Find the one missing number in range

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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