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Minimum deletions to convert given integer to an odd number whose sum of digits is even | Set 2
  • Last Updated : 03 Feb, 2021

Given a positive integer N, the task is to find the minimum number of digits required to be removed to convert N to an odd number whose sum of digits is even. If it is impossible to do so, then print “Not Possible”.

Examples:

Input: N = 12345
Output: 1
Explanation:
Removing the digit 3 modifies N to 1245. 
Since the sum of digits of N is 14 and N is odd, the required output is 1.

Input: N = 222
Output: Not Possible

Approach: The idea is to use the fact that summation of even count of odd numbers results in an even number. Follow the steps below to solve the problem:



  • Count total odd and even digits present in the integer N and store them in variables, say Odd and Even.
  • If Odd = 0, then one cannot convert the integer to an odd integer by deleting any number of digits. Hence, print “Not Possible”.
  • Otherwise, if Odd = 1, then to make the sum of digits even, one will have to delete that odd digit, which results in an even number. Therefore, print “Not Possible”.
  • Now, count the number of digits to be deleted after the last occurrence of an odd digit and store it in a variable, say ans.
  • If Odd is odd, then increment the count of ans by one, as one will have to delete an odd digit to make the sum even.
  • Finally, print ans if none of the above cases satisfies.

Below is the implementation of the above approach:

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum count of digits
// required to be remove to make N odd and
// the sum of digits of N even
void minOperations(string& N)
{
 
    // Stores count of even digits
    int even = 0;
 
    // Stores count of odd digits
    int odd = 0;
 
    // Iterate over the digits of N
    for (auto it : N) {
 
        // If current digit is even
        if ((it - '0') % 2 == 0) {
 
            // Update even
            even++;
        }
 
        // Otherwise
        else {
 
            // Update odd
            odd++;
        }
    }
 
    // Base conditions
    if (odd == 0 || odd == 1) {
        cout << "Not Possible"
             << "\n";
    }
 
    else {
 
        // Stores count of digits required to be
        // removed to make N odd and the sum of
        // digits of N even
        int ans = 0;
 
        // Iterate over the digits of N
        for (auto it : N) {
 
            // If current digit is even
            if ((it - '0') % 2 == 0) {
 
                // Update ans
                ans++;
            }
 
            // Otherwise,
            else {
 
                // Update ans
                ans = 0;
            }
        }
 
        // If count of odd digits is odd
        if (odd % 2) {
 
            // Update ans
            ans++;
        }
 
        // Finally print the ans
        cout << ans << endl;
    }
}
 
// Driver code
int main()
{
 
    // Input string
    string N = "12345";
 
    // Function call
    minOperations(N);
}

Java




// Java implementation of the above approach
import java.util.*;
class GFG
{
 
// Function to find minimum count of digits
// required to be remove to make N odd and
// the sum of digits of N even
static void minOperations(String N)
{
 
    // Stores count of even digits
    int even = 0;
 
    // Stores count of odd digits
    int odd = 0;
 
    // Iterate over the digits of N
    for (int it : N.toCharArray())
    {
 
        // If current digit is even
        if ((it - '0') % 2 == 0)
        {
 
            // Update even
            even++;
        }
 
        // Otherwise
        else
        {
 
            // Update odd
            odd++;
        }
    }
 
    // Base conditions
    if (odd == 0 || odd == 1)
    {
        System.out.print("Not Possible"
            + "\n");
    }
 
    else
    {
 
        // Stores count of digits required to be
        // removed to make N odd and the sum of
        // digits of N even
        int ans = 0;
 
        // Iterate over the digits of N
        for (int it : N.toCharArray())
        {
 
            // If current digit is even
            if ((it - '0') % 2 == 0)
            {
 
                // Update ans
                ans++;
            }
 
            // Otherwise,
            else
            {
 
                // Update ans
                ans = 0;
            }
        }
 
        // If count of odd digits is odd
        if (odd % 2 != 0)
        {
 
            // Update ans
            ans++;
        }
 
        // Finally print the ans
        System.out.print(ans +"\n");
    }
}
 
// Driver code
public static void main(String[] args)
{
 
    // Input String
    String N = "12345";
 
    // Function call
    minOperations(N);
}
}
 
// This code is contributed by shikhasingrajput.

Python3




# Python implementation of the above approach
 
# Function to find minimum count of digits
# required to be remove to make N odd and
# the sum of digits of N even
def minOperations(N):
   
    # Stores count of even digits
    even = 0;
 
    # Stores count of odd digits
    odd = 0;
 
    # Iterate over the digits of N
    for it in  N:
 
        # If current digit is even
        if (int(ord(it) - ord('0')) % 2 == 0):
 
            # Update even
            even += 1;
 
        # Otherwise
        else:
 
            # Update odd
            odd += 1;
 
    # Base conditions
    if (odd == 0 or odd == 1):
        print("Not Possible" + "");
 
    else:
 
        # Stores count of digits required to be
        # removed to make N odd and the sum of
        # digits of N even
        ans = 0;
 
        # Iterate over the digits of N
        for it in N:
 
            # If current digit is even
            if (int(ord(it) - ord('0')) % 2 == 0):
 
                # Update ans
                ans += 1;
 
            # Otherwise,
            else:
 
                # Update ans
                ans = 0;
 
        # If count of odd digits is odd
        if (odd % 2 != 0):
           
            # Update ans
            ans += 1;
 
        # Finally prthe ans
        print(ans, end=" ");
 
# Driver code
if __name__ == '__main__':
   
    # Input String
    N = "12345";
 
    # Function call
    minOperations(N);
 
# This code is contributed by shikhasingrajput

C#




// C# implementation of the above approach
using System;
class GFG
{
 
// Function to find minimum count of digits
// required to be remove to make N odd and
// the sum of digits of N even
static void minOperations(String N)
{
 
    // Stores count of even digits
    int even = 0;
 
    // Stores count of odd digits
    int odd = 0;
 
    // Iterate over the digits of N
    foreach (int it in N.ToCharArray())
    {
 
        // If current digit is even
        if ((it - '0') % 2 == 0)
        {
 
            // Update even
            even++;
        }
 
        // Otherwise
        else
        {
 
            // Update odd
            odd++;
        }
    }
 
    // Base conditions
    if (odd == 0 || odd == 1)
    {
        Console.Write("Not Possible"
            + "\n");
    }
    else
    {
 
        // Stores count of digits required to be
        // removed to make N odd and the sum of
        // digits of N even
        int ans = 0;
 
        // Iterate over the digits of N
        foreach (int it in N.ToCharArray())
        {
 
            // If current digit is even
            if ((it - '0') % 2 == 0)
            {
 
                // Update ans
                ans++;
            }
 
            // Otherwise,
            else
            {
 
                // Update ans
                ans = 0;
            }
        }
 
        // If count of odd digits is odd
        if (odd % 2 != 0)
        {
 
            // Update ans
            ans++;
        }
 
        // Finally print the ans
        Console.Write(ans +"\n");
    }
}
 
// Driver code
public static void Main(String[] args)
{
 
    // Input String
    String N = "12345";
 
    // Function call
    minOperations(N);
}
}
 
// This code is contributed by shikhasingrajput
Output: 
1

 

Time Complexity: O(log10(N))
Auxiliary Space: O(1)

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