Given two positive integers N and K, the task is to check if the given integer N can be expressed as the product of K consecutive integers or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Examples:
Input: N = 210, K = 3
Output: Yes
Explanation: 210 can be expressed as 5 * 6 * 7.Input: N = 780, K =4
Output: No
Approach: The given problem can be solved by using Sliding Window Technique. Follow the steps below to solve the problem:
- Initialize two integers, say Kthroot and product, to store the Kth root of the integer N and the product of K consecutive integers respectively.
- Store the product of integers over the range [1, K] in the variable product.
- Otherwise, iterate over the range [2, Kthroot] and perform the following steps:
- If the value of the product is equal to N, then print “Yes” and break out of the loop.
- Update the value of product as (product*(i + K – 1)) / (i – 1).
- After completing the above steps, if none of the above cases satisfy, then print “No” as N cannot be expressed as the product of K consecutive integers.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if N can be expressed // as the product of K consecutive integers string checkPro( int n, int k)
{ double exp = 1.0 / k;
// Stores the K-th root of N
int KthRoot = ( int ) pow (n, exp );
// Stores the product of K
// consecutive integers
int product = 1;
// Traverse over the range [1, K]
for ( int i = 1; i < k + 1; i++)
{
// Update the product
product = product * i;
}
// If product is N, then return "Yes"
if (product == n)
return "Yes" ;
else
{
// Otherwise, traverse over
// the range [2, Kthroot]
for ( int j = 2; j < KthRoot + 1; j++)
{
// Update the value of product
product = product * (j + k - 1);
product = product / (j - 1);
// If product is equal to N
if (product == n)
return "Yes" ;
}
}
// Otherwise, return "No"
return "No" ;
} // Driver code int main()
{ int N = 210;
int K = 3;
cout << checkPro(N, K);
return 0;
} // This code is contributed by avijitmondal1998 |
Java
// Java program for the above approach public class GFG {
// Function to check if N can be expressed
// as the product of K consecutive integers
static String checkPro( int n, int k){
double exp = 1.0 / k ;
// Stores the K-th root of N
int KthRoot = ( int )Math.pow(n, exp);
// Stores the product of K
// consecutive integers
int product = 1 ;
// Traverse over the range [1, K]
for ( int i = 1 ; i < k + 1 ; i++){
// Update the product
product = product * i;
}
// If product is N, then return "Yes"
if (product == n)
return "Yes" ;
else {
// Otherwise, traverse over
// the range [2, Kthroot]
for ( int j = 2 ; j < KthRoot + 1 ; j++) {
// Update the value of product
product = product * (j + k - 1 ) ;
product = product / (j - 1 ) ;
// If product is equal to N
if (product == n)
return "Yes" ;
}
}
// Otherwise, return "No"
return "No" ;
}
// Driver Code
public static void main (String[] args) {
int N = 210 ;
int K = 3 ;
System.out.println(checkPro(N, K));
}
} // This code is contributed by AnkThon |
Python3
# Python3 program for the above approach # Function to check if N can be expressed # as the product of K consecutive integers def checkPro(n, k):
# Stores the K-th root of N
KthRoot = int (n * * ( 1 / k))
# Stores the product of K
# consecutive integers
product = 1
# Traverse over the range [1, K]
for i in range ( 1 , k + 1 ):
# Update the product
product = product * i
print (product)
# If product is N, then return "Yes"
if (product = = N):
return ( "Yes" )
# Otherwise, traverse over
# the range [2, Kthroot]
for i in range ( 2 , KthRoot + 1 ):
# Update the value of product
product = product * (i + k - 1 )
product = product / (i - 1 )
print (product)
# If product is equal to N
if (product = = N):
return ( "Yes" )
# Otherwise, return "No"
return ( "No" )
# Driver Code N = 210
K = 3
# Function Call print (checkPro(N, K))
|
C#
// C# program for the above approach using System;
class GFG{
// Function to check if N can be expressed // as the product of K consecutive integers static string checkPro( int n, int k)
{ double exp = 1.0 / k ;
// Stores the K-th root of N
int KthRoot = ( int )Math.Pow(n, exp);
// Stores the product of K
// consecutive integers
int product = 1 ;
// Traverse over the range [1, K]
for ( int i = 1; i < k + 1; i++)
{
// Update the product
product = product * i;
}
// If product is N, then return "Yes"
if (product == n)
return "Yes" ;
else
{
// Otherwise, traverse over
// the range [2, Kthroot]
for ( int j = 2; j < KthRoot + 1; j++)
{
// Update the value of product
product = product * (j + k - 1);
product = product / (j - 1);
// If product is equal to N
if (product == n)
return "Yes" ;
}
}
// Otherwise, return "No"
return "No" ;
} // Driver Code static public void Main()
{ int N = 210;
int K = 3;
Console.WriteLine(checkPro(N, K));
} } // This code is contributed by sanjoy_62 |
Javascript
<script> // JavaScript program for the above approach // Function to check if N can be expressed
// as the product of K consecutive integers
function checkPro(n , k) {
var exp = 1.0 / k;
// Stores the K-th root of N
var KthRoot = parseInt( Math.pow(n, exp));
// Stores the product of K
// consecutive integers
var product = 1;
// Traverse over the range [1, K]
for (i = 1; i < k + 1; i++) {
// Update the product
product = product * i;
}
// If product is N, then return "Yes"
if (product == n)
return "Yes" ;
else {
// Otherwise, traverse over
// the range [2, Kthroot]
for (j = 2; j < KthRoot + 1; j++) {
// Update the value of product
product = product * (j + k - 1);
product = product / (j - 1);
// If product is equal to N
if (product == n)
return "Yes" ;
}
}
// Otherwise, return "No"
return "No" ;
}
// Driver Code
var N = 210;
var K = 3;
document.write(checkPro(N, K));
// This code contributed by Rajput-Ji </script> |
Output:
Yes
Time Complexity: O(K + N(1/K))
Auxiliary Space: O(1)