Find The Missing Number

FIND THE MISSING NUMBER

You are given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in list. One of the integers is missing in the list. Write an efficient code to find the missing integer.

Example:
I/P    [1, 2, 4, ,6, 3, 7, 8]
O/P    5

METHOD 1(Use sum formula)
Algorithm:

1. Get the sum of numbers 
       total = n*(n+1)/2
2  Subtract all the numbers from sum and
   you will get the missing number.

Code:





#include<stdio.h>



/* getMissingNo takes array and size of array as arguments*/

int getMissingNo (int a[], int n)

{

    int i, total;

    total  = (n+1)*(n+2)/2;  

    for ( i = 0; i< n; i++)

       total -= a[i];

    return total;

}



/*program to test above function */

int main()

{

    int a[] = {1,2,4,5,6};

    int miss = getMissingNo(a,5);

    printf("%d", miss);

    getchar();

}






Time Complexity: O(n)





METHOD 2(Use XOR)
  1) XOR all the array elements, let the result of XOR be X1.
  2) XOR all numbers from 1 to n, let XOR be X2.
  3) XOR of X1 and X2 gives the missing number.

Code:






#include<stdio.h>



/* getMissingNo takes array and size of array as arguments*/

int getMissingNo(int a[], int n)

{

    int i;

    int x1 = a[0]; /* For xor of all the elemets in arary */

    int x2 = 1; /* For xor of all the elemets from 1 to n+1 */

     

    for (i = 1; i< n; i++)

        x1 = x1^a[i];

            

    for ( i = 2; i <= n+1; i++)

        x2 = x2^i;        

    

    return (x1^x2);

}



/*program to test above function */

int main()

{

    int a[] = {1, 2, 4, 5, 6};

    int miss = getMissingNo(a, 5);

    printf("%d", miss);

    getchar();

}






Time Complexity: O(n)