The solution that doesn't involve repetitive additions or maybe the n(n+1)/2 formula doesn't get to you at an interview time for instance.

You have to use an array of 4 ints (32 bits) or 2 ints (64 bits). Initialize the last int with (-1 & ~(1 << 31)) >> 3. (the bits that are above 100 are set to 1) Or you may set the bits above 100 using a for loop.

- Go through the array of numbers and set 1 for the bit position corresponding to the number (e.g. 71 would be set on the 3rd int on the 7th bit from left to right)
- Go through the array of 4 ints (32 bit version) or 2 ints(64 bit version)

```
public int MissingNumber(int a[])
{
int bits = sizeof(int) * 8;
int i = 0;
int no = 0;
while(a[i] == -1)//this means a[i]'s bits are all set to 1, the numbers is not inside this 32 numbers section
{
no += bits;
i++;
}
return no + bits - Math.Log(~a[i], 2);//apply NOT (~) operator to a[i] to invert all bits, and get a number with only one bit set (2 at the power of something)
}
```

Example: (32 bit version) lets say that the missing number is 58. That means that the 26th bit (left to right) of the second integer is set to 0.

The first int is -1 (all bits are set) so, we go ahead for the second one and add to "no" the number 32. The second int is different from -1 (a bit is not set) so, by applying the NOT (~) operator to the number we get 64. The possible numbers are 2 at the power x and we may compute x by using log on base 2; in this case we get log2(64) = 6 => 32 + 32 - 6 = 58.

Hope this helps.