Check if a given integer is the product of K consecutive integers
Last Updated :
23 Jul, 2025
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 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 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# 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
<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>
Time Complexity: O(K + N(1/K))
Auxiliary Space: O(1)
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