Find 2 missing numbers in an array of integers with two missing values

Question

How do you do this? The values are unsorted but are of [1..n] Example array [3,1,2,5,7,8]. Answer: 4, 6

I saw this solution in

Answer 1

This method is not advisable as it suffers from integer overflow problems. So use XOR method to find the two numbers, which is highly performant. If you are interested i can explain.

As per the request from @ordinary below, i am explaining the algorithm:

EDIT

Suppose the maximum element of the array a[] is B i.e. suppose a[]={1,2,4} and here 3 and 5 are not present in a[] so max element is B=5.

• xor all the elements of the array a to X
• xor all the elements from 1 to B to x
• find the left most bit set of x by x = x &(~(x-1));
• Now if a[i] ^ x == x than xor a[i] to p else xor with q
• Now for all k from 1 to B if k ^ x == x than xor with p else xor with q
• Now print p and q

proof:

Let a = {1,2,4} and B is 5 i.e. from 1 to 5 the missing numbers are 3 and 5

Once we XOR elements of a and the numbers from 1 to 5 we left with XOR of 3 and 5 i.e. x.

Now when we find the leftmost bit set of x it is nothing but the left most different bit among 3 and 5. (3--> 011, 5 --> 101 and x = 010 where x = 3 ^ 5)

After this we are trying to divide into two groups according to the bit set of x so the two groups will be:

p = 2 , 2 , 3 (all has the 2nd last bit set)

q = 1, 1, 4, 4, 5 (all has the 2nd last bit unset)

if we XOR the elements of p among themselves we will find 3 and similarly if we xor all the elements of q among themselves than we will get 5. Hence the answer.

code in java

public void findNumbers(int[] a, int B){
    int x=0;
    for(int i=0; i