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.

## C

#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(); }

## Java

// Java program to find missing Number class Main { // Function to ind missing number static 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 */ public static void main(String args[]) { int a[] = {1,2,4,5,6}; int miss = getMissingNo(a,5); System.out.println(miss); } }

## Python

# getMissingNo takes list as argument def getMissingNo(A): n = len(A) total = (n+1)*(n+2)/2 sum_of_A = sum(A) return total - sum_of_A # Driver program to test above function A = [1, 2, 4, 5, 6] miss = getMissingNo(A) print(miss) # This code is contributed by Pratik Chhajer

Output:

3

Time Complexity: O(n)

There can be overflow if n is large. In order to avoid Integer Overflow, we can pick one number from known numbers and subtract one number from given numbers. This way we wont have Integer Overflow ever. Thanks to Sahil Rally for suggesting this improvement.

**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.

## C

#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 elements in array */ int x2 = 1; /* For xor of all the elements 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(); }

## Java

// Java program to find missing Number // using xor class Main { // Function to find missing number static int getMissingNo (int a[], int n) { int i; int x1 = a[0]; int x2 = 1; /* For xor of all the elements in array */ for (i = 1; i< n; i++) x1 = x1^a[i]; /* For xor of all the elements from 1 to n+1 */ for ( i = 2; i <= n+1; i++) x2 = x2^i; return (x1^x2); } /* program to test above function */ public static void main(String args[]) { int a[] = {1,2,4,5,6}; int miss = getMissingNo(a,5); System.out.println(miss); } }

## Python

# getMissingNo takes list as argument def getMissingNo(A): n = len(A) x1 = A[0] # For xor of all the elements in array x2 = 1 # For xor of all the elements from 1 to n+1 for i in range(n): x1 ^= A[i] for i in range(n+2): x2 ^= i return x1^x2 # Driver program to test above function A = [1, 2, 4, 5, 6] miss = getMissingNo(A) print(miss) # This code is contributed by Pratik Chhajer

Output:

3

Time Complexity: O(n)

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