CareerCup

The output length is n(n+1)/2 = n + (n-1) + (n-2) + ... + 2 + 1.

int k = 0;
for(int i = 0 ; i < n(n+1) / 2; i++)
 for(int j = 0; j < n-k; j++)
  output[i++] = input[j];

- S July 29, 2010 | Flag Reply

I forgot to decrease k after the second for. {{{ int k = 0; for(int i = 0 ; i < n(n+1) / 2; i++) { for(int j = 0; j < n-k; j++) output[i++] = input[j]; k--; } - S July 29, 2010 | Flag Reply

Didnt follow the question :(

- Anonymous July 29, 2010 | Flag Reply

So you just repeat the word using the first character as the last character?

- Anonymous July 30, 2010 | Flag Reply

Recursion could be used on substring of the string until length == 0

- addict July 30, 2010 | Flag Reply

#include<iostream> #include<string>

void print(std::string input,int len)
{
if(len == 0) {
return;
}
else {
std::cout << input.substr(0,len);
return print(input,len-1);
}
}
int main()
{
std::string a = "max";

print(a,a.length());
std::cout << "\n";
return 0;
}

- Anonymous July 30, 2010 | Flag Reply

void StrEcho(const char * str, char * output)
{
   // assume that there is enough memory at output: n(n+1)/2+1, n=strlen(str)
   for (size_t n = strlen(str); n; output += n--)
      strncpy(output, str, n);
   *output = '\0';
}

- fiddler.g July 30, 2010 | Flag Reply

This shoudl work:
char *a="cat";
int size=3;
int itersize=3;
int k=0;
for (int i=0; i<size*(size+1)/2; i++)
{
cout<<a[k];
k++;
if (k>=itersize)
{
itersize--;
k=0;
}
}

- guest August 10, 2010 | Flag Reply

1. Get the length of the input string.
2. Put the string into a queue.
3. If Queue is not empty,
.....Extract, print, and insert the elements of queue upto length - 1 times.
...Else
.....EXIT
4. Extract next element. //it will be the last element
5. Decrement length, goto 3

- D August 15, 2010 | Flag Reply

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
        int i,len=0;
        char input[80];
        memset(input,0,(80));
        scanf("%s",input);

        len = strlen(input);

        while(len)
        {
                for(i=0;i<len;i++)
                        printf("%c",input[i]);
                len--;
        }
        printf("\n");
        return 0;
}

- D August 15, 2010 | Flag Reply

string echo(string str)
{
  if(str.length() == 1)
     return str;
  retrun str + echo(str.substr(0,str.length()-1));
}

- ahj August 26, 2010 | Flag Reply

Simple dudes...
Push the characters of the string into the stack.
Ex: C(1st) A(2nd) T(3rd)
Now print the entire stack from bottom, perform a pop and print again. continue this till the stack is empty.
tats it lol!....

- Sunil August 27, 2010 | Flag Reply

The above logic is using DS.
ELSE u can go with logic of "ahj". Its pretty simple in Java

- Sunil August 27, 2010 | Flag Reply

What about, something like this?

public static void printChain(String w){
        for(int i=w.length();i>0;i--){
            String sub=w.substring(0,i);
            System.out.print(sub);
        }
    }

- markos September 28, 2010 | Flag Reply

String print(String s, int end){
    if(end==1)
	return s.substring(0, end);
    else
        return s.substring(0, end) + print(s,end-1);
}	

String echo(String s){
     return print(s, s.length());
}

- arv October 11, 2010 | Flag Reply

Eyes-closed in Python: return inputstring + inputstring[0:2] + inputstring[0]

- Anon January 28, 2011 | Flag Reply

#In Perl
use strict;
my $word = shift;
my $out = '';
while($word)
{
    $out = $out.$word;
    chop $word;
}
print $out;

- Odi December 29, 2011 | Flag Reply

a c# recursive implemntation

public static string func(string str)
        {
            if (str == "")
            {
                return str;
            }
            return str + func(str.Substring(0, str.Length - 1));
        }

- shahar November 11, 2012 | Flag Reply

String input = "cat";
		int k = 0, N = input.length();
		
		for(int i = N; i > 0; i--)
			for(int j = 0; j < i; j++)
				System.out.print(input.substring(j, j+1));

- vkapko February 06, 2015 | Flag Reply