CareerCup

Add both arrays and substract them.

take min and max element of the second array and make array of make an array of (max-min) size u can also use bit array if max -min is ahuge number because of space complexity
then traverse the second array and according to each digit set bits on bit array after that traverse first array and in which, which bit not set that is a missing number ...

// assume a.length == b.length + 1;
// XOR all the elements in both a and b arrays.

int getMissingInteger(int[] a, int b[]) {
    int ret = 0;
    for (int i=0; i<a.length-1; i++) {
         ret ^= a[i];
         ret ^= b[i];
    }
    return ret ^= a[i];
}

1)hash or bit map
2)sum(array1)-sum(array2)=x-y, square(array1) - square(array2) = x^2 - y^2, with this we will get x and y

This is very easy because the foreign element in the second array is 0, it would have been harder if it wasn't.
So just compute the total of each array and substruct them, you will get exactly the missing number.

int[] sat1 = {5,15,2,20,30,40,8,1};
int[] sat2 = {2,20,15,30,1,40,0,8};
for(int i=0;i<sat2.length;i++){
int count = 0;
for(int j=0;j<sat1.length;j++){
if(sat2[i]==sat1[j]){

}else{
count++;
}
}
// System.out.print(count + " ");
if(count==8){
System.out.print(sat2[i]);
}
}

}

public static ArrayList<Integer> getMissingNumbers(int[] array1, int[] array2) {
		HashSet<Integer> set2 =  new HashSet<Integer>();
		for(int num:array2){
			set2.add(num);
		}
		ArrayList<Integer> missingNumbers = new ArrayList<Integer>();
		for(int num:array1){
			if (set2.add(num)) {
				missingNumbers.add(num);
			}
		}
		return missingNumbers;
	}

C = [x for x in B if x not in A]

This solution only works when 0 is the number which is introduced in place of missing numbers.

int getMissingNumber(int arr1[], int arr2[], int sz)
{
  int xor1=0;
  int xor2=0;
  for(int i = 0; i < sz; i++) {
       xor1 ^= arr1[i];
       xor2 ^= arr2[i];
  }
  return xor1^xor2;
}

public static int findMissing(int[] arr1,int[] arr2)
{
Set<Integer> set = new HashSet<Integer>();
for(int i : arr2)
{
set.add(i);
}
for(int i=0;i<arr1.length;i++)
{
if(!set.contains(arr1[i]))
{
return arr1[i];
}
}
return -1;
}

public static int findMissing(int[] arr1,int[] arr2)
	{
		Set<Integer> set = new HashSet<Integer>();
		for(int i : arr2)
		{
			set.add(i);
		}
		for(int i=0;i<arr1.length;i++)
		{
			if(!set.contains(arr1[i]))
			{
				return arr1[i];
			}
		}
		return -1;
	}

Employing 2 loops:

#include <p30fxxxx.h>
#include <stdint.h>
#include <stdio.h>
#include <string.h>
/*Write a program to find the missing element in second array(Array2):
Array1:
5 15 2 20 30 40 8 1
Array2:
2 20 15 30 1 40 0 8 */
5 0 */

volatile int ara[8] = {5, 15, 2, 20, 30, 40, 8, 1};
volatile int arb[8] = {2, 20, 15, 30, 1, 40, 0, 8};
volatile int x;

int main (void)
{
	int *ptr;
	volatile int i, j, count; 
	for (i=0; i<8; i++)
	{ 	count =0x0;
		for (j=0; j<8; j++)
		{ 	
			if (arb[i]== ara[j])
				count++; 
		}
		if (!count)
				printf ("Missing element is %d\n",arb[i]);
	}
	x++;
	return 0;
}