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Count of factors of length K in the number itself

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  • Difficulty Level : Hard
  • Last Updated : 27 Jan, 2023
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Given an integer N and a value K, the task is to find the number of factors of length K that are present in the number itself.

Examples:

Input: N = 120, K = 2
Output: 2
Explanation: 12 (120), 20 (120). Both are factors of 120.

Input: N = 21, K = 2
Output: 1

Approach: The problem can be solved greedily using the following idea:

Find all the K-length numbers that can be generated from the number and check how many of them are factors of N.

Follow the steps mentioned below to implement the idea:

  • Iterate the number.
  • Initialize the start index with 0.
  • Slice the number for K size from the start index.
  • Then check if the sliced part of the number is a factor of the given number.
    • If it is a factor, increment the counter.
    • Else, do nothing.
  • Increment the start index to the next index.
  • Continue the above process until the start index is N-K.

Below is the implementation of the above approach.

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the number of factors
int countFactorsinNum(int num, int k)
{
    string str = to_string(num);
    int N = str.size();
    int ctr = 0;
 
    int i = 0;
    while (i <= N - k) {
 
        // substr(from where, nos of chars)
        string s2 = str.substr(i, k);
        int val = stoi(s2);
 
        // val!=0 to avoid div by 0 error
        if (val != 0 && num % val == 0) {
            ctr++;
        }
 
        i = i + 1;
    }
 
    return ctr;
}
 
// Driver code
int main()
{
    int N = 120, K = 2;
 
    // Function call
    cout << countFactorsinNum(N, K) << "\n";
 
    return 0;
}


Java




// Java code to implement the approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to find the number of factors
    static int countFactorsinNum(int num, int k)
    {
        String str = Integer.toString(num);
        int N = str.length();
        int ctr = 0;
 
        int i = 0;
        while (i <= N - k) {
 
            // substr(from where, nos of chars)
            String s2 = str.substring(i, i + k);
            int val = Integer.parseInt(s2);
 
            // val!=0 to avoid div by 0 error
            if (val != 0 && num % val == 0) {
                ctr++;
            }
 
            i = i + 1;
        }
 
        return ctr;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 120, K = 2;
 
        // Function call
        System.out.println(countFactorsinNum(N, K));
    }
}
 
// This code is contributed by lokesh


Python3




#Python code for the above approach
 
#Function to find the number of factors
def countFactorsinNum(num, k):
    # Convert num to a string
    str_num = str(num)
    N = len(str_num)
    ctr = 0
 
    # Iterate k times starting from the first digit of str_num
    for i in range(k):
        # Extract a substring of length k from str_num starting at position i
        s2 = str_num[i:i+k]
 
        # Convert s2 to an integer
        val = int(s2)
 
        # Check if val is a factor of num
        if val != 0 and num % val == 0:
            ctr += 1
 
    return ctr
 
# Driver code
N = 120
K = 2
 
# Function call
print(countFactorsinNum(N, K))
 
#This code is contributed by Potta Lokesh


Javascript




// Javascript code to implement the approach
 
// Function to find the number of factors
function countFactorsinNum(num, k)
{
    let str = num.toString();
    let N = str.length;
    let ctr = 0;
 
    let i = 0;
    while (i <= N - k) {
 
        // substr(from where, nos of chars)
        let s2 = str.substring(i, k);
        let val = parseInt(s2);
 
        // val!=0 to avoid div by 0 error
        if (val != 0 && num % val == 0) {
            ctr++;
        }
 
        i = i + 1;
    }
 
    return ctr;
}
 
// Driver code
 
    let N = 120, K = 2;
 
    // Function call
    console.log(countFactorsinNum(N, K));
 
// This code is contributed by poojaagarwal2.


C#




// C# code to implement the approach
 
using System;
using System.Collections.Generic;
 
public class Gfg {
    static int countFactorsinNum(int num, int k)
    {
        string str = num.ToString();
        int N = str.Length;
        int ctr = 0;
     
        int i = 0;
        while (i <= N - k) {
     
            // substr(from where, nos of chars)
            string s2 = str.Substring(i, k);
            int val = int.Parse(s2);
     
            // val!=0 to avoid div by 0 error
            if (val != 0 && num % val == 0) {
                ctr++;
            }
     
            i = i + 1;
        }
     
        return ctr;
    }
     
    // Driver code
    public static void Main(string[] args)
    {
        int N = 120, K = 2;
     
        // Function call
        Console.WriteLine(countFactorsinNum(N, K)+"\n");
     
    }
}


Output

2

Time Complexity: O(d) where d is the number of digits
Auxiliary Space: O(1)

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