Given a large number in form of string **str**. The task is to find the smallest odd number whose sum of digits is even by removing zero or more characters from the given string **str**, where the digits can be rearranged.

**Examples**

Input:str = “15470”

Output:15

Explanation:

Two smallest odd digits are 1 & 5. Hence the required number is 15.

Input:str = “124”

Output:-1

Explanation:

There is no smallest odd digit other than 1. Hence the required number can’t be form.

**Approach:**

On observing closely, by intuition, it can be understood that the number of digits in the smallest odd number possible is 2. And every digit in this number is odd because the sum of two odd digits is always even. Therefore, the idea to solve this problem is to iterate through the given string and store every odd number in an array. This array can be sorted and the first two digits together form the smallest odd number whose sum of its digits is even.

Below is the implementation of the above approach.

## C++

`// C++ program to find the smallest odd number ` `// with even sum of digits from the given number N ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the smallest odd number ` `// whose sum of digits is even from the given string ` `int` `smallest(string s) ` `{ ` ` ` `// Converting the given string ` ` ` `// to a list of digits ` ` ` `vector<` `int` `> a(s.length()); ` ` ` `for` `(` `int` `i = 0; i < s.length(); i++) ` ` ` `a[i] = s[i]-` `'0'` `; ` ` ` ` ` `// An empty array to store the digits ` ` ` `vector<` `int` `> b; ` ` ` ` ` `// For loop to iterate through each digit ` ` ` `for` `(` `int` `i = 0; i < a.size(); i++) ` ` ` `{ ` ` ` ` ` `// If the given digit is odd then ` ` ` `// the digit is appended to the array b ` ` ` `if` `((a[i]) % 2 != 0) ` ` ` `b.push_back(a[i]); ` ` ` `} ` ` ` ` ` `// Sorting the list of digits ` ` ` `sort(b.begin(),b.end()); ` ` ` ` ` `// If the size of the list is greater than 1 ` ` ` `// then a 2 digit smallest odd number is returned ` ` ` `// Since the sum of two odd digits is always even ` ` ` `if` `(b.size() > 1) ` ` ` `return` `(b[0])*10 + (b[1]); ` ` ` ` ` `// Else, -1 is returned ` ` ` `return` `-1; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `cout << (smallest(` `"15470"` `)); ` `} ` ` ` `// This code is contributed by Surendra_Gangwar ` |

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## Python

`# Python program to find the smallest odd number ` `# with even sum of digits from the given number N ` ` ` `# Function to find the smallest odd number ` `# whose sum of digits is even from the given string ` `def` `smallest(s): ` ` ` ` ` `# Converting the given string ` ` ` `# to a list of digits ` ` ` `a ` `=` `list` `(s) ` ` ` ` ` `# An empty array to store the digits ` ` ` `b ` `=` `[] ` ` ` ` ` `# For loop to iterate through each digit ` ` ` `for` `i ` `in` `range` `(` `len` `(a)): ` ` ` ` ` `# If the given digit is odd then ` ` ` `# the digit is appended to the array b ` ` ` `if` `(` `int` `(a[i])` `%` `2` `!` `=` `0` `): ` ` ` `b.append(a[i]) ` ` ` ` ` `# Sorting the list of digits ` ` ` `b ` `=` `sorted` `(b) ` ` ` ` ` `# If the size of the list is greater than 1 ` ` ` `# then a 2 digit smallest odd number is returned ` ` ` `# Since the sum of two odd digits is always even ` ` ` `if` `(` `len` `(b)>` `1` `): ` ` ` `return` `int` `(b[` `0` `])` `*` `10` `+` `int` `(b[` `1` `]) ` ` ` ` ` `# Else, -1 is returned ` ` ` `return` `-` `1` ` ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `print` `(smallest(` `"15470"` `)) ` |

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**Output:**

15

**Time Complexity:** O(N) where N = length of string.

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