Skip to content
Related Articles

Related Articles

Find largest valued even integer which is a non-empty substring of S

Improve Article
Save Article
Like Article
  • Last Updated : 31 Jan, 2022

Given a string S of size N, representing a large integer. The task is to find the largest valued even integer which is a non-empty substring of S. If no even integer can be made then return an empty string.

Examples:

Input: S = “4206”
Output: “4206”
Explanation: “4206” is already an even number.

Input: S = “23”
Output: “2”
Explanation: “The only non-empty substrings are “2”, “3”, and “23”. “2” is the only even number.

Input: S = “17”
Output: “”
Explanation: There is no even valued substring in the given string

 

Approach: The task can be solved by finding the first even character from the right, let’s say, it is found at an index ‘idx‘.
The resultant largest even valued non-empty substring would be the substring of S in the range [0, idx].

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the largest even valued
// substring
void get(string& s)
{
    int N = s.length();
    int idx = -1;
 
    // Finding the rightmost even character
    for (int i = N - 1; i >= 0; i--) {
        if ((s[i] - '0') % 2 == 0) {
            idx = i;
            break;
        }
    }
 
    if (idx == -1)
        cout << "";
    else
        cout << s.substr(0, idx + 1);
}
 
// Driver Code
int main()
{
    string S = "4206";
    get(S);
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find the largest even valued
  // substring
  static void get(String s)
  {
    int N = s.length();
    int idx = -1;
 
    // Finding the rightmost even character
    for (int i = N - 1; i >= 0; i--) {
      if ((s.charAt(i) - '0') % 2 == 0) {
        idx = i;
        break;
      }
    }
 
    if (idx == -1)
      System.out.print("");
    else
      System.out.print(s.substring(0, idx + 1));
  }
 
  // Driver Code
  public static void main (String[] args) {
    String S = "4206";
    get(S);
  }
}
 
// This code is contributed by hrithikgarg03188.


Python3




# Python code for the above approach
 
# Function to find the largest even valued
# substring
def get(s):
    N = len(s);
    idx = -1;
 
    # Finding the rightmost even character
    for i in range(N - 1, 0, -1):
        if ((ord(s[i]) - ord('0')) % 2 == 0):
            idx = i;
            break;
 
    if (idx == -1):
        print("");
    else:
        print(s[0: idx + 1]);
 
# Driver Code
S = "4206";
get(S);
 
# This code is contributed by gfgking


C#




// C# program for the above approach
using System;
class GFG {
 
  // Function to find the largest even valued
  // substring
  static void get(string s)
  {
    int N = s.Length;
    int idx = -1;
 
    // Finding the rightmost even character
    for (int i = N - 1; i >= 0; i--) {
      if ((s[i] - '0') % 2 == 0) {
        idx = i;
        break;
      }
    }
 
    if (idx == -1)
      Console.Write("");
    else
      Console.Write(s.Substring(0, idx + 1));
  }
 
  // Driver Code
  public static void Main () {
    string S = "4206";
    get(S);
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
    // JavaScript code for the above approach
 
    // Function to find the largest even valued
    // substring
    function get(s)
    {
        let N = s.length;
        let idx = -1;
 
        // Finding the rightmost even character
        for (let i = N - 1; i >= 0; i--)
        {
            if ((s[i].charCodeAt(0) - '0'.charCodeAt(0)) % 2 == 0)
            {
                idx = i;
                break;
            }
        }
 
        if (idx == -1)
            document.write("");
        else
            document.write(s.slice(0, idx + 1));
    }
 
    // Driver Code
    let S = "4206";
    get(S);
 
   // This code is contributed by Potta Lokesh
</script>


 
 

Output

4206

 

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

 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!