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Assuming your sequence start at 1.

int correct_sum = n*(n-1)/2;
int actual_sum = 0;
for (int i =1; i<=n;i++)
actual_sum += i;
int missing num = correct_sum - actual_sum
Time complexity: O(n) <- for the calculating the real sum.

If it does not start at 1 then might need some extra calculation.
i.e. sequence: 6,7,8,9
5+1, 5+2, 5+3,5+4 => 5*4 + (1+2+3+4) => 30.

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If the array is sorted, simply run a binary search of log(n) time.
Otherwise, the best run time has to be linear.

Sum Solution (might get integer overflow):
1) Get the sum of all numbers n * (n + 1) / 2
2) Subtract all the numbers in array from sum and you will get the missing number.

int getMissingNo (int a[], int n)
{
    int i, total;
    total  = (n+1)*(n+2)/2;   
    for ( i = 0; i< n; i++)
       total -= a[i];
    return total;
}

Bit Manipulation Solution:
1) XOR all the array elements, let the result of XOR be X1.
2) XOR all numbers from 1 to n, let XOR be X2.
3) XOR of X1 and X2 gives the missing number.

int getMissingNo(int a[], int n)
{
    int i;
    int x1 = a[0]; 
    int x2 = 1;
     
    for (i = 1; i< n; i++)
        x1 = x1^a[i];
            
    for ( i = 2; i <= n+1; i++)
        x2 = x2^i;         
    
    return (x1^x2);
}