j++) if(a[i]>a[j]){char aux=a[i]; a[i]=a[j]; a[j]=aux; } } } I think i did the same thig twice in the midle there ... but u get the picture... it works :banana: 10x for your help __________________ Don't take life too seriously, it's not permanent !

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Programming Forums > Application Development > Java

Could Be Simple