Product of Array Except Self
Last Updated : 13 Jan, 2025
Given an array arr[] of n integers, construct a product array res[] (of the same size) such that res[i] is equal to the product of all the elements of arr[] except arr[i].
Example:
Input: arr[] = [10, 3, 5, 6, 2]
Output: [180, 600, 360, 300, 900]
Explanation:
- For i=0, res[i] = 3 * 5 * 6 * 2 is 180.
- For i = 1, res[i] = 10 * 5 * 6 * 2 is 600.
- For i = 2, res[i] = 10 * 3 * 6 * 2 is 360.
- For i = 3, res[i] = 10 * 3 * 5 * 2 is 300.
- For i = 4, res[i] = 10 * 3 * 5 * 6 is 900.
Input: arr[] = [12, 0]
Output: [0, 12]
Explanation:
- For i = 0, res[i] = 0.
- For i = 1, res[i] = 12.
[Naive Approach] Using Nested Loops - O(n^2) Time and O(1) Space
The idea is simple, for every element arr[i], we compute product of all elements except itself. To compute the product, we run an inner loop.
C++ #include <iostream> #include <vector> using namespace std; // Function to calculate the product of all // elements except the current element vector<int> productExceptSelf(vector<int>& arr) { int n = arr.size(); // Fill result array with 1 vector<int> res(n, 1); for (int i = 0; i < n; i++) { // Compute product of all elements except arr[i] for (int j = 0; j < n; j++) { if (i != j) res[i] *= arr[j]; } } return res; } int main() { vector<int> arr = {10, 3, 5, 6, 2}; vector<int> res = productExceptSelf(arr); for (int val : res) cout << val << " "; return 0; } #include <stdio.h> // Function to calculate the product of all // elements except the current element void productExceptSelf(int arr[], int n, int res[]) { // Initialize result array as 1 for (int i = 0; i < n; i++) { res[i] = 1; } for (int i = 0; i < n; i++) { // Compute product of all elements except arr[i] for (int j = 0; j < n; j++) { if (i != j) { res[i] *= arr[j]; } } } } int main() { int arr[] = {10, 3, 5, 6, 2}; int n = sizeof(arr) / sizeof(arr[0]); int res[n]; productExceptSelf(arr, n, res); for (int i = 0; i < n; i++) { printf("%d ", res[i]); } return 0; } import java.util.Arrays; class GfG { // Function to calculate the product of all // elements except the current element static int[] productExceptSelf(int[] arr) { int n = arr.length; // Initialize the result array as 1 int[] res = new int[n]; Arrays.fill(res, 1); for (int i = 0; i < n; i++) { // Compute the product of all except arr[i] for (int j = 0; j < n; j++) { if (i != j) { res[i] *= arr[j]; } } } return res; } public static void main(String[] args) { int[] arr = {10, 3, 5, 6, 2}; int[] res = productExceptSelf(arr); for (int val : res) { System.out.print(val + " "); } } } # Function to calculate the product of all # elements except the current element def productExceptSelf(arr): n = len(arr) # Initialize the result list as 1 res = [1] * n for i in range(n): # Compute the product of all except arr[i] for j in range(n): if i != j: res[i] *= arr[j] return res if __name__ == "__main__": arr = [10, 3, 5, 6, 2] res = productExceptSelf(arr) print(" ".join(map(str, res))) using System; class GfG { // Function to calculate the product of all // elements except the current element static int[] productExceptSelf(int[] arr) { int n = arr.Length; int[] res = new int[n]; Array.Fill(res, 1); // Compute product of all elements except arr[i] for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (i != j) { res[i] *= arr[j]; } } } return res; } static void Main(string[] args) { int[] arr = {10, 3, 5, 6, 2}; int[] res = productExceptSelf(arr); foreach (int val in res) { Console.Write(val + " "); } } } // Function to calculate the product of all // elements except the current element function productExceptSelf(arr) { let n = arr.length; let res = new Array(n).fill(1); // Compute product of all elements except arr[i] for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { if (i !== j) { res[i] *= arr[j]; } } } return res; } // Driver code let arr = [10, 3, 5, 6, 2]; let res = productExceptSelf(arr); console.log(res.join(" ")); Output180 600 360 300 900
[Better approach] Using Prefix and Suffix Array - O(n) Time and O(n) Space
The above approach can be optimized by avoiding the repetitive calculation of products of elements. The idea is to precompute the prefix and suffix products and store them in two arrays. Now we can find the product of array except i-th element, by using these precomputed arrays in constant time.
product of array except i-th element = prefProduct[i] * suffProduct[i]
prefProduct[i] stores product of all elements before i-th index in the array.
suffProduct[i] stores product of all elements after i-th index in the array.
C++ #include <iostream> #include <vector> using namespace std; // Function to calculate the product of all // elements except the current element vector<int> productExceptSelf(vector<int> &arr) { int n = arr.size(); vector<int> prefProduct(n), suffProduct(n), res(n); // Construct the prefProduct array prefProduct[0] = 1; for (int i = 1; i < n; i++) prefProduct[i] = arr[i - 1] * prefProduct[i - 1]; // Construct the suffProduct array suffProduct[n - 1] = 1; for (int j = n - 2; j >= 0; j--) suffProduct[j] = arr[j + 1] * suffProduct[j + 1]; // Construct the result array using // prefProduct[] and suffProduct[] for (int i = 0; i < n; i++) res[i] = prefProduct[i] * suffProduct[i]; return res; } int main() { vector<int> arr = {10, 3, 5, 6, 2}; vector<int> res = productExceptSelf(arr); for (int val : res) cout << val << " "; } import java.util.Arrays; class GfG { // Function to calculate the product of all // elements except the current element static int[] productExceptSelf(int[] arr) { int n = arr.length; int[] prefProduct = new int[n]; int[] suffProduct = new int[n]; int[] res = new int[n]; // Construct the prefProduct array prefProduct[0] = 1; for (int i = 1; i < n; i++) prefProduct[i] = arr[i - 1] * prefProduct[i - 1]; // Construct the suffProduct array suffProduct[n - 1] = 1; for (int j = n - 2; j >= 0; j--) suffProduct[j] = arr[j + 1] * suffProduct[j + 1]; // Construct the result array using // prefProduct[] and suffProduct[] for (int i = 0; i < n; i++) res[i] = prefProduct[i] * suffProduct[i]; return res; } public static void main(String[] args) { int[] arr = {10, 3, 5, 6, 2}; int[] res = productExceptSelf(arr); System.out.println(Arrays.toString(res)); } } # Function to calculate the product of all # elements except the current element def productExceptSelf(arr): n = len(arr) prefProduct = [1] * n suffProduct = [1] * n res = [0] * n # Construct the prefProduct array for i in range(1, n): prefProduct[i] = arr[i - 1] * prefProduct[i - 1] # Construct the suffProduct array for j in range(n - 2, -1, -1): suffProduct[j] = arr[j + 1] * suffProduct[j + 1] # Construct the result array using # prefProduct[] and suffProduct[] for i in range(n): res[i] = prefProduct[i] * suffProduct[i] return res if __name__ == '__main__': arr = [10, 3, 5, 6, 2] res = productExceptSelf(arr) print(res)
using System; class GFG { // Function to calculate the product of all // elements except the current element static int[] productExceptSelf(int[] arr) { int n = arr.Length; int[] prefProduct = new int[n]; int[] suffProduct = new int[n]; int[] res = new int[n]; // Construct the prefProduct array prefProduct[0] = 1; for (int i = 1; i < n; i++) prefProduct[i] = arr[i - 1] * prefProduct[i - 1]; // Construct the suffProduct array suffProduct[n - 1] = 1; for (int j = n - 2; j >= 0; j--) suffProduct[j] = arr[j + 1] * suffProduct[j + 1]; // Construct the result array using // prefProduct[] and suffProduct[] for (int i = 0; i < n; i++) res[i] = prefProduct[i] * suffProduct[i]; return res; } static void Main() { int[] arr = {10, 3, 5, 6, 2}; int[] res = productExceptSelf(arr); Console.WriteLine(string.Join(" ", res)); } } // Function to calculate the product of all // elements except the current element function productExceptSelf(arr) { const n = arr.length; const prefProduct = new Array(n).fill(1); const suffProduct = new Array(n).fill(1); const res = new Array(n); // Construct the prefProduct array for (let i = 1; i < n; i++) { prefProduct[i] = arr[i - 1] * prefProduct[i - 1]; } // Construct the suffProduct array for (let j = n - 2; j >= 0; j--) { suffProduct[j] = arr[j + 1] * suffProduct[j + 1]; } // Construct the result array using // prefProduct[] and suffProduct[] for (let i = 0; i < n; i++) { res[i] = prefProduct[i] * suffProduct[i]; } return res; } // Driver Code const arr = [10, 3, 5, 6, 2]; const res = productExceptSelf(arr); console.log(res.join(" ")); Output180 600 360 300 900
[Efficient Approach] Using Product Array - O(n) Time and O(1) Space
The idea is to handle two special cases of the input array: when it contains zero(s) and when it doesn't.
If the array has no zeros, product of array at any index (excluding itself) can be calculated by dividing the total product of all elements by the current element.
However, division by zero is undefined, so if there are zeros in the array, the logic changes. If there is exactly one zero, the product for that index will be the product of all other non-zero elements, while the elements in rest of the indices will be zero.
If there are more than one zero, the product for all indices will be zero, since multiplying by zero results in zero.
C++ #include <iostream> #include <vector> using namespace std; // Function to calculate the product of all elements // except the current element vector<int> productExceptSelf(vector<int> &arr) { int zeros = 0, idx = -1; int prod = 1; // Count zeros and track the index of the zero for (int i = 0; i < arr.size(); ++i) { if (arr[i] == 0) { zeros++; idx = i; } else { prod *= arr[i]; } } vector<int> res(arr.size(), 0); // If no zeros, calculate the product for all elements if (zeros == 0) { for (int i = 0; i < arr.size(); i++) res[i] = prod / arr[i]; } // If one zero, set product only at the zero's index else if (zeros == 1) res[idx] = prod; return res; } int main() { vector<int> arr = {10, 3, 5, 6, 2}; vector<int> res = productExceptSelf(arr); for (int val : res) cout << val << " "; } import java.util.Arrays; class GfG { // Function to calculate the product of all elements // except the current element static int[] productExceptSelf(int[] arr) { int zeros = 0, idx = -1, prod = 1; int n = arr.length; // Count zeros and track the index of the zero for (int i = 0; i < n; i++) { if (arr[i] == 0) { zeros++; idx = i; } else { prod *= arr[i]; } } int[] res = new int[n]; Arrays.fill(res, 0); // If no zeros, calculate the product for all elements if (zeros == 0) { for (int i = 0; i < n; i++) res[i] = prod / arr[i]; } // If one zero, set product only at the zero's index else if (zeros == 1) res[idx] = prod; return res; } public static void main(String[] args) { int[] arr = {10, 3, 5, 6, 2}; int[] res = productExceptSelf(arr); for (int val : res) System.out.print(val + " "); } } # Function to calculate the product of all elements # except the current element def productExceptSelf(arr): zeros = 0 idx = -1 prod = 1 # Count zeros and track the index of the zero for i in range(len(arr)): if arr[i] == 0: zeros += 1 idx = i else: prod *= arr[i] res = [0] * len(arr) # If no zeros, calculate the product for all elements if zeros == 0: for i in range(len(arr)): res[i] = prod // arr[i] # If one zero, set product only at the zero's index elif zeros == 1: res[idx] = prod return res if __name__ == "__main__": arr = [10, 3, 5, 6, 2] res = productExceptSelf(arr) print(" ".join(map(str, res))) using System; class GfG { // Function to calculate the product of all elements // except the current element static int[] productExceptSelf(int[] arr) { int zeros = 0, idx = -1, prod = 1; int n = arr.Length; // Count zeros and track the index of the zero for (int i = 0; i < n; i++) { if (arr[i] == 0) { zeros++; idx = i; } else { prod *= arr[i]; } } int[] res = new int[n]; Array.Fill(res, 0); // If no zeros, calculate the product for all elements if (zeros == 0) { for (int i = 0; i < n; i++) res[i] = prod / arr[i]; } // If one zero, set product only at the zero's index else if (zeros == 1) res[idx] = prod; return res; } static void Main(string[] args) { int[] arr = {10, 3, 5, 6, 2}; int[] res = productExceptSelf(arr); Console.WriteLine(string.Join(" ", res)); } } // Function to calculate the product of all elements // except the current element function productExceptSelf(arr) { let zeros = 0, idx = -1, prod = 1; // Count zeros and track the index of the zero for (let i = 0; i < arr.length; i++) { if (arr[i] === 0) { zeros++; idx = i; } else { prod *= arr[i]; } } let res = new Array(arr.length).fill(0); // If no zeros, calculate the product for all elements if (zeros === 0) { for (let i = 0; i < arr.length; i++) res[i] = Math.floor(prod / arr[i]); } // If one zero, set product only at the zero's index else if (zeros === 1) res[idx] = prod; return res; } // Driver Code let arr = [10, 3, 5, 6, 2]; let res = productExceptSelf(arr); console.log(res.join(" ")); Output180 600 360 300 900
Other Mathematical Approaches:
Please refer A product array puzzle for more approaches that use log and pow to compute the required result without division operator.
Related Problem:
Construct an Array from XOR of all elements of array except element at same index
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