CareerCup

Use XOR

public class Missing_and_Duplicate {

/**
* @param args
*/
public static void main(String[] args)
{
// TODO Auto-generated method stub
int[] intArray2={2,4,7,5,3,1}; // 6 is missing
find_missing(intArray2);

}

public static void find_missing(int[] intArray)
{
int answer = intArray[0];
for(int i=1;i<intArray.length;i++)
{
answer = answer ^ intArray[i];
}
System.out.println("answer="+answer);
}

}

public class hello {
public static void main(String args[]){
int [] arr = {2,4,5,1,7,9,6,8,10,3,12,13, 11, 15}; // 1to15, 14 missing
findmissing(arr);
}

private static void findmissing(int[] arr) 
{
	int []temp = new int [arr.length+2];
	for(int data : arr)
   {
	temp[data]= 1;
   }
	
  for(int i = 1; i < temp.length; i++)
 { 
		if (temp[i]==0)
            {
		System.out.println(i);
	    }
 }

}
	
	
	
}

I don't like using a temp array, I was in a hurry to solve, further optimizations can be done.

Thanks

Sum all number in the array and subtract from (N*(N+1))/2 if 'N-1' is the size of the array.

we can use xor..:)

n(n+1)/2 - Sum of all elements in the array.

Mahesh is correct!

or keep diving every number by N!. Residue is the number which did not appear.

or keep dividing every number by N!. Residue is the number which did not appear.

or keep dividing every number from N!. Residue is the number which did not appear.

how can we use XOR

@sg

how can we use XOR ..?

//Elements missing in a range
int [] arr = {2,4,19, 18, 15,-15};
Arrays.sort(arr);
for (int i=arr[0],j=1;i<arr[arr.length-1];i++) {
if (i < arr[j-1]) {
System.out.println("Missing Element"+ i);
}
else {
j++;
}
}

Int overflow is a problem

The most efficient algorithm to find the missing number in the array is to use the concept of xor we will xor all the array elements and store them in one variable and we will xor all the numbers from 1 to n and store them in one variable and then next we will xor both of them which will return our missing element.