CareerCup

Here is my version of code, Please suggest if anyone has a better approach -

public static void main(String[] args) {
String input = "abc";
int length = input.length();
StringBuilder str = new StringBuilder();
for (int i = 0; i < length; i++) {
for (int j = 0; j < i+1; j++) {
str.append(input.charAt(i));
}
}
System.out.println(str.toString()+str.reverse().toString());
}

Here is the code:

char *pattern_convertion(char *str)
{
	char *ret = NULL;
	int l = strlen(str);
	if(l == 0) return NULL;
	if((ret = (char *)malloc(l*sizeof(char))) == NULL) return NULL;
	
	int c = 0, i,j;
	for(i = 0; i<l; i++)
	{
		for(j = 0; j<i+1; j++, c++)
			ret[c] = str[i];
	}

	for(i = l-1; i>=0; i--)
	{
		for(j = 0; j<=l; j--, c++)
			ret[c] = str[i];
	}
	str[c] = '\0';
	return str;

}

char *pattern_convertion(char *str)
{
	char *ret = NULL;
	int l = strlen(str);
	if(l == 0) return NULL;
	if((ret = (char *)malloc(l*sizeof(char))) == NULL) return NULL;
	
	int c = 0, i,j;
	for(i = 0; i<l; i++)
	{
		for(j = 0; j<i+1; j++, c++)
			ret[c] = str[i];
	}

	for(i = l-1; i>=0; i--)
	{
		for(j = 0; j<=l; j--, c++)
			ret[c] = str[i];
	}
	ret[c] = '\0';
	return ret;
}

is it better to use char array? m converting string to char array and then working with it.

public static void getOutput(String str)
	{
		System.out.println("\n Input="+str);
		char[] c=str.toCharArray();
		int k=0;	
		int i;
		for (i=0; i<c.length ; i++)
		{
		k=0;
		while(k<=i)
			{	
			System.out.print(c[i]);
			k++;
			}	
		}
	
		while(k>0)
		{
		k--;
	        i=k;	
		while(i>=0)
			{			
			System.out.print(c[k]);
			i--;
			}
		}
	}

Algo:
1.take a loop to traverse string from zero to string length.
2.one more loop inside first loop which will responsible for no. of times a particular char to be printed.
3.same concept with little variation to print remaining half string.
------------------------------
for(i=0;i<length;i++)
{
for(j=0;j<=i;j++)
{
system.out.println(str.charAt(i));
}
}
while(i!=0)
{
for(j=0;j<=i;j++)
{
system.out.println(str.charAt(i));
}
i--;
}

can anyone post a complete working solutioon with no mistakes

void construct_pattern(string& input, string& out)
{

string t;

for(int i=0; i<input.length(); i++)
{
for (int j=0; j<=i; j++)
{
out.push_back(input.at(i));
}
}

for(string::reverse_iterator it=out.rbegin(); it!= out.rend(); it++)
t.push_back(*it);

out = out + t;
}

Here is my recursive solution using C#. Guys please discuss the complexities

//call using : print("abc", "", 0, 0, 0, 12);
 void print(string input, string output, int cp, int op, int closeCount, int total) 
        {
            if (closeCount >= 12)
            {
                Console.WriteLine(output);
                return;
            }

            if (closeCount < total / 2)
            {
                if (cp == op)
                {
                    cp = cp + 1;
                    op = 0;
                }                
                print(input, output + input[cp - 1], cp, op + 1, closeCount + 1, total);
            }
            else
            {
                if (cp > 0 && cp == op )
                {
                    cp = cp - 1;
                    op = -1;
                }
                print(input, output + input[cp], cp, op + 1, closeCount + 1, total);
            }



        }

void pattern()
{
      int j,k=1;
      char  *c="abc";

      while(*c)
      {   
                 for(j=0;j<k;j++)
                 cout<<*c;
                 *c++;
                 k++;
      }
      *c--;
      k--;
      while(*c)
      {   
                 for(j=0;j<k;j++)
                 cout<<*c;
                 *c--;
                 k--;
      }
      }

char *str = "ABCD";
int nlen = strlen (str);
int j = 1, k = 0;
for(int i = 0; i < (nlen*2); i++)
{
j = 1;
if (i < nlen)
{
k++;
while (j <= k)
{
cout<<*(str+i);
j++;
}
}
else
{
while (j <= k)
{
cout<<*(str+k-1);
j++;
}
k--;
}
}

char *str = "ABCD";
	int nlen = strlen (str);
	int j = 1, k = 0;
	for(int i = 0; i < (nlen*2); i++)
	{
		j = 1;
		if (i < nlen)
		{
			k++;
			while (j <= k)
			{
				cout<<*(str+i);
				j++;
			}
			cout<<"-";
		}
		else
		{
			while (j <= k)
			{
				cout<<*(str+k-1);
				j++;
			}
			cout<<"-";
			k--;
		}
	}

public static void main(String[] args)
{
char[] chracters = args[0].toCharArray();

StringBuffer buffer = new StringBuffer();

for(int i=0, j=0; i<chracters.length; i++)
{
while(j <= i)
{
buffer.append(chracters[i]);
j++;
}

j=0;
}

System.out.println(buffer.toString());

System.out.println(buffer.toString() + buffer.reverse());

}//End of Main

Praveen have u teated it

public static void repetetivePalindrome(String x){
		char in [] = x.toCharArray();
		StringBuilder sb = new StringBuilder();
		for(int i=0 ; i<in.length; i++){
			for(int j=0;j<=i; j++){
				sb.append(in[i]);
			}
		}
		
		String result  = sb.toString() + sb.reverse().toString();
		System.out.println(result);

int main()
{
	string s = "abc";
	string s1, resStr;
	int i, j, k = 0 ;
	bool done = false;	
	for( i= 0; i < s.length(); ++i)
	{
		for( j = 0; j < i+1; ++j )
		{
			s1 += s[i];
		}
				
	}
	resStr += s1;
	resStr.append(strrev( const_cast<char*>( s1.c_str() )) );

	cout << resStr <<endl;	
	P;
	return 0;
}

package abc;

import java.util.Stack;

public class ABCSpecial {

public static void main(String[] args){
MyStack stack = new MyStack();
char[] myCharString = args[0].toCharArray();

int length = 0;
while(length!=myCharString.length){
stack.push(myCharString[length]);
length++;
}

while(length!=0){
stack.pop();
length--;
}
}

}

class MyStack extends Stack<Character>{

static int i = 1;

@Override
public Character push(Character obj){
Character myChar = super.push(obj);
for(int y=0;y<i;y++){
System.out.print(myChar);
}
i++;
return myChar;
}

@Override
public Character pop(){
Character myChar = super.pop();
for(int y=1;y<i;y++){
System.out.print(myChar);
}
i--;
return myChar;
}
}

Here we can do it in a very simple way, because there is a simple recurring pattern .
ie, first letter 1 time second 2 time third 3 time... nth n time , nth n time, n-1 th n-1 time,..
3rd 3 time 2 nd 2 time 1 st 1 time .
so the code snip will be like this,

int  main ()
{
char str[] = " abcd";
int l = strlen(str);
int j;
for ( int i =0; i<l; i++)
{
  j=0;
   while (j<=i) { cout << str[i]; j++}
  }
for ( int i=l-1;i>=0;i--)
{
 j=0;
while(j>=i)
{  cout<<str[i]; j++;}

}
 this  has  the time  complexity O(n)

}
}

#include<iostream>
#include<cstring>
 using namespace std;
 
int main()
 {
 	char a[10];
 	int j;
 	int i;
 	int flag=0;
 	
 	cout<<"Enter the input\n";
 	gets(a);
 	
 	for(i=0; a[i]!='\0'; ++i)
 	{
 	  j=0;
 	 
 	 while(j<=i)
 	 {
 	 	cout<<a[i];
 	 	
 	 	j++; 	 	
 	 
			  }
		}
			  
		for(;i>=0;--i)
		{
			j=0;
			
			while(j<=i)
			{
				cout<<a[i];
				j++;
			}
		}
		  }

#include <bits/stdc++.h>

using namespace std;

int main() {
string str;
cin>>str;
string newSt="";
int l=str.length();
for(int i=0;i<l;i++)
{
newSt+=string(i+1,str[i]);
}
string rev=newSt;
reverse(rev.begin(),rev.end());
newSt+=rev;
cout<<newSt;
return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main() {
string str;
cin>>str;
string newSt="";
int l=str.length();
for(int i=0;i<l;i++)
{
newSt+=string(i+1,str[i]);
}
string rev=newSt;
reverse(rev.begin(),rev.end());
newSt+=rev;
cout<<newSt;
return 0;
}