Check if the array has an element which is equal to product of remaining elements
Given an array of N elements, the task is to check if the array has an element which is equal to the product of all the remaining elements.
Examples:
Input: arr[] = {1, 2, 12, 3, 2}
Output: YES
12 is the product of all the remaining elements
i.e. 1 * 2 * 3 * 2 = 12
Input: arr[] = {1, 2, 3}
Output: NO
Method-1:
- First, take the product of all the element of the array.
- Now traverse the whole array again.
- For any element a[i] check if it is equal to the product of all elements divided by that element.
- Print Yes if at least one such element is found.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to Check if the array // has an element which is equal to // product of all the remaining elements bool CheckArray(int arr[], int n) { int prod = 1; // Calculate the product of all the elements for (int i = 0; i < n; ++i) prod *= arr[i]; // Return true if any such element is found for (int i = 0; i < n; ++i) if (arr[i] == prod / arr[i]) return true; // If no element is found return false; } int main() { int arr[] = { 1, 2, 12, 3, 2 }; int n = sizeof(arr) / sizeof(arr[0]); if (CheckArray(arr, n)) cout << "YES"; else cout << "NO"; return 0; } |
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Java
// Java implementation of the above approach import java.io.*; class GFG { // Function to Check if the array // has an element which is equal to // product of all the remaining elements static boolean CheckArray(int arr[], int n) { int prod = 1; // Calculate the product of all the elements for (int i = 0; i < n; ++i) prod *= arr[i]; // Return true if any such element is found for (int i = 0; i < n; ++i) if (arr[i] == prod / arr[i]) return true; // If no element is found return false; } public static void main (String[] args) { int arr[] = { 1, 2, 12, 3, 2 }; int n =arr.length; if (CheckArray(arr, n)) System.out.println("YES"); else System.out.println("NO"); } } // This code is contributed by shs.. |
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Python3
# Python 3 implementation of the above approach # Function to Check if the array # has an element which is equal to # product of all the remaining elements def CheckArray(arr, n): prod = 1 # Calculate the product of all # the elements for i in range(0, n, 1): prod *= arr[i] # Return true if any such element # is found for i in range(0, n, 1): if (arr[i] == prod / arr[i]): return True # If no element is found return False # Driver code if __name__ == '__main__': arr = [1, 2, 12, 3, 2] n = len(arr) if (CheckArray(arr, n)): print("YES") else: print("NO") # This code is contributed by # Surendra_Gangwar |
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C#
// C# implementation of the above approach class GFG { // Function to Check if the array // has an element which is equal to // product of all the remaining elements static bool CheckArray(int[] arr, int n) { int prod = 1; // Calculate the product of // all the elements for (int i = 0; i < n; ++i) prod *= arr[i]; // Return true if any such // element is found for (int i = 0; i < n; ++i) if (arr[i] == prod / arr[i]) return true; // If no element is found return false; } // Driver Code public static void Main () { int[] arr = new int[] { 1, 2, 12, 3, 2 }; int n = arr.Length; if (CheckArray(arr, n)) System.Console.WriteLine("YES"); else System.Console.WriteLine("NO"); } } // This code is contributed by mits |
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PHP
<?php // PHP implementation of the above approach // Function to Check if the array // has an element which is equal to // product of all the remaining elements function CheckArray($arr, $n) { $prod = 1; // Calculate the product of // all the elements for ($i = 0; $i < $n; ++$i) $prod *= $arr[$i]; // Return true if any such element // is found for ($i = 0; $i < $n; ++$i) if ($arr[$i] == $prod / $arr[$i]) return true; // If no element is found return false; } // Driver Code $arr = array(1, 2, 12, 3, 2); $n = sizeof($arr); if (CheckArray($arr, $n)) echo "YES"; else echo "NO"; // This code is contributed // by Akanksha Rai |
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Output:
YES
Method-2: The approach is to find the product of all the elements of the array and check if it is a perfect square or not. If it is a perfect square then check if the square root of the product exists in the array or not. If exists then print Yes else print No.
According to the problem statement, a * b = N
where b is the product of all the remaining elements of the array except a i.e arr[i]
And it is also mentioned that find the index such that a = b.
So, it simply means that a*a = N i.e. N is a perfect square. and a is its square root.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to Check if the array // has an element which is equal to // product of all the remaining elements bool CheckArray(int arr[], int n) { int prod = 1; // Storing frequency in map unordered_set<int> freq; // Calculate the product of all the elements for (int i = 0; i < n; ++i) { freq.insert(arr[i]); prod *= arr[i]; } int root = sqrt(prod); // If the prod is a perfect square if (root * root == prod) // then check if its square root // exist in the array or not if (freq.find(root) != freq.end()) return true; return false; } // Driver code int main() { int arr[] = { 1, 2, 12, 3, 2 }; int n = sizeof(arr) / sizeof(arr[0]); if (CheckArray(arr, n)) cout << "YES"; else cout << "NO"; return 0; } |
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Java
import java.util.ArrayList; // Java implementation of the above approach class GFG { // Function to Check if the array // has an element which is equal to // product of all the remaining elements static boolean CheckArray(int arr[], int n) { int prod = 1; // Storing frequency in map ArrayList<Integer> freq = new ArrayList<>(); // Calculate the product of all the elements for (int i = 0; i < n; ++i) { freq.add(arr[i]); prod *= arr[i]; } int root = (int) Math.sqrt(prod); // If the prod is a perfect square if (root * root == prod) // then check if its square root // exist in the array or not { if (freq.contains(root) & freq.lastIndexOf(root) != (freq.size())) { return true; } } return false; } // Driver code public static void main(String[] args) { int arr[] = {1, 2, 12, 3, 2}; int n = arr.length; if (CheckArray(arr, n)) { System.out.println("YES"); } else { System.out.println("NO"); } } } //This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the above approach import math # Function to Check if the array # has an element which is equal to # product of all the remaining elements def CheckArray( arr, n): prod = 1 # Storing frequency in map freq = [] # Calculate the product of all the elements for i in range(n) : freq.append(arr[i]) prod *= arr[i] root = math.sqrt(prod) # If the prod is a perfect square if (root * root == prod): # then check if its square root # exist in the array or not if root in freq: return True return False # Driver code if __name__ == "__main__": arr = [1, 2, 12, 3, 2 ] n = len(arr) if (CheckArray(arr, n)): print ("YES") else: print ("NO") |
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C#
// C# implemenatation of above approach using System; using System.Collections; class GFG { // Function to Check if the array // has an element which is equal to // product of all the remaining elements static bool CheckArray(int []arr, int n) { int prod = 1; // Storing frequency in map ArrayList freq = new ArrayList(); // Calculate the product of all the elements for (int i = 0; i < n; ++i) { freq.Add(arr[i]); prod *= arr[i]; } int root = (int) Math.Sqrt(prod); // If the prod is a perfect square if (root * root == prod) // then check if its square root // exist in the array or not { if (freq.Contains(root) & freq.LastIndexOf(root) != (freq.Count)) { return true; } } return false; } // Driver code public static void Main() { int []arr = {1, 2, 12, 3, 2}; int n = arr.Length; if (CheckArray(arr, n)) { Console.WriteLine("YES"); } else { Console.WriteLine("NO"); } } } /* This code contributed by PrinciRaj1992 */ |
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PHP
<?php // PHP implementation of the above approach // Function to Check if the array // has an element which is equal to // product of all the remaining elements function CheckArray($arr, $n) { $prod = 1; // Storing frequency in map $freq = array(); // Calculate the product of all the elements for ($i = 0; $i < $n; ++$i) { array_push($freq, $arr[$i]); $prod *= $arr[$i]; } $freq = array_unique($freq); $root = (int)(sqrt($prod)); // If the prod is a perfect square if ($root * $root == $prod) // then check if its square root // exist in the array or not if (in_array($root, $freq)) return true; return false; } // Driver code $arr = array( 1, 2, 12, 3, 2 ); $n = count($arr); if (CheckArray($arr, $n)) echo "YES"; else echo "NO"; // This code is contributed by mits ?> |
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Output:
YES
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