# Minimum number greater than the maximum of array which cannot be formed using the numbers in the array

Given an array arr[] of integers, the task is to find the minimum number greater than the maximum element from the array which cannot be formed using the numbers in the array (adding elements to form some other number). If no such element exists then print -1.

Examples:

Input: arr[] = {2, 6, 9}
Output: -1
There is no such number greater than 9
that cannot be formed using 2, 6 and 9.

Input: arr[] = {6, 7, 15}
Output: 16
16 is the smallest number greater than 15 that
cannot be formed using 6, 7 and 15.

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

Approach: The problem is similar to the minimum coin change problem with slight modification. First sort the array in ascending order and find the maximum element max that will be the number present at the last index of the array. Check for the numbers in the range (max, 2 * max) for the answer. If a number in this range cannot be formed using the elements of the array then that number is the answer, but if all the numbers can be formed in this range then there exists no such number that cannot be formed using the elements of the array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns the minimum ` `// number greater than maximum of the ` `// array that cannot be formed using the ` `// elements of the array ` `int` `findNumber(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Sort the given array ` `    ``sort(arr, arr + n); ` ` `  `    ``// Maximum number in the array ` `    ``int` `max = arr[n - 1]; ` ` `  `    ``// table[i] will store the minimum number of ` `    ``// elements from the array to form i ` `    ``int` `table[(2 * max) + 1]; ` ` `  `    ``table = 0; ` ` `  `    ``for` `(``int` `i = 1; i < (2 * max) + 1; i++) ` `        ``table[i] = INT_MAX; ` ` `  `    ``int` `ans = -1; ` ` `  `    ``// Calculate the minimum number of elements ` `    ``// from the array required to form ` `    ``// the numbers from 1 to (2 * max) ` `    ``for` `(``int` `i = 1; i < (2 * max) + 1; i++) { ` `        ``for` `(``int` `j = 0; j < n; j++) { ` `            ``if` `(arr[j] <= i) { ` `                ``int` `res = table[i - arr[j]]; ` `                ``if` `(res != INT_MAX && res + 1 < table[i]) ` `                    ``table[i] = res + 1; ` `            ``} ` `        ``} ` ` `  `        ``// If there exists a number greater than ` `        ``// the maximum element of the array that can be ` `        ``// formed using the numbers of array ` `        ``if` `(i > arr[n - 1] && table[i] == INT_MAX) { ` `            ``ans = i; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 7, 15 }; ` `    ``int` `n = (``sizeof``(arr) / ``sizeof``(``int``)); ` ` `  `    ``cout << findNumber(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.Arrays; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function that returns the minimum ` `// number greater than maximum of the ` `// array that cannot be formed using the ` `// elements of the array ` `static` `int` `findNumber(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Sort the given array ` `    ``Arrays.sort(arr); ` ` `  `    ``// Maximum number in the array ` `    ``int` `max = arr[n - ``1``]; ` ` `  `    ``// table[i] will store the minimum number of ` `    ``// elements from the array to form i ` `    ``int` `table[] = ``new` `int``[(``2` `* max) + ``1``]; ` ` `  `    ``table[``0``] = ``0``; ` ` `  `    ``for` `(``int` `i = ``1``; i < (``2` `* max) + ``1``; i++) ` `        ``table[i] = Integer.MAX_VALUE; ` ` `  `    ``int` `ans = -``1``; ` ` `  `    ``// Calculate the minimum number of elements ` `    ``// from the array required to form ` `    ``// the numbers from 1 to (2 * max) ` `    ``for` `(``int` `i = ``1``; i < (``2` `* max) + ``1``; i++) ` `    ``{ ` `        ``for` `(``int` `j = ``0``; j < n; j++) ` `        ``{ ` `            ``if` `(arr[j] <= i) ` `            ``{ ` `                ``int` `res = table[i - arr[j]]; ` `                ``if` `(res != Integer.MAX_VALUE && res + ``1` `< table[i]) ` `                    ``table[i] = res + ``1``; ` `            ``} ` `        ``} ` ` `  `        ``// If there exists a number greater than ` `        ``// the maximum element of the array that can be ` `        ``// formed using the numbers of array ` `        ``if` `(i > arr[n - ``1``] && table[i] == Integer.MAX_VALUE) ` `        ``{ ` `            ``ans = i; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = { ``6``, ``7``, ``15` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(findNumber(arr, n)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function that returns the minimum ` `# number greater than Maximum of the ` `# array that cannot be formed using the ` `# elements of the array ` `def` `findNumber(arr, n): ` ` `  `    ``# Sort the given array ` `    ``arr ``=` `sorted``(arr) ` ` `  `    ``# Maximum number in the array ` `    ``Max` `=` `arr[n ``-` `1``] ` ` `  `    ``# table[i] will store the minimum number of ` `    ``# elements from the array to form i ` `    ``table ``=` `[``10``*``*``9` `for` `i ``in` `range``((``2` `*` `Max``) ``+` `1``)] ` ` `  `    ``table[``0``] ``=` `0` ` `  `    ``ans ``=` `-``1` ` `  `    ``# Calculate the minimum number of elements ` `    ``# from the array required to form ` `    ``# the numbers from 1 to (2 * Max) ` `    ``for` `i ``in` `range``(``1``, ``2` `*` `Max` `+` `1``): ` `        ``for` `j ``in` `range``(n): ` `            ``if` `(arr[j] <``=` `i): ` `                ``res ``=` `table[i ``-` `arr[j]] ` `                ``if` `(res !``=` `10``*``*``9` `and` `res ``+` `1` `< table[i]): ` `                    ``table[i] ``=` `res ``+` `1` `             `  `        ``# If there exists a number greater than ` `        ``# the Maximum element of the array that can be ` `        ``# formed using the numbers of array ` `        ``if` `(i > arr[n ``-` `1``] ``and` `table[i] ``=``=` `10``*``*``9``): ` `            ``ans ``=` `i ` `            ``break` `         `  `    ``return` `ans ` ` `  `# Driver code ` `arr ``=` `[``6``, ``7``, ``15``] ` `n ``=` `len``(arr) ` ` `  `print``(findNumber(arr, n)) ` `  `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function that returns the minimum ` `// number greater than maximum of the ` `// array that cannot be formed using the ` `// elements of the array ` `static` `int` `findNumber(``int``[] arr, ``int` `n) ` `{ ` ` `  `    ``// Sort the given array ` `    ``Array.Sort(arr); ` ` `  `    ``// Maximum number in the array ` `    ``int` `max = arr[n - 1]; ` ` `  `    ``// table[i] will store the minimum number of ` `    ``// elements from the array to form i ` `    ``int``[] table = ``new` `int``[(2 * max) + 1]; ` ` `  `    ``table = 0; ` ` `  `    ``for` `(``int` `i = 1; i < (2 * max) + 1; i++) ` `        ``table[i] = ``int``.MaxValue; ` ` `  `    ``int` `ans = -1; ` ` `  `    ``// Calculate the minimum number of elements ` `    ``// from the array required to form ` `    ``// the numbers from 1 to (2 * max) ` `    ``for` `(``int` `i = 1; i < (2 * max) + 1; i++) ` `    ``{ ` `        ``for` `(``int` `j = 0; j < n; j++) ` `        ``{ ` `            ``if` `(arr[j] <= i) ` `            ``{ ` `                ``int` `res = table[i - arr[j]]; ` `                ``if` `(res != ``int``.MaxValue && res + 1 < table[i]) ` `                    ``table[i] = res + 1; ` `            ``} ` `        ``} ` ` `  `        ``// If there exists a number greater than ` `        ``// the maximum element of the array that can be ` `        ``// formed using the numbers of array ` `        ``if` `(i > arr[n - 1] && table[i] == ``int``.MaxValue) ` `        ``{ ` `            ``ans = i; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main()  ` `{ ` `    ``int``[] arr = { 6, 7, 15 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(findNumber(arr, n)); ` `} ` `} ` ` `  `/* This code contributed by Code_Mech */`

## PHP

 ` ``\$arr``[``\$n` `- 1] && ``\$table``[``\$i``] == PHP_INT_MAX) ` `        ``{ ` `            ``\$ans` `= ``\$i``; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``return` `\$ans``; ` `} ` ` `  `// Driver code ` `{ ` `    ``\$arr` `= ``array``(6, 7, 15 ); ` `    ``\$n` `= sizeof(``\$arr``); ` ` `  `    ``echo``(findNumber(``\$arr``, ``\$n``)); ` `} ` ` `  ` `  `/* This code contributed by Code_Mech*/`

Output:

```16
```

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