Count Subarray of size K in given Array with given LCM Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given an array arr[] of length N, the task is to find the number of subarrays where the least common multiple (LCM) of the subarray is equal to K and the size of that subarray is S. Examples: Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 6, S = 2Output: 1Explanation: {1, 2, 3 }, {2, 3}, {6}There are 3 subarrays that can be generated from the main array with each having its LCM as 6. Out of which only {2, 3} is the length of 2. Input: arr[] = {3, 6, 2, 8, 4}, K = 6, S = 2Output: 2Explanation: {3, 6}, {6, 2}There are only 2 subarrays having LCM as 6 and length as 2. Approach: Implement the idea below to solve the problem Maintain two loops, so as to calculate LCM starting from each index of arr[]. When the LCM get's equal to K, check the length of the subarray. Steps were taken to solve the problem: Initialize count = 0, to count the number of subarrays.Maintain a loop to iterate through each index of array arr[].Initialize LCM = arr[i], then again start a loop from that index to the end of array arr[].Find the LCM of the lcm calculated till now for the current subarray and arr[j] as LCM =( a * b ) / GCD(a, b).When LCM gets equal to given K and the size of the subarray is equal to S, increment in count variable by 1.Whenever LCM gets greater than k, Break the inner loop. As the LCM is going to increase or stay same, it is never going to decrease. Below is the implementation of the above approach. C++ // Code to implement the approach #include <bits/stdc++.h> using namespace std; // Calculate LCM between a and b int LCM(int a, int b) { long prod = a * b; return prod / __gcd(a, b); } // Function to calculate number of subarrays // where LCM is equal to k and // size of subarray is S int subarrayEqualsLCMSize(int arr[], int k, int S, int N) { // Initialize variable to store number of subarrays int count = 0; // Generating all subarrays for (int i = 0; i < N; i++) { int lcm = arr[i]; for (int j = i; j < N; j++) { // Function call to calculate lcm lcm = LCM(lcm, arr[j]); // Check the conditions given if (lcm == k && j - i + 1 == S) count++; // If LCM becomes larger than k, // break as LCM is never going to // decrease if (lcm > k) break; } } // Return the count of subarrays return count; } // Driver Code int main() { int arr[] = { 3, 2, 6, 8, 4 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 6, S = 2; // Function call cout << subarrayEqualsLCMSize(arr, K, S, N); return 0; } Java // Java code to implement the approach import java.io.*; class GFG { // Function to calculate GCD static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } // Calculate LCM between a and b static int LCM(int a, int b) { int prod = a * b; return prod / gcd(a, b); } // Function to calculate number of subarrays // where LCM is equal to k and // size of subarray is S static int subarrayEqualsLCMSize(int[] arr, int k, int S, int N) { // Initialize variable to store number of subarrays int count = 0; // Generating all subarrays for (int i = 0; i < N; i++) { int lcm = arr[i]; for (int j = i; j < N; j++) { // Function call to calculate lcm lcm = LCM(lcm, arr[j]); // Check the conditions given if (lcm == k && j - i + 1 == S) { count++; } // If LCM becomes larger than k, // break as LCM is never going to // decrease if (lcm > k) { break; } } } // Return the count of subarrays return count; } public static void main(String[] args) { int[] arr = { 3, 2, 6, 8, 4 }; int N = arr.length; int K = 6, S = 2; // Function call System.out.print( subarrayEqualsLCMSize(arr, K, S, N)); } } Python # Python code to implement the approach # Function to calculate GCD def gcd(a, b): if (b == 0): return a return gcd(b, a % b) # Calculate LCM between a and b def LCM(a, b): prod = a * b return prod / gcd(a, b) # Function to calculate number of subarrays # where LCM is equal to k and # size of subarray is S def subarrayEqualsLCMSize(arr, k, S, N): # Initialize variable to store number of subarrays count = 0 # Generating all subarrays for i in range(0, N): lcm = arr[i] for j in range(i, N): # Function call to calculate lcm lcm = LCM(lcm, arr[j]) # Check the conditions given if (lcm == k and j - i + 1 == S): count += 1 # If LCM becomes larger than k, # break as LCM is never going to # decrease if (lcm > k): break # Return the count of subarrays return count # Driver Code arr = [3, 2, 6, 8, 4] N = len(arr) K = 6 S = 2 # Function call print(subarrayEqualsLCMSize(arr, K, S, N)) # This code is contributed by Samim Hossain Mondal. C# // C# implementation using System; public class GFG { static int GCD(int num1, int num2) { int Remainder; while (num2 != 0) { Remainder = num1 % num2; num1 = num2; num2 = Remainder; } return num1; } // Calculate LCM between a and b public static int LCM(int a, int b) { long prod = a * b; return (int)(prod / GCD(a, b)); } // Function to calculate number of subarrays // where LCM is equal to k and // size of subarray is S public static int subarrayEqualsLCMSize(int[] arr, int k, int S, int N) { // Initialize variable to store number of subarrays int count = 0; // Generating all subarrays for (int i = 0; i < N; i++) { int lcm = arr[i]; for (int j = i; j < N; j++) { // Function call to calculate lcm lcm = LCM(lcm, arr[j]); // Check the conditions given if (lcm == k && j - i + 1 == S) count++; // If LCM becomes larger than k, // break as LCM is never going to // decrease if (lcm > k) break; } } // Return the count of subarrays return count; } static public void Main() { int[] arr = { 3, 2, 6, 8, 4 }; int N = arr.Length; int K = 6, S = 2; // Function call Console.WriteLine( subarrayEqualsLCMSize(arr, K, S, N)); } } // This code is contributed by ksam24000 JavaScript function GCD(num1, num2) { let Remainder; while (num2 != 0) { Remainder = num1 % num2; num1 = num2; num2 = Remainder; } return num1; } // Calculate LCM between a and b function LCM( a, b) { let prod = a * b; return Math.floor(prod / GCD(a, b)); } // Function to calculate number of subarrays // where LCM is equal to k and // size of subarray is S function subarrayEqualsLCMSize(arr, k, S, N) { // Initialize variable to store number of subarrays let count = 0; // Generating all subarrays for (let i = 0; i < N; i++) { let lcm = arr[i]; for (let j = i; j < N; j++) { // Function call to calculate lcm lcm = LCM(lcm, arr[j]); // Check the conditions given if (lcm == k && j - i + 1 == S) count++; // If LCM becomes larger than k, // break as LCM is never going to // decrease if (lcm > k) break; } } // Return the count of subarrays return count; } let arr = [ 3, 2, 6, 8, 4 ]; let N = arr.length; let K = 6, S = 2; // Function call console.log(subarrayEqualsLCMSize(arr, K, S, N)); // This code is contributed by akashish__ Output2 Time Complexity: O(N2 * Log N)Auxiliary Space: O(1) Related Articles: Introduction to Arrays - Data Structures and Algorithms TutorialsProgram to find LCM of two numbers Comment More infoAdvertise with us Next Article Types of Asymptotic Notations in Complexity Analysis of Algorithms R rbkraj000 Follow Improve Article Tags : Technical Scripter DSA Arrays Technical Scripter 2022 GCD-LCM subarray +2 More Practice Tags : Arrays Similar Reads Basics & PrerequisitesTime Complexity and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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