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result = greater number - smaller number

In the result check if there are consecutive zeros of length equal to the length of smaller number(from right) , if so return the index where the last zero is present in the result

- Rajesh January 23, 2013 | Flag Reply

Obviously cannot use library functions.

- sjosh.theonlyone January 21, 2013 | Flag Reply

In C++ :

int lengthOfNumber (int num) {
	
	int temp, i = 0;
	temp = num;
	
	while (temp > 0) {
		temp = temp/10;
		i++;
	}
	
	return i;
}

int returnIndex(int a, int b)  {
	
	int length_a = lengthOfNumber(a);
	int length_b = lengthOfNumber(b);
	int divisor = 10^length_a;
	int counter = 0;
	int rem = b, div = b;

	while (div != 0) {
		rem = rem%divisor;
		if (rem == a)
			return length_b - counter - length_a;
		div = div/10;
		counter++;
	}
	
	return -1; //a does not exist in b
}

- Maneet & Arjun January 21, 2013 | Flag Reply

divisor in while is constant
so in first itr
rem = 45
second
0
third 0
and so on.....

I am missing something here...?I didnt try it with a compiler..

PS: apologies for reposting....I dnt know what I am doing...time for coffee..

- sjosh.theonlyone January 21, 2013 | Flag Reply

#include <iostream>
#include <stdio.h>

using namespace std;

int main() {

    int numa = 126547;
    int numb = 654;
    int temp = numb;
    int index = 0;
    int traversal = 0;


    if (numa < numb) {
        temp = numa;
        numa = numb;
        numb = temp;
        temp = numb;
    }

    while (numa > 0) {
        if (temp > 0) {
            if (numa % 10 == temp % 10) {
                temp = temp / 10;
                if (temp == 0)
                    index = traversal;
            } else {
                temp = numb;
            }
        }
        numa = numa / 10;
        traversal++;
    }

    if (temp > 0)
        cout << " NOt found ";
    else
        cout << " Found at " << traversal - index;
}

- isandesh7 January 22, 2013 | Flag Reply

In JAVA :

public class Test1 {
    
    public static void main(String args[]) {
        
        int a = 12345;
        String sa = Integer.toString(a);
        int saLen = sa.length();
        
        int b = 34;
        String sb = Integer.toString(b);
        int sbLen = sb.length();
        
        int match;
        
        for( int i = 0 ; i < saLen ; i++ ) {
            
            match = 1;
            
            if( sa.charAt(i) == sb.charAt(0) ) {
            
                 for( int j = 0 ; j < sbLen ; j++ ) {
                     if( sa.charAt(i+j) != sb.charAt(j) ) {
                         match = 0;
                         break;
                     }
                 }
                 
                 if( match == 1 ) {
                     System.out.println(" Index Found at i : " + i);
                     System.exit(0);
                 }
            
            }
            
        }
        
    }
    
}

- Another Coder January 22, 2013 | Flag Reply

There is an algorithm for pattern matching, KMP algorithm which solves the problem in O(n) time

- Anonymous January 22, 2013 | Flag Reply

put the digits of num1 and num2 in 2 arrays number and pattern respectively. Apply KMP algorithm now. This will be O(m+n) time.

- Anonymous January 22, 2013 | Flag Reply

int NumberInNumber(int n1, int n2)
{
  int temp1 = n1;
  int index = 0;
  int temp2 = n2;
  
  while(temp1)
    {
      while(temp1%10 == temp2%10 && temp2)
	{
	  temp1/=10;
	  temp2/=10;
	}
      if(!temp2)
	{
	while(temp1)
	  {
	    index++;
	    temp1/=10;
	  }
	return index;
	}
      temp1/=10;
      temp2=n2;
    }
  
  return -1;
}

- Anonymous January 22, 2013 | Flag Reply

int intIndex(int a, int b)
{
   // assume a and b are positive, otherwise use abs
   int power = 1;
   int a_len = 1, b_len = 1;
   while (power < a)
   {
      a_len++;
      power *= 10;
   }

   int index = -1;
   for (int i = 0; b > 0; i++, b /= 10)
   {
      if (b % power == a)
      {
         index = i;
      }
      b_len++;
   }

   return (index > 0) ? (b_len - index - a_len) : -1;
}

- Anonymous January 22, 2013 | Flag Reply

#include <stdio.h> #if 0 /* logic here */ pattern: 12345 find: 23 count = 2 /* 23 has two integers */ so divide with 10*10 first iteration: 12345%100 = 45 second iteration: 1234%100 = 34 third ..... : 123%100 = 23 /* so found here */ #endif int main() {{{ int pattern = 456789101; int find = 10; int count = 0; int i, flag, divide_with = 1; /* count the integers in the find */ i = find; do {{{ i = i/10; count++; }}}while(i); /* get the 'count' number of integers from pattern */ i = pattern; while(count) {{{ divide_with = divide_with*10; count--; }}} flag = 0; do {{{ i = pattern%divide_with; pattern = pattern/10; if(i == find) {{{ flag = 1; break; }}} count++; }}}while(pattern); if(flag) printf("found\n"); else printf("not found\n"); return 0; }}} - anish[1] January 22, 2013 | Flag Reply

Thank you all. I did something similar. Cheers. :)

- sjosh.theonlyone January 22, 2013 | Flag Reply

Just correct the first answer.
BTW, the "^" symbol is not power in c++. It's bitwise XOR.

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

int length (int num) {

int temp, i = 0;
temp = num;

while (temp > 0) {
temp = temp/10;
i++;
}

return i;
}

int index(int a, int b) {

int len_a = length(a);
int len_b = length(b);
int divisor = static_cast<int>(pow(10,length_a));
int counter = 0;
int rem = b, div = b;

while (div != 0) {
int index = rem%divisor;
rem = rem/10;
if (index == a)
return len_b - counter - len_a;
div = div/10;
counter++;
}

return -1; //a does not exist in b
}

int main()
{
int a = 34, b = 12345;
int index = index(a, b);
printf("The index is: %d\n", index);
return 0;
}

- Anonymous January 22, 2013 | Flag Reply