Find the largest odd number in String

Given a string S, representing a large integer, the task is to return the largest-valued odd integer (as a string) that is a substring of the given string S.

Note: A substring is a contiguous sequence of characters within a string. A null string (“”) is also a substring.

Examples:

Input: S = “504”
Output: “5”
Explanation: The only substring “5” is an odd number.

Input: S = “2042”
Output: “”
Explanation: All the possible non-empty substrings have even values.

Approach: To solve the problem follow the below idea:

The idea is to check odd number from the last in the string just to return the largest odd number.

Follow the steps to solve the problem:

• Iterate through i = s.length – 1 till 0:
  • Check if s[i] is odd then return the substring from 0 to i+1.
• Return an empty string in case of no odd number.

Below is the implementation for the above approach:

C++

// C++ code for the above approach:
#include <iostream>
using namespace std;

string maxOdd(string s)
{

    for (int i = s.length() - 1; i >= 0; i--) {

        if (s[i] % 2 != 0) {

            string s1 = s.substr(0, i + 1);

            return s1;
        }
    }

    return "";
}

// Drivers code
int main()
{

    string s = "504";
    string ans = maxOdd(s);

    // Function Call
    cout << ans;
    return 0;
}

Java

// Java code for the above approach:
import java.io.*;

class GFG {

  // returns a substring that contains the maximum odd number at its end
  public static String maxOdd(String s)
  {
    // Loop through the string backwards, starting from the end
    for (int i = s.length() - 1; i >= 0; i--)
    {

      // Check if the current character is odd
      if (s.charAt(i) % 2 != 0)
      {

        // If it is, return the substring that contains
        // all the characters up to and including the current character
        String s1 = s.substring(0, i + 1);
        return s1;
      }
    }

    // If no odd number is found in the string, return an empty string
    return "";
  }

  // driver functiom
  public static void main(String[] args)
  {

    String s = "504";

    // function call
    String ans = maxOdd(s);
    System.out.println(ans);
  }
}

Python

def max_odd(s):
    for i in range(len(s) - 1, -1, -1):
        if int(s[i]) % 2 != 0:
            return s[:i+1]
    return ""

# Driver code
s = "504"
ans = max_odd(s)

# Function call
print(ans)

C#

// C# code for the above approach:
using System;

class GFG {

// returns a substring that contains the maximum odd number at its end
public static string maxOdd(string s)
{
    // Loop through the string backwards, starting from the end
    for (int i = s.Length - 1; i >= 0; i--)
    {

    // Check if the current character is odd
    if (s[i] % 2 != 0)
    {

        // If it is, return the substring that contains
        // all the characters up to and including the current character
        string s1 = s.Substring(0, i + 1);
        return s1;
    }
    }

    // If no odd number is found in the string, return an empty string
    return "";
}

// driver functiom
public static void Main()
{

    string s = "504";

    // function call
    string ans = maxOdd(s);
    Console.WriteLine(ans);
}
}

// This code is contributed by Pushpesh Raj

Javascript

// Javascript code for the above approach

// returns a substring that contains the maximum odd number at its end
function maxOdd(s) {

  // Loop through the string backwards, starting from the end
  for (let i = s.length - 1; i >= 0; i--) {

// Check if the current character is odd
if (parseInt(s.charAt(i)) % 2 !== 0) {

  // If it is, return the substring that contains
  // all the characters up to and including the current character
  return s.substring(0, i + 1);
}
  }

  // If no odd number is found in the string, return an empty string
  return '';
}

  // driver function

  const s = '504';
  // function call
  const ans = maxOdd(s);
  console.log(ans);

  // This code is contributed by Vaibhav Nandan.

Output

5

Time Complexity: O(|S|).
Auxiliary Space: O(1).