Print all numbers up to N having product of digits equal to K

• Difficulty Level : Expert
• Last Updated : 18 Nov, 2021

Given two integers N and K, the task is to print all the numbers from the range [1, N] whose product of digits is equal to K. If no such number is found, then print “-1”.

Examples:

Input: N =100, K = 25
Output: 55
Explanation: There is only a single number 55 whose product of digits is equal to K.

Input: N = 500, K = 10
Output: 25 52 125 152 215 251

Approach: Follow the steps below to solve the problem:

• Initialize a variable, say flag, to store whether any number satisfying the given conditions exists or not.
• Declare a function, productOfDigits(), to find the product of digits of the number.
• Iterate over the range [1, N]:
• If the product of digits of arr[i] is equal to K, print that number and set flag = 1.
• If flag is equal to 0, which means that no such number is found in the range [1, N]. Therefore, print “-1”.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include using namespace std; // Function to find the product// of digits of a numberint productOfDigits(int N){    // Stores the product of    // digits of a number    int product = 1;     while (N != 0) {        product = product * (N % 10);        N = N / 10;    }     // Return the product    return product;} // Function to print all numbers upto// N having product of digits equal to Kvoid productOfDigitsK(int N, int K){    // Stores whether any number satisfying    // the given conditions exists or not    int flag = 0;     // Iterate over the range [1, N]    for (int i = 1; i <= N; ++i) {         // If product of digits of        // arr[i] is equal to K or not        if (K == productOfDigits(i)) {             // Print that number            cout << i << " ";            flag = 1;        }    }     // If no numbers are found    if (flag == 0)        cout << "-1";} // Driver Codeint main(){    // Given value of N & K    int N = 500, K = 10;     // Function call to print all numbers    // from [1, N] with product of digits K    productOfDigitsK(N, K);}

Java

 // Java Program to implement// the above approachimport java.io.*;import java.util.*; class GFG {   // Function to find the product  // of digits of a number  static int productOfDigits(int N)  {         // Stores the product of    // digits of a number    int product = 1;     while (N != 0) {      product = product * (N % 10);      N = N / 10;    }     // Return the product    return product;  }   // Function to print all numbers upto  // N having product of digits equal to K  static void productOfDigitsK(int N, int K)  {    // Stores whether any number satisfying    // the given conditions exists or not    int flag = 0;     // Iterate over the range [1, N]    for (int i = 1; i <= N; ++i) {       // If product of digits of      // arr[i] is equal to K or not      if (K == productOfDigits(i)) {         // Print that number        System.out.print(i + " ");        flag = 1;      }    }     // If no numbers are found    if (flag == 0)      System.out.println(-1);  }   // Driver Code  public static void main(String[] args)  {         // Given value of N & K    int N = 500, K = 10;     // Function call to print all numbers    // from [1, N] with product of digits K    productOfDigitsK(N, K);  }} // This code is contribute by Kingash.

Python3

 # Python3 program for the above approach # Function to find the product# of digits of a numberdef productOfDigits(N) :     # Stores the product of    # digits of a number    product = 1;     while (N != 0) :        product = product * (N % 10);        N = N // 10;     # Return the product    return product; # Function to print all numbers upto# N having product of digits equal to Kdef productOfDigitsK(N, K) :     # Stores whether any number satisfying    # the given conditions exists or not    flag = 0;     # Iterate over the range [1, N]    for i in range(1, N + 1) :         # If product of digits of        # arr[i] is equal to K or not        if (K == productOfDigits(i)) :             # Print that number            print(i, end =" ");            flag = 1;     # If no numbers are found    if (flag == 0) :        print("-1"); # Driver Codeif __name__ == "__main__" :         # Given value of N & K    N = 500; K = 10;     # Function call to print all numbers    # from [1, N] with product of digits K    productOfDigitsK(N, K);         # This code is contributed by AnkThon

C#

 // C# program for the above approachusing System;class GFG{   // Function to find the product  // of digits of a number  static int productOfDigits(int N)  {     // Stores the product of    // digits of a number    int product = 1;     while (N != 0) {      product = product * (N % 10);      N = N / 10;    }     // Return the product    return product;  }   // Function to print all numbers upto  // N having product of digits equal to K  static void productOfDigitsK(int N, int K)  {     // Stores whether any number satisfying    // the given conditions exists or not    int flag = 0;     // Iterate over the range [1, N]    for (int i = 1; i <= N; ++i) {       // If product of digits of      // arr[i] is equal to K or not      if (K == productOfDigits(i)) {         // Print that number        Console.Write(i + " ");        flag = 1;      }    }     // If no numbers are found    if (flag == 0)      Console.WriteLine(-1);  }    // Driver Code  static public void Main()  {     // Given value of N & K    int N = 500, K = 10;     // Function call to print all numbers    // from [1, N] with product of digits K    productOfDigitsK(N, K);  }} // This code is contributed by jana_sayantan.

Javascript


Output
25 52 125 152 215 251

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

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