Given an array **arr[]** of size **10** where **arr[i]** represents the frequency of the digit **i**. The task is to find the smallest positive number which can not be represented by the given digits.**Examples:**

Input:arr[] = {2, 1, 1, 4, 0, 6, 3, 2, 2, 2}Output:4

Since the count of 4th digit is 0. So 4 can not be made.Input :arr[] = {0, 1, 1, 1, 1, 1, 1, 1, 1, 1}Output :10

Since the count of 0th digit is 0. So smallest positive integer that can not be made is 10.Input :arr[] = {2, 2, 1, 2, 1, 1, 3, 1, 1, 1}Output :22

To make 22 we require two 2’s but here the count of 2 is only 1.

**Approach: **

- Calculate the min value in the array starting from 1st index and also store it’s index.
- Now compare the min value with the value at 0th index.
- If its value at 0th index is less than the min value than either 10, 100, 1000 … can’t be expressed.
- If its value is greater than min value than iterate the loop up to first min value index and simply print it index value.

Below is the implementation of the above approach:

## C++

`// CPP program to find the smallest positive` `// number which can not be represented by given digits` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the smallest positive` `// number which can not be represented by given digits` `void` `expressDigit(` `int` `arr[], ` `int` `n){` ` ` ` ` `int` `min = 9, index = 0, temp = 0;` ` ` ` ` `// Storing the count of 0 digit` ` ` `// or store the value at 0th index` ` ` `temp = arr[0];` ` ` ` ` `// Calculates the min value in the array starting` ` ` `// from 1st index and also store it index.` ` ` `for` `(` `int` `i = 1; i < 10; i++){` ` ` ` ` `if` `(arr[i] < min){` ` ` `min = arr[i];` ` ` `index = i;` ` ` `}` ` ` `}` ` ` ` ` `// Now compare the min value with the` ` ` `// value at 0th index.` ` ` ` ` `// If its value at 0th index is less than min value` ` ` `// than either 10, 100, 1000 ... can't be expressed` ` ` `if` `(temp < min)` ` ` `{` ` ` `cout << 1;` ` ` `for` `(` `int` `i = 1; i <= temp + 1; i++)` ` ` `cout << 0;` ` ` `}` ` ` ` ` ` ` `// If it value is greater than min value than` ` ` `// iterate the loop upto first min value index` ` ` `// and simply print it index value.` ` ` `else` ` ` `{` ` ` `for` `(` `int` `i = 0; i < min; i++)` ` ` `cout << index;` ` ` ` ` `cout << index;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = {2, 2, 1, 2, 1, 1, 3, 1, 1, 1};` ` ` ` ` `// Value of N is always 10 as we take digit from 0 to 9` ` ` `int` `N = 10;` ` ` ` ` `// Calling function` ` ` `expressDigit(arr, N);` ` ` ` ` `return` `0;` ` ` `}` |

## Java

`// Java program to find the smallest positive` `// number which can not be represented by given digits` `import` `java.util.*;` `class` `GFG` `{` `// Function to find the smallest positive` `// number which can not be represented by given digits` `static` `void` `expressDigit(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `min = ` `9` `, index = ` `0` `, temp = ` `0` `;` ` ` ` ` `// Storing the count of 0 digit` ` ` `// or store the value at 0th index` ` ` `temp = arr[` `0` `];` ` ` ` ` `// Calculates the min value in the array starting` ` ` `// from 1st index and also store it index.` ` ` `for` `(` `int` `i = ` `1` `; i < ` `10` `; i++)` ` ` `{` ` ` `if` `(arr[i] < min)` ` ` `{` ` ` `min = arr[i];` ` ` `index = i;` ` ` `}` ` ` `}` ` ` ` ` `// Now compare the min value with the` ` ` `// value at 0th index.` ` ` ` ` `// If its value at 0th index is less than min value` ` ` `// than either 10, 100, 1000 ... can't be expressed` ` ` `if` `(temp < min)` ` ` `{` ` ` `System.out.print(` `1` `);` ` ` `for` `(` `int` `i = ` `1` `; i <= temp + ` `1` `; i++)` ` ` `System.out.print(` `0` `);` ` ` `}` ` ` ` ` `// If it value is greater than min value than` ` ` `// iterate the loop upto first min value index` ` ` `// and simply print it index value.` ` ` `else` ` ` `{` ` ` `for` `(` `int` `i = ` `0` `; i < min; i++)` ` ` `System.out.print(index);` ` ` ` ` `System.out.print(index);` ` ` `}` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = {` `2` `, ` `2` `, ` `1` `, ` `2` `, ` `1` `, ` `1` `, ` `3` `, ` `1` `, ` `1` `, ` `1` `};` ` ` ` ` `// Value of N is always 10` ` ` `// as we take digit from 0 to 9` ` ` `int` `N = ` `10` `;` ` ` ` ` `// Calling function` ` ` `expressDigit(arr, N);` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 program to find the` `# smallest positive number which` `# can not be represented by given digits` `# Function to find the smallest` `# positive number which can not be` `# represented by given digits` `def` `expressDigit(arr, n):` ` ` `min` `=` `9` ` ` `index ` `=` `0` ` ` `temp ` `=` `0` ` ` `# Storing the count of 0 digit` ` ` `# or store the value at 0th index` ` ` `temp ` `=` `arr[` `0` `]` ` ` `# Calculates the min value in` ` ` `# the array starting from` ` ` `# 1st index and also store its index.` ` ` `for` `i ` `in` `range` `(` `1` `, ` `10` `):` ` ` `if` `(arr[i] < ` `min` `):` ` ` `min` `=` `arr[i]` ` ` `index ` `=` `i` ` ` `# Now compare the min value with the` ` ` `# value at 0th index.` ` ` `# If its value at 0th index is` ` ` `# less than min value than either` ` ` `# 10, 100, 1000 ... can't be expressed` ` ` `if` `(temp < ` `min` `):` ` ` `print` `(` `1` `, end ` `=` `"")` ` ` `for` `i ` `in` `range` `(` `1` `, temp ` `+` `1` `):` ` ` `print` `(` `0` `, end ` `=` `"")` ` ` `# If its value is greater than` ` ` `# min value then iterate the loop` ` ` `# upto first min value index and` ` ` `# simply print its index value.` ` ` `else` `:` ` ` `for` `i ` `in` `range` `(` `min` `):` ` ` `print` `(index, end ` `=` `"")` ` ` `print` `(index)` `# Driver code` `arr ` `=` `[` `2` `, ` `2` `, ` `1` `, ` `2` `, ` `1` `,` ` ` `1` `, ` `3` `, ` `1` `, ` `1` `, ` `1` `]` `# Value of N is always 10` `# as we take digit from 0 to 9` `N ` `=` `10` `# Calling function` `expressDigit(arr, N)` `# This code is contributed by Mohit Kumar` |

## C#

`// C# program to find the smallest positive` `// number which can not be represented by given digits` `using` `System;` `class` `GFG` `{` `// Function to find the smallest positive` `// number which can not be represented by given digits` `static` `void` `expressDigit(` `int` `[]arr, ` `int` `n)` `{` ` ` `int` `min = 9, index = 0, temp = 0;` ` ` ` ` `// Storing the count of 0 digit` ` ` `// or store the value at 0th index` ` ` `temp = arr[0];` ` ` ` ` `// Calculates the min value in the array starting` ` ` `// from 1st index and also store it index.` ` ` `for` `(` `int` `i = 1; i < 10; i++)` ` ` `{` ` ` `if` `(arr[i] < min)` ` ` `{` ` ` `min = arr[i];` ` ` `index = i;` ` ` `}` ` ` `}` ` ` ` ` `// Now compare the min value with the` ` ` `// value at 0th index.` ` ` ` ` `// If its value at 0th index is less than min value` ` ` `// than either 10, 100, 1000 ... can't be expressed` ` ` `if` `(temp < min)` ` ` `{` ` ` `Console.Write(1);` ` ` `for` `(` `int` `i = 1; i <= temp + 1; i++)` ` ` `Console.Write(0);` ` ` `}` ` ` ` ` `// If it value is greater than min value than` ` ` `// iterate the loop upto first min value index` ` ` `// and simply print it index value.` ` ` `else` ` ` `{` ` ` `for` `(` `int` `i = 0; i < min; i++)` ` ` `Console.Write(index);` ` ` ` ` `Console.Write(index);` ` ` `}` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[]arr = {2, 2, 1, 2, 1, 1, 3, 1, 1, 1};` ` ` ` ` `// Value of N is always 10` ` ` `// as we take digit from 0 to 9` ` ` `int` `N = 10;` ` ` ` ` `// Calling function` ` ` `expressDigit(arr, N);` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// javascript program to find the smallest positive` `// number which can not be represented by given digits` ` ` `// Function to find the smallest positive` ` ` `// number which can not be represented by given digits` ` ` `function` `expressDigit(arr, n)` ` ` `{` ` ` `var` `min = 9, index = 0, temp = 0;` ` ` `// Storing the count of 0 digit` ` ` `// or store the value at 0th index` ` ` `temp = arr[0];` ` ` `// Calculates the min value in the array starting` ` ` `// from 1st index and also store it index.` ` ` `for` `(i = 1; i < 10; i++)` ` ` `{` ` ` `if` `(arr[i] < min)` ` ` `{` ` ` `min = arr[i];` ` ` `index = i;` ` ` `}` ` ` `}` ` ` `// Now compare the min value with the` ` ` `// value at 0th index.` ` ` `// If its value at 0th index is less than min value` ` ` `// than either 10, 100, 1000 ... can't be expressed` ` ` `if` `(temp < min)` ` ` `{` ` ` `document.write(1);` ` ` `for` `(i = 1; i <= temp + 1; i++)` ` ` `document.write(0);` ` ` `}` ` ` `// If it value is greater than min value than` ` ` `// iterate the loop upto first min value index` ` ` `// and simply prvar it index value.` ` ` `else` `{` ` ` `for` `(i = 0; i < min; i++)` ` ` `document.write(index);` ` ` `document.write(index);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `var` `arr = [ 2, 2, 1, 2, 1, 1, 3, 1, 1, 1 ];` ` ` `// Value of N is always 10` ` ` `// as we take digit from 0 to 9` ` ` `var` `N = 10;` ` ` `// Calling function` ` ` `expressDigit(arr, N);` `// This code is contributed by gauravrajput1` `</script>` |

**Output:**

22

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