Answer / nikita

add the sum of elements in array a and then that of array
b.. subtract the sums of the two arrays a and b. ul get the
missing element

Answer / sumanth

firt do sum of n elements in first array a i.e., n(n+1)/2
next the sum of n-1 elements in second array b i.e., n(n-1)/2...
from diff of these 2 sums we will get the missing element in b

Answer / brijesh

the question says that we have to find the element which is
not there in array b but there in array a. So we can just
compare each element in a with all the elements in b and if
we don't find a match then that is the desired element.

Answer / vadivel t

#include<stdio.h>
#include<math.h>
void main()
{
int a1[5] = {2,5,3,1,4}, a2[4] = {1,2,4,5};
int i, n = 5, sum1= 0, sum2 = 0;
for(i = 0; i <= n-1; i++)
{
sum1 = sum1 + a1[i];
if(i <= n-2)
{
sum2 = sum2 + a2[i];
}
}
printf("MISSED NO IS: %d",(sum1 - sum2));
_getch();
}

Answer / rasika

The answers are only applicable to array of numbers.. what
if the array is of strings..?

Answer / ajay c.

Take a temp variable of datatype of the array element.
Place a[0] in temp.

now for each element in array a compare with all elements
of array b. if any element is not found in array b, that is
the missing element.

what i guess though is that the interviewer may be trying
to look for a short-cut method to achieve the above. So, i
googled it and found a very nice answer:
http://stackoverflow.com/questions/1235033/java-comparing-
two-string-arrays-and-removing-elements-that-exist-in-both-
arrays

Answer / furquan

@Rasika : It will work even for array of characters i.e.
string. It will calculate on the ascii value.

int main()
{
char a[] = "abcghs", b[] = "abchs";
int sum = 0,i;

int n = strlen(a);
for (i=0;i<n-1;i++){
sum += a[i] - b[i];
}
sum += a[i];

printf("%c\n",sum);
return 0;
}

Answer / venu

use XOR association operation.
I mean: a = b^a^b = a^b^b = a;[bec a^a =0 always]

int main()
{
int a[3] = {3,4,5};
int b[4] = {7,4,5,3};
int r,i;
int n=4;

r1 = b[3];

for(i=0;i<n-1;i++)
{
r1 = r1^a[i]^b[i];
}

printf("number = %d \n",r);
}

Answer / ajnr

if and only if it is a sorted array which consists of nos
and the elements compare have the same index no. then the
element missing in array 'b' will obviously be the 'nth'
element of array 'a'...... but this is true if the above
conditions are satisfied

Answer / reshma

class My
{

public static void main(String[] args) {

int a1[] = {7,6,3,1,4};
int a2[] = {4,3,7,6};

int sum1= 0;
int sum2 = 0;
for(int i=0 ;i<a1.length;i++)
{
sum1 = sum1 + a1[i];
}

for(int j=0;j<a2.length;j++)
{
sum2 = sum2 + a2[j];

}
System.out.println("MISSED NO IS: "+(sum1 - sum2));
}
}

prog for 1st five prime numbers in 2^x - 1

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