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.

Program:

#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.

#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)

In method 1, if the sum of the numbers goes beyond maximum allowed integer, then there can be integer overflow and we may not get correct answer. Method 2 has no such problems.