Largest number made up of X and Y with count of X divisible by Y and of Y by X

Given three integers X, Y and N, the task is to find the largest number possible of length N consisting only of X and Y as its digits, such that, the count of X‘s in it is divisible by Y and vice-versa. If no such number can be formed, print -1.
Examples:

Input: N = 3, X = 5, Y = 3 
Output: 555 
Explanation:
Count of 5’s = 3, which is divisible by 3 
Count of 3’s = 0
Input: N = 4, X = 7, Y = 5 
Output: -1 

Approach:
Follow the steps below to solve the problem: 

• Consider the larger of X and Y as X and smaller as Y.
• Since, the number needs to of length N, perform the following two steps until N ≤ 0: 
  • If N is divisible by Y, append X, N times to the answer and reduce N to zero.
  • Otherwise, reduce N by X and append Y, X times to the answer.
• After completion of the above step, if N <0, then a number of required type is not possible. Print -1.
• Otherwise, print the answer.

Below is the implementation of the above approach:

C++

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

// Function to generate and return
// the largest number
void largestNumber(int n, int X, int Y)
{
    int maxm = max(X, Y);

    // Store the smaller in Y
    Y = X + Y - maxm;

    // Store the larger in X
    X = maxm;

    // Stores respective counts
    int Xs = 0;
    int Ys = 0;

    while (n > 0) {

        // If N is divisible by Y
        if (n % Y == 0) {

            // Append X, N times to
            // the answer
            Xs += n;

            // Reduce N to zero
            n = 0;
        }
        else {

            // Reduce N by X
            n -= X;

            // Append Y, X times
            // to the answer
            Ys += X;
        }
    }

    // If number can be formed
    if (n == 0) {
        while (Xs-- > 0)
            cout << X;

        while (Ys-- > 0)
            cout << Y;
    }

    // Otherwise
    else
        cout << "-1";
}

// Driver Code
int main()
{
    int n = 19, X = 7, Y = 5;
    largestNumber(n, X, Y);
    return 0;
}

Java

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

class GFG{

// Function to generate and return
// the largest number
public static void largestNumber(int n, int X,
                                        int Y)
{
    int maxm = Math.max(X, Y);

    // Store the smaller in Y
    Y = X + Y - maxm;

    // Store the larger in X
    X = maxm;

    // Stores respective counts
    int Xs = 0;
    int Ys = 0;

    while (n > 0)
    {

        // If N is divisible by Y
        if (n % Y == 0)
        {

            // Append X, N times to
            // the answer
            Xs += n;

            // Reduce N to zero
            n = 0;
        }
        else
        {
            // Reduce N by X
            n -= X;

            // Append Y, X times
            // to the answer
            Ys += X;
        }
    }

    // If number can be formed
    if (n == 0)
    {
        while (Xs-- > 0)
            System.out.print(X);

        while (Ys-- > 0)
            System.out.print(Y);
    }

    // Otherwise
    else
        System.out.print("-1");
}

// Driver code
public static void main (String[] args)
{
    int n = 19, X = 7, Y = 5;

    largestNumber(n, X, Y);
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to implement
# the above approach

# Function to generate and return
# the largest number
def largestNumber(n, X, Y):

    maxm = max(X, Y)

    # Store the smaller in Y
    Y = X + Y - maxm

    # Store the larger in X
    X = maxm

    # Stores respective counts
    Xs = 0
    Ys = 0

    while (n > 0):

        # If N is divisible by Y
        if (n % Y == 0):

            # Append X, N times to
            # the answer
            Xs += n

            # Reduce N to zero
            n = 0

        else:

            # Reduce N by x
            n -= X

            # Append Y, X times to
            # the answer
            Ys += X

    # If number can be formed
    if (n == 0):

        while (Xs > 0):
            Xs -= 1
            print(X, end = '')

        while (Ys > 0):
            Ys -= 1
            print(Y, end = '')

    # Otherwise
    else:
        print("-1")

# Driver code
n = 19
X = 7
Y = 5

largestNumber(n, X, Y)

# This code is contributed by himanshu77

C#

// C# program to implement the
// above approach
using System;
class GFG{

// Function to generate and return
// the largest number
public static void largestNumber(int n, int X,
                                        int Y)
{
    int maxm = Math.Max(X, Y);

    // Store the smaller in Y
    Y = X + Y - maxm;

    // Store the larger in X
    X = maxm;

    // Stores respective counts
    int Xs = 0;
    int Ys = 0;

    while (n > 0)
    {

        // If N is divisible by Y
        if (n % Y == 0)
        {

            // Append X, N times to
            // the answer
            Xs += n;

            // Reduce N to zero
            n = 0;
        }
        else
        {
            // Reduce N by X
            n -= X;

            // Append Y, X times
            // to the answer
            Ys += X;
        }
    }

    // If number can be formed
    if (n == 0)
    {
        while (Xs-- > 0)
            Console.Write(X);

        while (Ys-- > 0)
            Console.Write(Y);
    }

    // Otherwise
    else
        Console.Write("-1");
}

// Driver code
public static void Main (String[] args)
{
    int n = 19, X = 7, Y = 5;

    largestNumber(n, X, Y);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

// Javascript program to implement the
// above approach

// Function to generate and return
// the largest number
function largestNumber(n, X, Y)
{
    let maxm = Math.max(X, Y);

    // Store the smaller in Y
    Y = X + Y - maxm;

    // Store the larger in X
    X = maxm;

    // Stores respective counts
    let Xs = 0;
    let Ys = 0;

    while (n > 0)
    {

        // If N is divisible by Y
        if (n % Y == 0)
        {

            // Append X, N times to
            // the answer
            Xs += n;

            // Reduce N to zero
            n = 0;
        }
        else
        {
            // Reduce N by X
            n -= X;

            // Append Y, X times
            // to the answer
            Ys += X;
        }
    }

    // If number can be formed
    if (n == 0)
    {
        while (Xs-- > 0)
            document.write(X);

        while (Ys-- > 0)
            document.write(Y);
    }

    // Otherwise
    else
        document.write("-1");
}

// Driver Code

    let n = 19, X = 7, Y = 5;

    largestNumber(n, X, Y);

 // This code is contributed by splevel62.
</script>

Output:

7777755555555555555

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