# Find the missing number in a sorted array of limited range

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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).

## C++

 `// C++ program to find missing Number in ` `// a sorted array of size n and distinct ` `// elements. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find missing number ` `int` `getMissingNo(``int` `a[], ``int` `n) ` `{ ` `    ``for` `(``int` `i=0; i

## Java

 `// Java program to find missing Number in ` `// a sorted array of size n and distinct ` `// elements. ` `class` `Main ` `{ ` `    ``// Function to find missing number ` `    ``static` `int` `getMissingNo(``int` `a[]) ` `    ``{ ` `        ``int` `n = a.length; ` `        ``for` `(``int` `i=``0``; i

## Python

 `# 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

 ` `

Output :

`3`

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