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<a.length;i++){
        x=x^a[i];
    }
    for(int i=1;i<=B;i++){
        x=x^i;
    }
    x = x &(~(x-1));
    int p=0, q=0;
    for(int i=0;i<a.length;i++){
        if((a[i] & x) == x){
            p=p^a[i];
        }
        else{
            q=q^a[i];
        }   
    }
    for(int i=1;i<=B;i++){
        if((i & x) == x){
            p=p^i;
        }
        else{
            q=q^i;
        }
    }

    System.out.println("p: "+p+" : "+q);
}

Answer 2

Let x and y be the roots of a quadratic equation.

• Sum of the roots, SUM = x + y
• Product of the roots, PRODUCT = x * y

If we know the sum and the product, we can reconstruct the quadratic equation as:

z^2 - (SUM)z + (PRODUCT) = 0

In the algorithm you mentioned, we find the sum and the product, and from that, we reconstruct the quadratic equation using the above formula.

If you are interested in a detailed derivation, here is a reference. Read "Reconstruction of the quadratic equation from the sum and product of roots".

Answer 3

public class MissingNumber{

static int[] array = { 1, 3, 5 };

public static void getMissingNumber() {

for (int i = 0; i < array.length; i++)
    System.out.println(array[i] + " ");

System.out.println("The Missing Number is:");
 int j = 0;
for (int i = 1; i <= 5; i++) {
    if (j < array.length && i == array[j])
    j++;
    else
    System.out.println(i + " ");
}

}
public static void main(String[] args) {
getMissingNumber();
}

}

Answer 4

Below is the generic answer in java code for any number of missing numbers in a given array
//assumes that there are no duplicates
a = [1,2,3,4,5]
b = [1,2,5]
a-b=[3,4]

public list<integer> find(int[] input){
  int[] a= new int[] {1,2,3,4,5};//create a new array without missing numbers
  List<Integer> l = new ArrayList<Integer>();//list for missing numbers
  Map<Integer,Integer> m = new HashMap<Integer>();

  //populate a hashmap with the elements in the new array
  for(int i=0;i<input.length;i++){  
   m.put(input[i], 1);
  }
//loop through the given input array and check for missing numbers
 for(int i=0;i<a.length;i++){
  if (!m.contains(input[i]))
   l.add(input[i]);
}
 return l;
}

Answer 5

I have an algorithm for above problem.

Suppose

Actual Series: 1 2 3 4 5 6          a:sum=21 product= 720
Missing Number series: 1 * 3 4 * 6  b:sum=14 product= 72

a+b=21-14= 7 , ab=720/72=10

Now we need to find a-b= sqrt[(a+b)^2 -4ab].

If you calculate:

a-b= 3

Now

a+b=7
a-b=3

Add both equations:

2a=10, a=5

then b=7-5=2 so, 2 and 5 are missing.

Answer 6

#include <stdio.h>
#include <math.h>

/*
    the sum should be 1+...+n = n(n+1)/2
    the sum of squares should be 1^2+...+n^2 = n(n+1)(2n+1)/6.
*/

void find_missing_2_numbers(int *arr, int n);

int main()
{
    int arr[] = {3,7,1,6,8,5};

    find_missing_2_numbers(arr, 8);

    return 0;
}

void find_missing_2_numbers(int *arr, int n)
{

    int i, size, a, b, sum, sum_of_sqr, a_p_b, as_p_bs, a_i_b, a_m_b;
    size = n - 2;

    sum = 0;
    sum_of_sqr = 0;
    for (i = 0; i < size; i++)
    {
        sum += arr[i];
        sum_of_sqr += (arr[i] * arr[i]);
    }

    a_p_b = (n*(n+1))/2 - sum;
    as_p_bs = (n*(n+1)*(2 * n + 1))/6 - sum_of_sqr;
    a_i_b = ((a_p_b * a_p_b) - as_p_bs ) / 2;
    a_m_b = (int) sqrt((a_p_b * a_p_b) - 4 * a_i_b);
    a = (a_p_b + a_m_b) / 2;
    b = a_p_b - a;

    printf ("A: %d, B: %d\n", a, b);
}