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Find the missing number in a sorted array of limited range

Difficulty Level : Basic
Last Updated : 16 Feb, 2023

Given a sorted array of size n and given that there are numbers from 1 to n+1 with one missing, the missing number is to be found. It may be assumed that array has distinct elements.

Examples:

Input : 1 3 4 5 6
Output : 2
Input : 1 2 3 4 5 7 8 9 10
Output : 6

We traverse all elements. For every element a[i], we check if it is equal to i+1 or not. If not, we return (i+1).

Implementation:

C++

// C++ program to find missing Number in
// a sorted array of size n and distinct
// elements.
#include<bits/stdc++.h>
using namespace std;
  
// Function to find missing number
int getMissingNo(int a[], int n)
{
    for (int i=0; i<n; i++)        
        if (a[i] != (i+1))
            return (i+1);
   
    // If all numbers from 1 to n
    // are present
    return n+1;
}
  
// Driver code
int main()
{
    int a[] = {1, 2, 4, 5, 6};
    int n = sizeof(a) / sizeof(a[0]);
    cout << getMissingNo(a, n);
    return 0;
}
  
// This code is contributed by Sachin Bisht

Java

// Java program to find missing Number in
// a sorted array of size n and distinct
// elements.
import java.io.*;
public class Main
{
    // Function to find missing number
    static int getMissingNo(int a[])
    {
        int n = a.length;
        for (int i=0; i<n; i++)        
            if (a[i] != (i+1))
                return (i+1);
  
        // If all numbers from 1 to n
        // are present
        return n+1;
    }
        
    /* program to test above function */
    public static void main(String args[])
    {
        int a[] = {1, 2, 4, 5, 6};
        System.out.println(getMissingNo(a));
    }
}

Python3

# Python program to find missing Number in
# a sorted array of size n and distinct
# elements.
  
# function to find missing number
def getMissingNo(a):
    n = len(a)
      
    for i in range(n):
        if(a[i] != i + 1):
            return i + 1
              
    # If all numbers from 1 to n
    # are present
    return n+1
  
# Driver code
a = [1, 2, 4, 5, 6]
print(getMissingNo(a))
  
# This code is contributed by Sachin Bisht

C#

// C# program to find missing Number 
// in a sorted array of size n and 
// distinct elements.
using System;
  
class GFG
{
      
// Function to find missing number
static int getMissingNo(int []a, int n)
{
    for (int i = 0; i < n; i++) 
        if (a[i] != (i + 1))
            return (i + 1);
  
    // If all numbers from
    // 1 to n are present
    return n + 1;
}
  
// Driver code
public static void Main()
{
    int []a = {1, 2, 4, 5, 6};
    int n = a.Length;
    Console.WriteLine(getMissingNo(a, n));
}
}
  
// This code is contributed by ihritik

PHP

<?php
// PHP program to find missing Number 
// in a sorted array of size n and
// distinct elements.
  
// Function to find missing number
function getMissingNo($a, $n)
{
    for ($i = 0; $i < $n; $i++) 
        if ($a[$i] != ($i + 1))
            return ($i + 1);
  
    // If all numbers from 
    // 1 to n are present
    return $n + 1;
}
  
// Driver code
$a = array(1, 2, 4, 5, 6);
$n = sizeof($a);
echo getMissingNo($a, $n);
  
// This code is contributed 
// by ihritik
?>

Javascript

<script>
// javascript program to find missing Number 
// in a sorted array of size n and 
// distinct elements.
      
// Function to find missing number
function getMissingNo(a,n)
{
    for (let i = 0; i < n; i++) 
        if (a[i] != (i + 1))
            return (i + 1);
  
    // If all numbers from
    // 1 to n are present
    return n + 1;
}
  
// Driver code
let a = [1, 2, 4, 5, 6]
let n = a.length;
document.write(getMissingNo(a, n));
  
// This code is contributed by mohit kumar 29.
</script>

Output

Time Complexity: O(N), Where N is the length of the given array.
Auxiliary Space: O(1)

Another Approach: (Use mathematical approach to solve this problem)

It is given, elements in the array are in the range of [1, n + 1], so first calculate the required total sum by adding all the numbers from 1 to n + 1, this can be calculated by using the formula of “sum of first N natural numbers” by sum = N*(N+1) / 2, where N is the first N natural numbers. After that subtract the required total sum by current sum of the given array. This will result in the missing number that are not present in the given array.