Check if subarray with given product exists in an array Last Updated : 21 Jul, 2024 Comments Improve Suggest changes Like Article Like Report Given an array of both positive and negative integers and a number K., The task is to check if any subarray with product K is present in the array or not.Examples: Input: arr[] = {-2, -1, 3, -4, 5}, K = 2Output: YESInput: arr[] = {3, -1, -1, -1, 5}, K = 3Output: YESApproach: The code provided seeks to determine if an array arr contains a contiguous subarray whose product of elements equals a particular number k. Using a brute-force method, the algorithm calculates the products of every conceivable subarray to see if any of them equal k.Below is the implementation of the above approach: C++ #include <bits/stdc++.h> using namespace std; // Function to check if there exists a subarray with a product equal to k bool hasSubarrayWithProduct(int* arr, int n, int k) { // Iterate over all possible starting points of subarrays for (int start = 0; start < n; ++start) { int product = 1; // Initialize the product for the current subarray // Iterate over all possible end points for subarrays starting at 'start' for (int end = start; end < n; ++end) { product *= arr[end]; // Update the product for the current subarray // Check if the current subarray product is equal to k if (product == k) { return true; // Return true if such subarray is found } } } return false; // Return false if no subarray with product k is found } int main() { int arr[] = {1, 2, -5, -4}; // Input array int product = -10; // Target product value int n = sizeof(arr) / sizeof(arr[0]); // Calculate the number of elements in the array // Check if there is a subarray with the given product and print the result if (hasSubarrayWithProduct(arr, n, product)) { cout << "YES" << endl; } else { cout << "NO" << endl; } return 0; // Exit the program } Java public class Main { // Function to check if there exists a subarray with a product equal to k public static boolean hasSubarrayWithProduct(int[] arr, int k) { int n = arr.length; // Iterate over all possible starting points of subarrays for (int start = 0; start < n; ++start) { int product = 1; // Initialize the product for the current subarray // Iterate over all possible end points for subarrays starting at 'start' for (int end = start; end < n; ++end) { product *= arr[end]; // Update the product for the current subarray // Check if the current subarray product is equal to k if (product == k) { return true; // Return true if such subarray is found } } } return false; // Return false if no subarray with product k is found } public static void main(String[] args) { int[] arr = {1, 2, -5, -4}; // Input array int product = -10; // Target product value // Check if there is a subarray with the given product and print the result if (hasSubarrayWithProduct(arr, product)) { System.out.println("YES"); } else { System.out.println("NO"); } } } Python def has_subarray_with_product(arr, k): n = len(arr) # Iterate over all possible starting points of subarrays for start in range(n): product = 1 # Initialize the product for the current subarray # Iterate over all possible end points for subarrays starting at 'start' for end in range(start, n): product *= arr[end] # Update the product for the current subarray # Check if the current subarray product is equal to k if product == k: return True # Return true if such subarray is found return False # Return false if no subarray with product k is found # Input array arr = [1, 2, -5, -4] # Target product value product = -10 # Check if there is a subarray with the given product and print the result if has_subarray_with_product(arr, product): print("YES") else: print("NO") C# using System; public class Program { // Function to check if there exists a subarray with a product equal to k public static bool HasSubarrayWithProduct(int[] arr, int k) { int n = arr.Length; // Iterate over all possible starting points of subarrays for (int start = 0; start < n; ++start) { int product = 1; // Initialize the product for the current subarray // Iterate over all possible end points for subarrays starting at 'start' for (int end = start; end < n; ++end) { product *= arr[end]; // Update the product for the current subarray // Check if the current subarray product is equal to k if (product == k) { return true; // Return true if such subarray is found } } } return false; // Return false if no subarray with product k is found } public static void Main() { int[] arr = {1, 2, -5, -4}; // Input array int product = -10; // Target product value // Check if there is a subarray with the given product and print the result if (HasSubarrayWithProduct(arr, product)) { Console.WriteLine("YES"); } else { Console.WriteLine("NO"); } } } Javascript function hasSubarrayWithProduct(arr, k) { const n = arr.length; // Iterate over all possible starting points of subarrays for (let start = 0; start < n; ++start) { let product = 1; // Initialize the product for the current subarray // Iterate over all possible end points for subarrays starting at 'start' for (let end = start; end < n; ++end) { product *= arr[end]; // Update the product for the current subarray // Check if the current subarray product is equal to k if (product === k) { return true; // Return true if such subarray is found } } } return false; // Return false if no subarray with product k is found } // Input array const arr = [1, 2, -5, -4]; // Target product value const product = -10; // Check if there is a subarray with the given product and print the result if (hasSubarrayWithProduct(arr, product)) { console.log("YES"); } else { console.log("NO"); } OutputYES Time Complexity: O(n2)Auxiliary Space: O(1)Efficient Approach: we use the two-pointer technique (sliding window), but adapt it to handle products. Here is a way to approach it:Initialize two pointers, start and end.Use a variable product to keep track of the product of elements in the current window.Expand the window by moving end to the right and update the product.If the product exceeds k, move start to the right until the product is less than or equal to k.Check if the product matches k.Below is the implementation of above idea. C++ #include <iostream> using namespace std; // Function to check if there exists a subarray with a product equal to k bool hasSubarrayWithProduct(int* arr, int n, int k) { int start = 0; int product = 1; for (int end = 0; end < n; ++end) { product *= arr[end]; // If the product becomes zero and k is not zero, move the start to skip zero while (start <= end && product == 0 && k != 0) { start++; product = 1; for (int i = start; i <= end; ++i) { product *= arr[i]; } } // If product exceeds k, move the start pointer to reduce the product while (start <= end && product != 0 && abs(product) > abs(k)) { product /= arr[start]; start++; } // Check if the current product is equal to k if (product == k) { return true; } } return false; // If no subarray with product k is found, return false } int main() { int arr[] = {1, 2, -5, -4}; // Input array int product = -10; // Target product value int n = sizeof(arr) / sizeof(arr[0]); // Calculate the number of elements in the array // Check if there is a subarray with the given product and print the result if (hasSubarrayWithProduct(arr, n, product)) { cout << "YES" << endl; } else { cout << "NO" << endl; } return 0; // Exit the program } OutputYES Time Complexity: O(n)Auxiliary Space: O(1) Comment More infoAdvertise with us B barykrg Follow Improve Article Tags : Misc DSA Arrays subarray Arrays +1 More Practice Tags : ArraysArraysMisc Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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