Number formed by deleting digits such that sum of the digits becomes even and the number odd

Given a non-negative number N, the task is to convert the number by deleting some digits of the number, such that the sum of the digits becomes even but the number is odd. In case there is no possible number, then print -1.

Note: There can be multiple numbers possible for a given N.

Examples:

Input: N = 18720
Output: 17
Explanation:
After Deleting 8, 2, 0 digits the number becomes 17 which is odd and the digit-sum is 8 which is even.

Input: N = 3
Output: -1
Explanation:
There is no possibility such that number becomes odd and the digit-sum is even.



Approach:
The idea is to use the fact that “Even number of odd digits will give the sum to even number”. So, If the digits in the number contain even count of odd digits then it is possible to convert the number otherwise converting such number is not possible.

Below is the implementation of the above approach.

C++

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// C++ implementation to convert
// a number into odd number such
// that digit-sum is odd
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to convert a number into
// odd number such that digit-sum is odd
void converthenumber(int n)
{
    string s = to_string(n);
    string res;
  
    // Loop to find any first two
    // odd number such that their
    // sum is even and number is odd
    for (int i = 0; i < s.length(); i++) {
        if (s[i] == '1' || s[i] == '3'
            || s[i] == '5' || s[i] == '7'
            || s[i] == '9')
            res += s[i];
        if (res.size() == 2)
            break;
    }
  
    // Print the result
    if (res.size() == 2)
        cout << res << endl;
    else
        cout << "-1" << endl;
}
  
// Driver Code
int main()
{
    int n = 18720;
    converthenumber(n);
  
    return 0;
}

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Java

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// Java implementation to convert 
// a number into odd number such 
// that digit-sum is odd 
import java.util.*;
import java.lang.*;
import java.io.*;
  
class Main 
  
// Function to convert a number into 
// odd number such that digit-sum is odd 
static void converthenumber(int n) 
    String s = Integer.toString(n); 
    String res = ""
  
    // Loop to find any first two 
    // odd number such that their 
    // sum is even and number is odd 
    for (int i = 0; i < s.length(); i++) 
    
        if (s.charAt(i) == '1' || s.charAt(i) == '3'
            || s.charAt(i) == '5' || s.charAt(i) == '7'
            || s.charAt(i) == '9'
            res += s.charAt(i); 
        if (res.length() == 2
            break
    
  
    // Print the result 
    if (res.length() == 2
        System.out.println(res); 
    else
        System.out.println(-1); 
  
// Driver code 
public static void main (String[] args) 
    int n = 18720
    converthenumber(n); 
  
// This code is contributed by Subhadeep Gupta

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Python3

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# Python3 implementation to convert 
# a number into odd number such 
# that digit-sum is odd 
  
# Function to convert a number into 
# odd number such that digit-sum is odd 
def converthenumber(n) :
    s = str(n);
    res = "";
      
    # Loop to find any first two
    # odd number such that their
    # sum is even and number is odd
    for i in range(len(s)) :
        if (s[i] == '1' or s[i] == '3'
        or s[i] == '5' or s[i] == '7'
        or s[i] == '9') :
            res += s[i];
              
        if (len(res) == 2) :
            break;
              
    # Print the result
    if (len(res) == 2) :
        print(res);
          
    else :
        print("-1"); 
  
# Driver Code 
if __name__ == "__main__"
  
    n = 18720
    converthenumber(n); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation to convert
// a number into odd number such
// that digit-sum is odd
using System; 
  
class GFG 
  
  
// Function to convert a number into
// odd number such that digit-sum is odd
static void converthenumber(int n)
{
    String s = n.ToString();
    String res = "";
  
    // Loop to find any first two
    // odd number such that their
    // sum is even and number is odd
    for (int i = 0; i < s.Length; i++)
    {
        if (s[i] == '1' || s[i] == '3'
            || s[i] == '5' || s[i] == '7'
            || s[i] == '9')
            res += s[i];
        if (res.Length == 2)
            break;
    }
  
    // Print the result
    if (res.Length == 2)
        Console.WriteLine(res);
    else
        Console.WriteLine(-1);
}
  
// Driver code 
public static void Main (String[] args) 
    int n = 18720;
    converthenumber(n);
  
// This code is contributed by Mohit kuamr 29

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Output:

17

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