## 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.

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