Smallest odd number with even sum of digits from the given number N

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++

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// 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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Java

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// Java program to find the smallest
// odd number with even sum of digits
// from the given number N 
import java.util.*; 
class GFG{
      
// Function to find the smallest 
// odd number whose sum of digits
// is even from the given string 
public static int smallest(String s) 
      
    // Converting the given string 
    // to a list of digits 
    int[] a = new int[s.length()];
    for(int i = 0; i < s.length(); i++) 
        a[i] = s.charAt(i) - '0';
      
    // An empty array to store the digits 
    Vector<Integer> b = new Vector<Integer>(); 
      
    // For loop to iterate through each digit 
    for(int i = 0; i < a.length; i++) 
    
          
        // If the given digit is odd 
        // then the digit is appended
        // to the array b 
        if(a[i] % 2 != 0
            b.add(a[i]); 
    
              
    // Sorting the list of digits 
    Collections.sort(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(b.size() > 1
        return (b.get(0)) * 10 + (b.get(1)); 
      
    // Else, -1 is returned 
    return -1
  
// Driver code
public static void main(String[] args)
{
    System.out.print(smallest("15470"));
}
}
  
// This code is contributed by divyeshrabadiya07

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Python

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# 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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