Given an integer array with duplicate elements. The task is to find the product of all distinct elements in the given array.
Examples:
Input : arr[] = {12, 10, 9, 45, 2, 10, 10, 45, 10};
Output : 97200
Here we take 12, 10, 9, 45, 2 for product
because these are the only distinct elementsInput : arr[] = {1, 10, 9, 4, 2, 10, 10, 45, 4};
Output : 32400
A Simple Solution is to use two nested loops. The outer loop picks an element one by one starting from the leftmost element. The inner loop checks if the element is present on the left side of it. If present, then ignores the element.
Algorithm:
- Define a function named productOfDistinct that takes an integer array arr and its size n as input.
- Initialize product variable to 1.
- Loop through every element of the array using a for loop.
- Set a boolean flag isDistinct to true.
- Loop through all the previous elements of the array using another for loop.
- If the current element arr[i] is equal to any previous element arr[j], set isDistinct to false and break the loop.
- If the element is distinct, multiply it to the product.
- Return the product of distinct elements.
- In the main function, define an integer array arr and its size n.
- Call the productOfDistinct function with arr and n as arguments.
- Print the product of distinct elements.
Below is the implementation of the approach:
C++
// C++ code for the approach#include <bits/stdc++.h>using namespace std;// Function to find the product of all// non-repeated elements in an arrayint productOfDistinct(int arr[], int n) { int product = 1; // Loop through every element of the array for (int i = 0; i < n; i++) { bool isDistinct = true; // Check if the element is present on the left side of it for (int j = 0; j < i; j++) { if (arr[i] == arr[j]) { isDistinct = false; break; } } // If the element is distinct, multiply it to the product if (isDistinct) { product *= arr[i]; } } // Return the product of distinct elements return product;}// Driver's codeint main() { // Input int arr[] = { 1, 2, 3, 1, 1, 4, 5, 6 }; int n = sizeof(arr) / sizeof(arr[0]); // Find the product of distinct elements in the array int product = productOfDistinct(arr, n); // Print the product cout << product << endl; return 0;} |
Java
// Java code for the approachimport java.util.HashSet;public class Main { // Function to find the product of all non-repeated elements in an array static int productOfDistinct(int[] arr, int n) { int product = 1; // Creating a set to track distinct elements HashSet<Integer> distinctElements = new HashSet<>(); // iterate over every element of the array for (int i = 0; i < n; i++) { // Check if the element is distinct if (!distinctElements.contains(arr[i])) { product *= arr[i]; distinctElements.add(arr[i]); } } // Return the product of distinct elements return product; } // Driver code public static void main(String[] args) { // Input int[] arr = { 1, 2, 3, 1, 1, 4, 5, 6 }; int n = arr.length; // Find the product of distinct elements in the array int product = productOfDistinct(arr, n); // Print the product System.out.println(product); }} |
Python3
# Function to find the product of all# non-repeated elements in an arraydef productOfDistinct(arr): product = 1 # Loop through every element of the array for i in range(len(arr)): isDistinct = True # Check if the element is present on the left side of it for j in range(i): if arr[i] == arr[j]: isDistinct = False break # If the element is distinct, multiply it to the product if isDistinct: product *= arr[i] # Return the product of distinct elements return product# Driver's codeif __name__ == '__main__': # Input arr = [1, 2, 3, 1, 1, 4, 5, 6] # Find the product of distinct elements in the array product = productOfDistinct(arr) # Print the product print(product) |
C#
using System;class Program{ // Function to find the product of all // non-repeated elements in an array static int ProductOfDistinct(int[] arr) { int product = 1; // Loop through every element of the array for (int i = 0; i < arr.Length; i++) { bool isDistinct = true; // Check if the element is present on the left side of it for (int j = 0; j < i; j++) { if (arr[i] == arr[j]) { isDistinct = false; break; } } // If the element is distinct, multiply it to the product if (isDistinct) { product *= arr[i]; } } // Return the product of distinct elements return product; } // Driver's code static void Main(string[] args) { // Input int[] arr = { 1, 2, 3, 1, 1, 4, 5, 6 }; // Find the product of distinct elements in the array int product = ProductOfDistinct(arr); // Print the product Console.WriteLine(product); }} |
Javascript
// Function to find the product of all non-repeated elements in an arrayfunction productOfDistinct(arr) { let product = 1; // Loop through every element of the array for (let i = 0; i < arr.length; i++) { let isDistinct = true; // Check if the element is present on the left side of it for (let j = 0; j < i; j++) { if (arr[i] === arr[j]) { isDistinct = false; break; } } // If the element is distinct, multiply it to the product if (isDistinct) { product *= arr[i]; } } // Return the product of distinct elements return product;}// Driver's codeconst arr = [1, 2, 3, 1, 1, 4, 5, 6];// Find the product of distinct elements in the arrayconst product = productOfDistinct(arr);// Print the productconsole.log(product); |
Output
720
Complexity Analysis:
- Time Complexity: O(N2)
- Auxiliary Space: O(1)
A Better Solution to this problem is to first sort all elements of the array in ascending order and find one by one distinct element in the array. Finally, find the product of all distinct elements.
Below is the implementation of this approach:
C++
// C++ program to find the product of all// non-repeated elements in an array#include <bits/stdc++.h>using namespace std;// Function to find the product of all// non-repeated elements in an arrayint findProduct(int arr[], int n){ // sort all elements of array sort(arr, arr + n); int prod = 1; for (int i = 0; i < n; i++) { if (arr[i] != arr[i + 1]) prod = prod * arr[i]; } return prod;}// Driver codeint main(){ int arr[] = { 1, 2, 3, 1, 1, 4, 5, 6 }; int n = sizeof(arr) / sizeof(int); cout << findProduct(arr, n); return 0;} |
Java
// Java program to find the product of all// non-repeated elements in an arrayimport java.util.Arrays;class GFG {// Function to find the product of all// non-repeated elements in an arraystatic int findProduct(int arr[], int n){ // sort all elements of array Arrays.sort(arr); int prod = 1 * arr[0]; for (int i = 0; i < n - 1; i++) { if (arr[i] != arr[i + 1]) { prod = prod * arr[i + 1]; } } return prod;}// Driver codepublic static void main(String[] args) { int arr[] = {1, 2, 3, 1, 1, 4, 5, 6}; int n = arr.length; System.out.println(findProduct(arr, n)); }}// This code is contributed by PrinciRaj1992 |
Python3
# Python 3 program to find the product # of all non-repeated elements in an array# Function to find the product of all# non-repeated elements in an arraydef findProduct(arr, n): # sort all elements of array sorted(arr) prod = 1 for i in range(0, n, 1): if (arr[i - 1] != arr[i]): prod = prod * arr[i] return prod;# Driver codeif __name__ == '__main__': arr = [1, 2, 3, 1, 1, 4, 5, 6] n = len(arr) print(findProduct(arr, n))# This code is contributed by # Surendra_Gangwar |
C#
// C# program to find the product of all// non-repeated elements in an arrayusing System;class GFG {// Function to find the product of all// non-repeated elements in an arraystatic int findProduct(int []arr, int n){ // sort all elements of array Array.Sort(arr); int prod = 1 * arr[0]; for (int i = 0; i < n - 1; i++) { if (arr[i] != arr[i + 1]) { prod = prod * arr[i + 1]; } } return prod;}// Driver codepublic static void Main() { int []arr = {1, 2, 3, 1, 1, 4, 5, 6}; int n = arr.Length; Console.WriteLine(findProduct(arr, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript program to find the product of all// non-repeated elements in an array // Function to find the product of all // non-repeated elements in an array function findProduct(arr , n) { // sort all elements of array arr.sort(); var prod = 1 * arr[0]; for (i = 0; i < n - 1; i++) { if (arr[i] != arr[i + 1]) { prod = prod * arr[i + 1]; } } return prod; } // Driver code var arr = [ 1, 2, 3, 1, 1, 4, 5, 6 ]; var n = arr.length; document.write(findProduct(arr, n));// This code contributed by gauravrajput1</script> |
Output
720
Complexity Analysis:
- Time Complexity: O(N * log N)
- Auxiliary Space: O(1)
An Efficient Solution is to traverse the array and keep a hash map to check if the element is repeated or not. While traversing if the current element is already present in the hash or not, if yes then it means it is repeated and should not be multiplied with the product, if it is not present in the hash then multiply it with the product and insert it into hash.
Below is the implementation of this approach:
CPP
// C++ program to find the product of all// non- repeated elements in an array#include <bits/stdc++.h>using namespace std;// Function to find the product of all// non-repeated elements in an arrayint findProduct(int arr[], int n){ int prod = 1; // Hash to store all element of array unordered_set<int> s; for (int i = 0; i < n; i++) { if (s.find(arr[i]) == s.end()) { prod *= arr[i]; s.insert(arr[i]); } } return prod;}// Driver codeint main(){ int arr[] = { 1, 2, 3, 1, 1, 4, 5, 6 }; int n = sizeof(arr) / sizeof(int); cout << findProduct(arr, n); return 0;} |
Java
// Java program to find the product of all// non- repeated elements in an arrayimport java.util.HashSet;class GFG { // Function to find the product of all // non-repeated elements in an array static int findProduct(int arr[], int n) { int prod = 1; // Hash to store all element of array HashSet<Integer> s = new HashSet<>(); for (int i = 0; i < n; i++) { if (!s.contains(arr[i])) { prod *= arr[i]; s.add(arr[i]); } } return prod; } // Driver code public static void main(String[] args) { int arr[] = {1, 2, 3, 1, 1, 4, 5, 6}; int n = arr.length; System.out.println(findProduct(arr, n)); }}/* This code contributed by PrinciRaj1992 */ |
Python
# Python3 program to find the product of all# non- repeated elements in an array# Function to find the product of all# non-repeated elements in an arraydef findProduct( arr, n): prod = 1 # Hash to store all element of array s = dict() for i in range(n): if (arr[i] not in s.keys()): prod *= arr[i] s[arr[i]] = 1 return prod# Driver codearr= [1, 2, 3, 1, 1, 4, 5, 6] n = len(arr)print(findProduct(arr, n))# This code is contributed by mohit kumar |
C#
// C# program to find the product of all// non- repeated elements in an arrayusing System;using System.Collections.Generic; class GFG { // Function to find the product of all // non-repeated elements in an array static int findProduct(int []arr, int n) { int prod = 1; // Hash to store all element of array HashSet<int> s = new HashSet<int>(); for (int i = 0; i < n; i++) { if (!s.Contains(arr[i])) { prod *= arr[i]; s.Add(arr[i]); } } return prod; } // Driver code public static void Main(String[] args) { int []arr = {1, 2, 3, 1, 1, 4, 5, 6}; int n = arr.Length; Console.WriteLine(findProduct(arr, n)); }}// This code is contributed by Princi Singh |
Javascript
<script>// JavaScript program to find the product of all// non- repeated elements in an array// Function to find the product of all// non-repeated elements in an arrayfunction findProduct(arr,n){ let prod = 1; // Hash to store all element of array let s = new Set(); for (let i = 0; i < n; i++) { if (!s.has(arr[i])) { prod *= arr[i]; s.add(arr[i]); } } return prod;}// Driver codelet arr=[1, 2, 3, 1, 1, 4, 5, 6];let n = arr.length;document.write(findProduct(arr, n));// This code is contributed by patel2127</script> |
Output
720
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
Another Approach (Using Built-in Python functions):
Steps to find the product of unique elements:
- Calculate the frequencies using the Counter() function
- Convert the frequency keys to the list.
- Calculate the product of the list.
- In JavaScript, the frequency is getting calculated by the single for loop
Below is the implementation of this approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// c++ program for the above approachint sumOfElements(vector<int>& arr, int n){ // loop to calculate frequency of elements of array map<int, int> freq; for (int i = 0; i < n; i++) { freq[arr[i]]++; } // Converting keys of freq dictionary to list vector<int> lis; for (auto x : arr) { lis.push_back(x); } // Return product of list int product = 1; for (int i = 0; i < lis.size(); i++) { product *= lis[i]; } return product;}// Driver codeint main(){ // Driver code vector<int> arr = { 1, 2, 3, 1, 1, 4, 5, 6 }; int n = arr.size(); cout << (sumOfElements(arr, n)); return 0;}// The code is contributed by Nidhi goel. |
Java
// Java program for the above approachimport java.util.*;public class Main { public static int sumOfElements(ArrayList<Integer> arr, int n) { // loop to calculate frequency of elements of array Map<Integer, Integer> freq = new HashMap<Integer, Integer>(); for (int i = 0; i < n; i++) { freq.put(arr.get(i), freq.getOrDefault(arr.get(i), 0) + 1); } // Converting keys of freq dictionary to list ArrayList<Integer> lis = new ArrayList<Integer>(arr); // Return product of list int product = 1; for (int i = 0; i < lis.size(); i++) { product *= lis.get(i); } return product; } public static void main(String[] args) { // Driver code ArrayList<Integer> arr = new ArrayList<Integer>(Arrays.asList(1, 2, 3, 1, 1, 4, 5, 6)); int n = arr.size(); System.out.println(sumOfElements(arr, n)); }}// This code is contributed by sdeadityasharma |
Python3
# Python program for the above approachfrom collections import Counter# Function to return the product of distinct elementsdef sumOfElements(arr, n): # Counter function is used to # calculate frequency of elements of array freq = Counter(arr) # Converting keys of freq dictionary to list lis = list(freq.keys()) # Return product of list product=1 for i in lis: product*=i return product# Driver codeif __name__ == "__main__": arr = [1, 2, 3, 1, 1, 4, 5, 6] n = len(arr) print(sumOfElements(arr, n))# This code is contributed by Pushpesh Raj |
C#
using System;using System.Collections.Generic;using System.Linq;class Program { static void Main(string[] args) { int[] arr = {1, 2, 3, 1, 1, 4, 5, 6}; int n = arr.Length; Console.WriteLine(sumOfElements(arr, n)); } static int sumOfElements(int[] arr, int n) { // Counter function is used to calculate frequency of elements of array Dictionary<int, int> freq = arr.GroupBy(x => x).ToDictionary(x => x.Key, x => x.Count()); // Converting keys of freq dictionary to list List<int> lis = freq.Keys.ToList(); // Return product of list int product = 1; foreach (int i in lis) { product *= i; } return product; }} |
Javascript
// JavaScript program for the above approachfunction sumOfElements(arr, n){ // loop to calculate frequency of elements of array let freq = new Map(); for (let i = 0; i < n; i++) { if (freq.has(arr[i])) { freq.set(arr[i], freq.get(arr[i])+1); } else { freq.set(arr[i], 1); } } // Converting keys of freq dictionary to list let lis = Array.from(freq.keys()); // Return product of list let product = 1; for (let i = 0; i < lis.length; i++) { product *= lis[i]; } return product;}// Driver codelet arr = [1, 2, 3, 1, 1, 4, 5, 6];let n = arr.length;console.log(sumOfElements(arr, n)); |
Output
720
Complexity analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
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