# Count subsets in an array having product K

• Last Updated : 20 Apr, 2021

Given an array arr[] of size N, the task is to find the count of all subsets from the given array whose product of is equal to K.

Examples:

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Input: arr[] = { 1, 1, 2, 2, 3 }, K = 4
Output:
Explanation:
Subsets with product equal to K(= 4) are: { { arr, arr, arr, arr }, { arr, arr, arr }, { arr, arr, arr }, { arr, arr } } .
Therefore, the required output is 4

Input: arr[] = { 1, 1, 2, 2, 3 }, K = 4
Output: 8

Approach: The problem can be solved using Dynamic Programming using the following recurrence relation:

cntSub(idx, prod) = cntSub(idx – 1, prod * arr[i]) + cntSub(idx – 1, prod)

idx: Stores index of an array element
prod: Stores product of elements of a subset
cntSub(i, prod): Stores the count of subsets from the subarray { arr[i], …, arr[N – 1] } whose product is equal to prod.

Follow the steps below to solve the problem:

• Initialize a 2D array, say dp[][], to store the overlapping subproblems of the above recurrence relation.
• Using the above recurrence relation, calculate the count of subsets whose product of elements is equal to K.
• Finally, print the value of dp[N – 1][K].

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `#define M 1000` `// Function to find the count of subsets``// whose product of elements is equal to K``int` `cntSub(``int` `arr[], ``int` `idx,``           ``int` `prod, ``int` `K, ``int` `dp[][M])``{``    ``// Base Case``    ``if` `(idx < 0) {` `        ``return` `(prod == K);``    ``}` `    ``// If an already computed``    ``// subproblem occurred``    ``if` `(dp[idx][prod] != -1) {` `        ``return` `dp[idx][prod];``    ``}` `    ``// Count subsets including idx-th``    ``// element in the subset``    ``int` `X = cntSub(arr, idx - 1,``                   ``prod * arr[idx], K, dp);` `    ``// Count subsets without including``    ``// idx-th element in the subset``    ``int` `Y = cntSub(arr, idx - 1,``                   ``prod, K, dp);` `    ``return` `dp[idx][prod] = X + Y;``}` `// Utility function to count subsets in``// an array whose product is equal to K``int` `UtilcntSub(``int` `arr[], ``int` `N, ``int` `K)``{``    ``// dp[i][j]: Stores numberof subsets``    ``// with product j up to the i-th element``    ``int` `dp[N][M];` `    ``// Initialize dp[][] to -1``    ``memset``(dp, -1, ``sizeof``(dp));` `    ``cout << cntSub(arr, N - 1, 1, K, dp);``}` `// Driver Code``int` `main()``{` `    ``int` `arr[] = { 1, 1, 2, 2, 3, 4 };` `    ``int` `K = 4;` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``UtilcntSub(arr, N, K);``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``  ` `class` `GFG{``      ` `static` `int` `M = ``1000``;`` ` `// Function to find the count of subsets``// whose product of elements is equal to K``static` `int` `cntSub(``int` `arr[], ``int` `idx,``                  ``int` `prod, ``int` `K, ``int` `dp[][])``{``    ` `    ``// Base Case``    ``if` `(idx < ``0``)``    ``{``        ``if` `(prod == K)``            ``return` `1``;``        ``else``            ``return` `0``;``    ``}`` ` `    ``// If an already computed``    ``// subproblem occurred``    ``if` `(dp[idx][prod] != -``1``)``    ``{``        ``return` `dp[idx][prod];``    ``}`` ` `    ``// Count subsets including idx-th``    ``// element in the subset``    ``int` `X = cntSub(arr, idx - ``1``,``                   ``prod * arr[idx], K, dp);`` ` `    ``// Count subsets without including``    ``// idx-th element in the subset``    ``int` `Y = cntSub(arr, idx - ``1``,``                   ``prod, K, dp);`` ` `    ``return` `dp[idx][prod] = X + Y;``}`` ` `// Utility function to count subsets in``// an array whose product is equal to K``static` `void` `UtilcntSub(``int` `arr[], ``int` `N, ``int` `K)``{``    ` `    ``// dp[i][j]: Stores numberof subsets``    ``// with product j up to the i-th element``    ``int``[][] dp = ``new` `int``[N][M];`` ` `    ``// Initialize dp[][] to -1``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``for``(``int` `j = ``0``; j < M; j++)``        ``{``            ``dp[i][j] = -``1``;``        ``}``    ``}` `    ``System.out.print(cntSub(arr, N - ``1``, ``1``, K, dp));``}``  ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int``[] arr = { ``1``, ``1``, ``2``, ``2``, ``3``, ``4` `};`` ` `    ``int` `K = ``4``;`` ` `    ``int` `N = arr.length;`` ` `    ``UtilcntSub(arr, N, K);``}``}` `// This code is contributed by code_hunt`

## Python3

 `# Python program to implement``# the above approach``M ``=` `1000` `# Function to find the count of subsets``# whose product of elements is equal to K``def` `cntSub(arr, idx, prod, K):``    ``global` `dp``    ` `    ``# Base Case``    ``if` `(idx < ``0``):``        ``return` `(prod ``=``=` `K)` `    ``# If an already computed``    ``# subproblem occurred``    ``if` `(dp[idx][prod] !``=` `-``1``):``        ``return` `dp[idx][prod]` `    ``# Count subsets including idx-th``    ``# element in the subset``    ``X ``=` `cntSub(arr, idx ``-` `1``, prod ``*` `arr[idx], K)` `    ``# Count subsets without including``    ``# idx-th element in the subset``    ``Y ``=` `cntSub(arr, idx ``-` `1``, prod, K)``    ``dp[idx][prod] ``=` `X ``+` `Y``    ``return` `dp[idx][prod]` `# Utility function to count subsets in``# an array whose product is equal to K``def` `UtilcntSub(arr, N, K):``  ` `    ``# dp[i][j]: Stores numberof subsets``    ``# with product j up to the i-th element``    ``print` `(cntSub(arr, N ``-` `1``, ``1``, K))``    ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``dp ``=` `[[``-``1` `for` `i ``in` `range``(``1000``)] ``for` `i ``in` `range``(``1000``)]``    ``arr ``=` `[``1``, ``1``, ``2``, ``2``, ``3``, ``4``]``    ``K ``=` `4``    ``N ``=` `len``(arr)``    ``UtilcntSub(arr, N, K)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement``// the above approach ``using` `System;``class` `GFG``{`` ` `static` `int` `M = 1000;``  ` `// Function to find the count of subsets``// whose product of elements is equal to K``static` `int` `cntSub(``int``[] arr, ``int` `idx,``                  ``int` `prod, ``int` `K, ``int``[,] dp)``{``     ` `    ``// Base Case``    ``if` `(idx < 0)``    ``{``        ``if` `(prod == K)``            ``return` `1;``        ``else``            ``return` `0;``    ``}``  ` `    ``// If an already computed``    ``// subproblem occurred``    ``if` `(dp[idx, prod] != -1)``    ``{``        ``return` `dp[idx, prod];``    ``}``  ` `    ``// Count subsets including idx-th``    ``// element in the subset``    ``int` `X = cntSub(arr, idx - 1,``                   ``prod * arr[idx], K, dp);``  ` `    ``// Count subsets without including``    ``// idx-th element in the subset``    ``int` `Y = cntSub(arr, idx - 1,``                   ``prod, K, dp);``  ` `    ``return` `dp[idx, prod] = X + Y;``}``  ` `// Utility function to count subsets in``// an array whose product is equal to K``static` `void` `UtilcntSub(``int``[] arr, ``int` `N, ``int` `K)``{``     ` `    ``// dp[i][j]: Stores numberof subsets``    ``// with product j up to the i-th element``    ``int``[,] dp = ``new` `int``[N, M];``  ` `    ``// Initialize dp[][] to -1``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``for``(``int` `j = 0; j < M; j++)``        ``{``            ``dp[i, j] = -1;``        ``}``    ``}`` ` `    ``Console.Write(cntSub(arr, N - 1, 1, K, dp));``}`` ` `// Driver code``public` `static` `void` `Main()``{``    ``int``[] arr = { 1, 1, 2, 2, 3, 4 };``    ``int` `K = 4; ``    ``int` `N = arr.Length; ``    ``UtilcntSub(arr, N, K);``}``}` `// This code is contributed by susmitakundugoaldanga`

## Javascript

 ``
Output:
`8`

Time Complexity: O(N * K)
Auxiliary Space: O(N * K)

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