Nth term of a sequence formed by sum of current term with product of its largest and smallest digit

Given two numbers N and K, where K represents the starting term of the sequence. The task is to find the Nth term of a sequence formed by sum of current term with product of largest and smallest digit of current term, i.e., 

A_N+1 = A_N + max(digits of A_N) * min(digits of A_N)

Examples:

Input: K = 1, N = 5 
Output: 50 
Explanation: 
A₁ = 1 
A₂ = A₁ + minDigit( A₁ ) * maxDigit( A₁ ) = 1 + min(1) * max(1) = 1 + 1*1 = 2 
A₃ = A₂ + minDigit( A₂ ) * maxDigit( A₂ ) = 2 + min(2) * max(2) = 2 + 2*2 = 6 
A₄ = A₃ + minDigit( A₃ ) * maxDigit( A₃ ) = 6 + min(6) * max(6) = 6 + 6*6 = 42 
A₅ = A₄ + minDigit( A₄ ) * maxDigit( A₄ ) = 42 + min(4, 2) * max(4, 2) = 42 + 2*4 = 50

Input: K = 487, N = 2 
Output: 519 
Explanation: 
A₁ = 487 
A₂ = A₁ + minDigit( A₁ ) * maxDigit( a₁ ) = 487 + min(4, 8, 7) * max(4, 8, 7) = 487 + 4*8 = 519 

Approach:
Let us try to see some observations, 

When K = 1, the sequence becomes: 1, 2, 6, 42, 50, 50, 50, … 
When K = 2, the sequence becomes: 2, 6, 42, 50, 50, 50, … 
. 
. 
When K = 5, the sequence becomes: 5, 30, 30, 30, 30, 30, … 
. 
. 
Similarly, When K = 10, the sequence becomes: 10, 10, 10, 10, 10, 10, … 

From the above examples, it can be observed that the sequence eventually stops increasing after an integer has at least one digit becomes 0. If any digit becomes 0, then the minimum digit would be always 0 and after that, all the integers in the sequence remain the same. 
So the approach is to find the terms of the sequence till any 0 is encountered in the digits of the current term,
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 integer
int find(int K, int N)
{

    // because 1st integer is K itself
    N--;

    while (N--) {
        int curr_term = K;

        // Initialize min_d and max_d
        int min_d = 9;
        int max_d = 0;

        while (curr_term > 0) {
            int r = curr_term % 10;

            // updating min_d and max_d
            min_d = min(min_d, r);
            max_d = max(max_d, r);

            curr_term = curr_term / 10;
        }

        // break if min digit is 0
        if (min_d == 0) {
            break;
        }

        K = K + min_d * max_d;
    }

    return K;
}

// Driver code
int main()
{
    int K = 487;
    int N = 2;

    cout << find(K, N) << endl;

    return 0;
}

Java

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

class GFG{

// Function to find integer
static int find(int K, int N)
{

    // Because 1st integer is K itself
    N--;

    while (N-- != 0)
    {
        int curr_term = K;

        // Initialize min_d and max_d
        int min_d = 9;
        int max_d = 0;

        while (curr_term > 0)
        {
            int r = curr_term % 10;

            // Updating min_d and max_d
            min_d = Math.min(min_d, r);
            max_d = Math.max(max_d, r);

            curr_term = curr_term / 10;
        }

        // Break if min digit is 0
        if (min_d == 0)
        {
            break;
        }
        K = K + min_d * max_d;
    }
    return K;
}

// Driver code
public static void main(String[] args)
{
    int K = 487;
    int N = 2;

    System.out.print(find(K, N) + "\n");
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach.

# Function to find integer
def find(K, N):

    # Because 1st integer is K itself
    N = N - 1

    for i in range(0, N):
        curr_term = K

        # Initialize min_d and max_d
        min_d = 9
        max_d = 0

        while curr_term > 0:
            r = int(curr_term % 10)

            # Updating min_d and max_d
            min_d = min(min_d, r)
            max_d = max(max_d, r)

            curr_term = int(curr_term / 10)

        # Break if min digit is 0
        if min_d == 0:
           break

        K = K + min_d * max_d
        return K

# Driver code
K = 487
N = 2

print(find(K, N))

# This code is contributed by ishayadav181

C#

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

class GFG{

// Function to find integer
static int find(int K, int N)
{

    // Because 1st integer is K itself
    N--;

    while (N-- != 0)
    {
        int curr_term = K;

        // Initialize min_d and max_d
        int min_d = 9;
        int max_d = 0;

        while (curr_term > 0)
        {
            int r = curr_term % 10;

            // Updating min_d and max_d
            min_d = Math.Min(min_d, r);
            max_d = Math.Max(max_d, r);

            curr_term = (int)(curr_term / 10);
        }

        // Break if min digit is 0
        if (min_d == 0)
        {
            break;
        }

        K = K + min_d * max_d;
    }
    return K;
}

// Driver code
public static void Main()
{
    int K = 487;
    int N = 2;

    Console.Write(find(K, N));
}
}

// This code is contributed by Code_Mech

Javascript

// Function to find integer
function find(K, N) {

    // Because 1st integer is K itself
    N--;

    while (N-- !== 0) {
        let curr_term = K;

        // Initialize min_d and max_d
        let min_d = 9;
        let max_d = 0;

        while (curr_term > 0) {
            let r = curr_term % 10;

            // Updating min_d and max_d
            min_d = Math.min(min_d, r);
            max_d = Math.max(max_d, r);

            curr_term = Math.floor(curr_term / 10);
        }

        // Break if min digit is 0
        if (min_d === 0) {
            break;
        }

        K = K + min_d * max_d;
    }
    return K;
}

// Driver code
let K = 487;
let N = 2;
console.log(find(K, N));
//This code is contributed by chinmaya121221

Output:

519

Time Complexity: O(N * log₁₀ K) , where N and K are the given integers.
Auxiliary Space: O(1), no extra space is required, so it is a constant.