Given an array of size n. It is also given that range of numbers is from smallestNumber to smallestNumber + n where **smallestNumber** is the smallest number in array. The array contains number in this range but one number is missing so the task is to find this missing number.

Examples:

Input : arr[] = {13, 12, 11, 15} Output : 14 Input : arr[] = {33, 36, 35, 34}; Output : 37

The problem is very close to find missing number.

There are many approaches to solve this problem.

A **simple approach** is to first find minimum, then one by one search all elements. Time complexity of this approach is O(n*n)

A **better solution** is to sort the array. Then traverse the array and find the first element which is not present. Time complexity of this approach is O(n Log n)

The** best solution **is to first XOR all the numbers. Then XOR this result to all numbers from smallest number to n+smallestNumber. The XOR is our result.

Example:-

arr[n] = {13, 12, 11, 15} smallestNumber = 11 first find the xor of this array 13^12^11^15 = 5 Then find the XOR first number to first number + n 11^12^13^14^15 = 11; Then xor these two number's 5^11 = 14 // this is the missing number

## C++

// CPP program to find missing // number in a range. #include <bits/stdc++.h> using namespace std; // Find the missing number // in a range int missingNum(int arr[], int n) { int minvalue = *min_element(arr, arr+n); // here we xor of all the number int xornum = 0; for (int i = 0; i < n; i++) { xornum ^= (minvalue) ^ arr[i]; minvalue++; } // xor last number return xornum ^ minvalue; } // Driver code int main() { int arr[] = { 13, 12, 11, 15 }; int n = sizeof(arr)/sizeof(arr[0]); cout << missingNum(arr, n); return 0; }

## Java

// Java program to find // missing number in a range. import java.io.*; import java.util.*; class GFG { // Find the missing number in a range static int missingNum(int arr[], int n) { List<Integer> list = new ArrayList<>(arr.length); for (int i :arr) { list.add(Integer.valueOf(i)); } int minvalue = Collections.min(list);; // here we xor of all the number int xornum = 0; for (int i = 0; i < n; i++) { xornum ^= (minvalue) ^ arr[i]; minvalue++; } // xor last number return xornum ^ minvalue; } public static void main (String[] args) { int arr[] = { 13, 12, 11, 15 }; int n = arr.length; System.out.println(missingNum(arr, n)); } } //This code is contributed by Gitanjali.

## Python3

# python3 program to check # missingnumber in a range # Find the missing number # in a range def missingNum( arr, n): minvalue = min(arr) # here we xor of all the number xornum = 0 for i in range (0,n): xornum ^= (minvalue) ^ arr[i] minvalue = minvalue+1 # xor last number return xornum ^ minvalue # Driver method arr = [ 13, 12, 11, 15 ] n = len(arr) print (missingNum(arr, n)) # This code is contributed by Gitanjali.

Output:

14

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