if array a conatins 'n' elements and array b conatins 'n-1'
elements.array b has all element which are present in array
a but one element is missing in array b. find that
element.
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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
| Is This Answer Correct ? | 48 Yes | 1 No |
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
| Is This Answer Correct ? | 30 Yes | 11 No |
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.
| Is This Answer Correct ? | 9 Yes | 3 No |
#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();
}
| Is This Answer Correct ? | 8 Yes | 2 No |
Answer / rasika
The answers are only applicable to array of numbers.. what
if the array is of strings..?
| Is This Answer Correct ? | 5 Yes | 1 No |
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
| Is This Answer Correct ? | 4 Yes | 0 No |
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;
}
| Is This Answer Correct ? | 3 Yes | 1 No |
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);
}
| Is This Answer Correct ? | 1 Yes | 0 No |
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
| Is This Answer Correct ? | 1 Yes | 0 No |
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));
}
}
| Is This Answer Correct ? | 1 Yes | 1 No |
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