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 numbervoid 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 Codeint main(){ int n = 19, X = 7, Y = 5; largestNumber(n, X, Y); return 0;} |
Java
// Java program to implement the// above approachimport java.util.*;class GFG{ // Function to generate and return// the largest numberpublic 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 codepublic 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 numberdef 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 coden = 19X = 7Y = 5largestNumber(n, X, Y)# This code is contributed by himanshu77 |
C#
// C# program to implement the// above approachusing System;class GFG{ // Function to generate and return// the largest numberpublic 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 codepublic 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 numberfunction 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)
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