Find the missing element in the array

Suppose we have an array A containing elements from 1 to n. An element is made zero and the resulting array is called B. Find the element that was made zero. Consider all the test cases and boundary conditions. Try to post multiple solutions. (This question was asked in Adobe written exam)

# include<iostream>
using namespace std;
main()
{int n,i,s1=0,s2=0;
cin>>n;
int a[n];
for(i=0;i<n;++i)
{cin>>a[i];
s2+=a[i];
}
s1=(n*(n+1))/2;
cout<<s1-s2;
system("pause");
}

for i <--- 1 to n
{do C[i] <- 0
}
for j <--- 1 to n
{do C[A[j]] <--- C[A[j]] + 1
}
for i <--- 1 to n
{if C[i] == 0
then Break
}
outpu i

missing_element=0
for i=1 to n{
missing_element^=(i^a[i]);
}
print missing element

How come this is a faster version???

faster not in terms of asymptotic notation but it will take less time as
bitwise operations are faster than addition/subtraction are faster than multiply/divide are faster than comparison.

What if n(n+1)/2 exceeds the limit of integer ???

Re think of your solution and try to give algo of the solution

ankurgutpa wrote:missing_element=0
for i=1 to n{
missing_element^=(i^a[i]);
}
print missing element

Too Good..

ya der was one problem.....the 0 wud hav created the problem.....
thanx.....
The algo tat'll work


for i <--- 0 to n
{do C[i] <- 0
}
for j <--- 1 to n
{do C[A[j]] <--- C[A[j]] + 1
}
for i <--- 0 to n
{if C[i] == 0
then Break
}
output i

question is same but try to do it by following constraint
you can only fetch jth bit of an element of array(assume in constant time) and nothing else
O(nlogn) solution exist for this problem but try to do it in O(n) ( as stated in CLRS 4-2) .