How to find missing number in array - c++

Question

#include <iostream>
#include <algorithm>
using std::cin;
using std::cout;
using std::endl;
using std::sort;
int main()
{
int x = 0;
int n; // Enter Size of 2 array
cin >> n; // enter 5
long long *ptr1 = new long long[n - 1]; // size of array must be less than 5 by one n-1
for (int x = 0; x < n - 1; x++)
{
cin >> ptr1[x];
}
sort(ptr1, ptr1 + (n - 1));
for (int z = 1; z < n; z++)
{
if (z != ptr1[x])
{
cout << z;
break;
}
x++;
}
return 0;
}
You're given all positive integers from 1,2,…,n except one integer. Find the missing integer.
Input
The first line of input contains an integer n (2≤n≤2×105).
The second line of input contains n−1 distinct integers from 1 to n (inclusive).
Output
Print the missing integer.
when i try to sumbit this code i get wrong in test 10 but i don't know why! and he didn't show the test so what is wrong?

Answer 1

I have a three-fold answer:
This program leaks memory
You included <algorithm> please use it. (Look on cppreference)
Spoilers vector, iota, mismatch
Also, you don't reset x before the second loop.
That's never going to work unless the missing integer is the last of the array, and not equal to 1
// for (... z)
if (z != ptr1[x] /* Here */) {
// print 1, end loop OR invoke undefined behavior
}
x++; // Now x is equal to (n - 1)

Answer 2

There are some problems with your code:
long long *ptr1 = new long long[n - 1];
You call new, without delete. This will create a memory leak.
You haven't reinitialized x, so any access to ptr1[x] is out-of-bounds.
Slove all of that, and your code will look like:
#include <iostream>
#include <algorithm>
#include <vector>
int main()
{
int n;
cin >> n;
std::vector<int> vec(n-1); // std::vector instead
for (int x = 0; x < n - 1; x++)
{
std::cin >> vec[x];
}
std::sort(vec.begin(), vec.end());
int x = 0; // use another variable instead of reused the old one.
for (int z = 1; z < n; z++)
{
if (z != vec[x])
{
std::cout << z;
break;
}
x++;
}
return 0;
}
But, this isn't the best approach anyway. As #molbdnilo suggest:
#include <iostream>
int main()
{
int n;
std::cin >> n;
int sum{};
for (int i = 0; i < n - 1; ++i)
{
int tmp;
std::cin >> tmp;
sum += tmp;
}
std::cout << (n + 1) * n / 2 - sum;
}

Answer 3

You can solve this in a more straightforward way if you subtract the numbers that were entered from the expected sum of all numbers 1, 2, ..., n (see comments by #molbdnilo, #Aconcagua and #marcus-müller):
#include <iostream>
int main() {
std::size_t n;
std::cin >> n;
std::size_t sum{ n*(n + 1)/2 };
for (std::size_t idx = 1; idx != n; ++idx) {
std::size_t thisNumber;
std::cin >> thisNumber;
sum -= thisNumber;
}
std::cout << "Missing number: " << sum << std::endl;
}

Answer 4

Building blocks:
The expected sum: int expected_sum = n * (n + 1) / 2;
The sum of all integers fed to the test after you've read n:
#include <iterator> // istream_iterator
#include <numeric> // accumulate
int sum = std::accumulate(std::istream_iterator<int>(std::cin),
std::istream_iterator<int>{}, 0);
Now, expected_sum - sum should give you the missing value.