Minimum operations required to make N even by reversing prefix of any length

Given an integer N, the task is to find the minimum number of operations required to make N even such that in one operation prefix of any length can be reversed. If it’s not possible to make N as even number print -1.

Examples:

Input: N = 376502
Output: 0
Explanation: 
– N is already divisible by 2, hence it is an even number so the total number of prefix swaps will be 0.

Input: N = 36543
Output: 2
Explanation: 
– N is not an even number so to make it even first swap the prefix of length 2. Now N=63543
– Now swap the prefix of length 5 and N=34536.

Approach:

There can be 2 cases.

• N is an even number
• N is an odd number.

Now,  

• If N is an even number, the number of minimum steps to make N an even number will be 0.
• If N is an odd number, there can be 3 cases.
  1. N does not contain any even digits. In this case, it is impossible to make N an even number so the answer will be -1.
  2. N contains even digits and the first digit of N is even. In this case, we can swap the whole number, so the minimum number of steps to make N an even number will be 1.
  3. N contains even digits and the first digit of N is not even. In this case, we can first swap the prefix till any even digit and then swap the whole number, so the minimum number of steps to make N an even number will be 2.

Follow the steps below to solve the problem:

• Initialize the string s[] as the string representation of N.
• Initialize the variable ans as -1 and len as the length of the string s[].
• If (s[len-1] – ‘0’)%2==0 the set ans as 0.
• Else, perform the following tasks:
  • If the first character of the string s[] is even then set ans as 1.
  • Iterate over the range [0, len) using the variable i and perform the following tasks:
    • If s[i] is even, then set ans as 2 and break.
• After performing the above steps, print the value of ans as the answer.

Below is the implementation for the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find minimum number
// of steps required to make N an even number
void MinSteps(int N)
{

    // Converting N into string
    string s = to_string(N);

    int ans = -1;

    // Number of digits in N
    int len = s.size();

    // If the number is even
    if ((s[len - 1] - '0') % 2 == 0) {
        ans = 0;
    }
    else {

        // If the first digit is even
        if ((s[0] - '0') % 2 == 0) {
            ans = 1;
        }
        else {

            // Checking if s contains
            // any even digits
            for (int i = 0; i < len; i++) {
                if ((s[i] - '0') % 2 == 0) {
                    ans = 2;
                    break;
                }
            }
        }
    }

    // Printing the minimum number
    // of steps to make N an even number
    cout << ans << "\n";
}

// Driver Code
int main()
{
    int N;
    N = 36543;

    MinSteps(N);
}

Java

// Java code for the above approach
import java.util.*;
class GFG
{

  // Function to find minimum number
  // of steps required to make N an even number
  static void MinSteps(int N)
  {

    // Converting N into String
    String s = String.valueOf(N);
    int ans = -1;

    // Number of digits in N
    int len = s.length();

    // If the number is even
    if ((s.charAt(len - 1) - '0') % 2 == 0) {
      ans = 0;
    }
    else {

      // If the first digit is even
      if ((s.charAt(0) - '0') % 2 == 0) {
        ans = 1;
      }
      else {

        // Checking if s contains
        // any even digits
        for (int i = 0; i < len; i++) {
          if ((s.charAt(i) - '0') % 2 == 0) {
            ans = 2;
            break;
          }
        }
      }
    }

    // Printing the minimum number
    // of steps to make N an even number
    System.out.print(ans+ "\n");
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N;
    N = 36543;

    MinSteps(N);
  }
}

// This code is contributed by 29AjayKumar

Python3

# python3 code for the above approach

# Function to find minimum number
# of steps required to make N an even number
def MinSteps(N):

    # Converting N into string
    s = str(N)
    ans = -1

    # Number of digits in N
    le = len(s)

    # If the number is even
    if ((ord(s[le - 1]) - ord('0')) % 2 == 0):
        ans = 0

    else:

        # If the first digit is even
        if ((ord(s[0]) - ord('0')) % 2 == 0):
            ans = 1

        else:

            # Checking if s contains
            # any even digits
            for i in range(0, le):
                if ((ord(s[i]) - ord('0')) % 2 == 0):
                    ans = 2
                    break

    # Printing the minimum number
    # of steps to make N an even number
    print(ans)

# Driver Code
if __name__ == "__main__":

    N = 36543

    MinSteps(N)

# This code is contributed by rakeshsahni

C#

// C# code for the above approach
using System;
class GFG
{

  // Function to find minimum number
  // of steps required to make N an even number
  static void MinSteps(int N)
  {

    // Converting N into String
    String s = Convert.ToString(N);
    int ans = -1;

    // Number of digits in N
    int len = s.Length;

    // If the number is even
    if ((s[len - 1] - '0') % 2 == 0)
    {
      ans = 0;
    }
    else
    {

      // If the first digit is even
      if ((s[0] - '0') % 2 == 0)
      {
        ans = 1;
      }
      else
      {

        // Checking if s contains
        // any even digits
        for (int i = 0; i < len; i++)
        {
          if ((s[i] - '0') % 2 == 0)
          {
            ans = 2;
            break;
          }
        }
      }
    }

    // Printing the minimum number
    // of steps to make N an even number
    Console.Write(ans + "\n");
  }

  // Driver Code
  public static void Main()
  {
    int N;
    N = 36543;

    MinSteps(N);
  }
}

// This code is contributed by Saurabh Jaiswal

Javascript

<script>

// JavaScript code for the above approach

// Function to find minimum number
// of steps required to make N an even number
const MinSteps = (N) => {

    // Converting N into string
    let s = N.toString();

    let ans = -1;

    // Number of digits in N
    let len = s.length;

    // If the number is even
    if ((s.charCodeAt(len - 1) -
       '0'.charCodeAt(0)) % 2 == 0)
    {
        ans = 0;
    }
    else
    {

        // If the first digit is even
        if ((s.charCodeAt(0) -
           '0'.charCodeAt(0)) % 2 == 0)
        {
            ans = 1;
        }
        else
        {

            // Checking if s contains
            // any even digits
            for(let i = 0; i < len; i++)
            {
                if ((s.charCodeAt(i) -
                   '0'.charCodeAt(0)) % 2 == 0)
                {
                    ans = 2;
                    break;
                }
            }
        }
    }

    // Printing the minimum number
    // of steps to make N an even number
    document.write(`${ans}<br/>`);
}

// Driver Code
let N = 36543;

MinSteps(N);

// This code is contributed by rakeshsahni

</script>

Output

2

Time Complexity: O(len), where len is the length of the string.
Auxiliary Space: O(len)