Longest Subsequence from a numeric String divisible by K

Given an integer K and a numeric string str, the task is to find the longest subsequence from the given string which is divisible by K.

Examples:

Input: str = “121400”, K = 8
Output: 121400
Explanation:
Since the whole string is divisible by 8, the entire string is the answer.

Input: str: “7675437”, K = 64
Output: 64
Explanation:
The longest subsequence from the string which is divisible by 64, is “64” itself.

Approach: The idea is to find all subsequences of the given string and for each subsequence, check if its integer representation is divisible by K or not. Follow the steps below to solve the problem:

• Traverse the string. For every character encountered, two possibilities exists.
• Either consider the current character in the subsequence or not. For both the cases, proceed to the next characters of the string and find the longest subsequence that is divisible by K.
• Compare the longest subsequences obtained above with the current maximum length of longest subsequence and update accordingly.
• Repeat the above steps for every character of the string and finally, print the maximum length obtained.

Below is the implementation of the above approach.

C++

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

// Function to if the integer representation
// of the current string is divisible by K
bool isdivisible(string& newstr, long long K)
{
    // Stores the integer
    // representation of the string
    long long num = 0;

    for (int i = 0; i < newstr.size(); i++) {
        num = num * 10 + newstr[i] - '0';
    }

    // Check if the num is
    // divisible by K
    if (num % K == 0)
        return true;

    else
        return false;
}

// Function to find the longest subsequence
// which is divisible by K
void findLargestNo(string& str, string& newstr,
                string& ans, int index,
                long long K)
{

    if (index == (int)str.length()) {

        // If the number is divisible by K
        if (isdivisible(newstr, K)) {

            // If current number is the
            // maximum obtained so far
            if ((ans < newstr
                && ans.length() == newstr.length())
                || newstr.length() > ans.length()) {
                ans = newstr;
            }
        }

        return;
    }

    string x = newstr + str[index];

    // Include the digit at current index
    findLargestNo(str, x, ans, index + 1, K);

    // Exclude the digit at current index
    findLargestNo(str, newstr, ans, index + 1, K);
}

// Driver Code
int main()
{

    string str = "121400";

    string ans = "", newstr = "";

    long long K = 8;

    findLargestNo(str, newstr, ans, 0, K);

    // Printing the largest number
    // which is divisible by K
    if (ans != "")
        cout << ans << endl;

    // If no number is found
    // to be divisible by K
    else
        cout << -1 << endl;
}

Java

// Java program for the
// above approach
import java.util.*;
import java.lang.*;

class GFG{

// Function to if the integer representation
// of the current string is divisible by K
static boolean isdivisible(StringBuilder newstr,
                           long K)
{

    // Stores the integer
    // representation of the string
    long num = 0;

    for(int i = 0; i < newstr.length(); i++)
    {
        num = num * 10 + newstr.charAt(i) - '0';
    }

    // Check if the num is
    // divisible by K
    if (num % K == 0)
        return true;
    else
        return false;
}

// Function to find the longest
// subsequence which is divisible
// by K
static void findLargestNo(String str, StringBuilder newstr,
                          StringBuilder ans, int index,
                          long K)
{
    if (index == str.length())
    {

        // If the number is divisible by K
        if (isdivisible(newstr, K))
        {

            // If current number is the
            // maximum obtained so far
            if ((newstr.toString().compareTo(ans.toString()) > 0 &&
                ans.length() == newstr.length()) ||
                newstr.length() > ans.length())
            {
                ans.setLength(0);
                ans.append(newstr);
            }
        }
        return;
    }

    StringBuilder x = new StringBuilder(
        newstr.toString() + str.charAt(index));

    // Include the digit at current index
    findLargestNo(str, x, ans, index + 1, K);

    // Exclude the digit at current index
    findLargestNo(str, newstr, ans, index + 1, K);
}

// Driver code
public static void main (String[] args)
{
    String str = "121400";

    StringBuilder ans = new StringBuilder(),
               newstr = new StringBuilder();

     long K = 8;

    findLargestNo(str, newstr, ans, 0, K);

    // Printing the largest number
    // which is divisible by K
    if (ans.toString() != "")
        System.out.println(ans);

    // If no number is found
    // to be divisible by K
    else
        System.out.println(-1);
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach

# Function to if the integer representation
# of the current string is divisible by K
def isdivisible(newstr, K):

    # Stores the integer
    # representation of the string
    num = 0;

    for i in range(len(newstr)):
        num = num * 10 + int(newstr[i])

    # Check if the num is
    # divisible by K
    if (num % K == 0):
        return True
    else:
        return False

# Function to find the longest subsequence
# which is divisible by K
def findLargestNo(str, newstr, ans, index, K):

    if (index == len(str)):

        # If the number is divisible by K
        if (isdivisible(newstr, K)):

            # If current number is the
            # maximum obtained so far
            if ((ans < newstr and len(ans) == len(newstr)) or len(newstr) > len(ans)):
                ans = newstr

        return ans

    x = newstr + str[index];

    # Include the digit at current index
    ans = findLargestNo(str, x, ans, index + 1, K);

    # Exclude the digit at current index
    ans = findLargestNo(str, newstr, ans, index + 1, K);

    return ans;

# Driver Code
str = "121400";
ans = ""
newstr = "";
K = 8;
ans = findLargestNo(str, newstr, ans, 0, K);

# Printing the largest number
# which is divisible by K
if (ans != "") :
    print(ans)

# If no number is found
# to be divisible by K
else:
    print( -1)

# This code is contributed by phasing17

C#

// C# program for the above approach
using System;
using System.Text;

class GFG{

// Function to if the integer representation
// of the current string is divisible by K
static bool isdivisible(StringBuilder newstr,
                        long K)
{

    // Stores the integer representation
    // of the string
    long num = 0;

    for(int i = 0; i < newstr.Length; i++)
    {
        num = num * 10 + newstr[i] - '0';
    }

    // Check if the num is
    // divisible by K
    if (num % K == 0)
        return true;
    else
        return false;
}

// Function to find the longest
// subsequence which is divisible
// by K
static void findLargestNo(String str, StringBuilder newstr,
                          StringBuilder ans, int index,
                          long K)
{
    if (index == str.Length)
    {

        // If the number is divisible by K
        if (isdivisible(newstr, K))
        {

            // If current number is the
            // maximum obtained so far
            if ((newstr.ToString().CompareTo(ans.ToString()) > 0 &&
                ans.Length == newstr.Length) ||
                newstr.Length > ans.Length)
            {
                ans.EnsureCapacity(0);//.SetLength(0);
                ans.Append(newstr);
            }
        }
        return;
    }

    StringBuilder x = new StringBuilder(
        newstr.ToString() + str[index]);

    // Include the digit at current index
    findLargestNo(str, x, ans, index + 1, K);

    // Exclude the digit at current index
    findLargestNo(str, newstr, ans, index + 1, K);
}

// Driver code
public static void Main(String[] args)
{
    String str = "121400";

    StringBuilder ans = new StringBuilder(),
               newstr = new StringBuilder();

     long K = 8;

    findLargestNo(str, newstr, ans, 0, K);

    // Printing the largest number
    // which is divisible by K
    if (ans.ToString() != "")
        Console.WriteLine(ans);

    // If no number is found
    // to be divisible by K
    else
        Console.WriteLine(-1);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// Javascript program for the above approach

// Function to if the integer representation
// of the current string is divisible by K
function isdivisible(newstr, K)
{

    // Stores the integer
    // representation of the string
    var num = 0;

    for(var i = 0; i < newstr.length; i++)
    {
        num = num * 10 + newstr[i].charCodeAt(0) -
                               '0'.charCodeAt(0);
    }

    // Check if the num is
    // divisible by K
    if (num % K == 0)
        return true;
    else
        return false;
}

// Function to find the longest subsequence
// which is divisible by K
function findLargestNo(str, newstr, ans, index, K)
{
    if (index == str.length)
    {

        // If the number is divisible by K
        if (isdivisible(newstr, K))
        {

            // If current number is the
            // maximum obtained so far
            if ((ans < newstr &&
                 ans.length == newstr.length) ||
                 newstr.length > ans.length)
            {
                ans = newstr;
            }
        }
        return ans;
    }

    var x = newstr + str[index];

    // Include the digit at current index
    ans = findLargestNo(str, x, ans,
                        index + 1, K);

    // Exclude the digit at current index
    ans = findLargestNo(str, newstr, ans,
                        index + 1, K);

    return ans;
}

// Driver Code
var str = "121400";
var ans = "", newstr = "";
var K = 8;
ans = findLargestNo(str, newstr, ans, 0, K);

// Printing the largest number
// which is divisible by K
if (ans != "")
    document.write(ans)

// If no number is found
// to be divisible by K
else
    document.write( -1)

// This code is contributed by itsok

</script>

Output:

121400

Time Complexity: O(N * 2^N)
Auxiliary Space: O(1)