You are given an array of n-elements, you have to find the number of subsets whose product of elements is less than or equal to a given integer k.

**Examples:**

Input : arr[] = {2, 4, 5, 3}, k = 12 Output : 8 Explanation : All possible subsets whose products are less than 12 are: (2), (4), (5), (3), (2, 4), (2, 5), (2, 3), (4, 3) Input : arr[] = {12, 32, 21 }, k = 1 Output : 0 Explanation : there is not any subset such that product of elements is less than 1

**Approach :** If we go through the basic approach to solve this problem, then we have to generate all possible 2^{n} subset and for each of then we have to calculate product of elements of subset and compare products value with given then. But the disadvantage of this approach is that its time complexity is too high i.e. O(n*2^{n}). Now, we can see that it is going to be exponential time complexity which should be avoided in case of competitive codings.

**Advance Approach :** We are going to use the concept of meet in the middle. By, using this concept we can reduce the complexity of our approach to O(n*2^{n/2}).

**How to use MEET IN THE MIDDLE Approach :**First of all we simply divide the given array into two equal parts and after that we generate all possible subsets for both parts of array and store value of elements product for each subset separately into two vectors (say subset1 & subset2). Now this will cost O(2^{n/2}) time complexity. Now if we sort these two vectors(subset1 & subset2) having (2^{n/2}) elements each then this will cost O(2^{n/2}*log2^{n/2}) ≈ O(n*(2^{n/2})) Time complexity. In next step we traverse one vector subset1 with 2^{n/2} elements and find the upper bound of k/subset1[i] in second vector which will tell us the count of total elements whose products will be less than or equal to k. And thus for each element in subset1 we will try to perform a binary search in form of upper_bound in subset2 resulting again a Time complexity of O(n*(2^{n/2})). So, if we try to compute our overall complexity for this approach we will have O(n*(2^{n/2}) + n*(2^{n/2}) + n*(2^{n/2})) ≈ O(n*(2^{n/2})) as our time complexity which is much efficient than our brute force approach.

**Algorithm :**

- Divide array into two equal parts.
- Generate all subsets and for each subset calculate product of elements and push this to a vector. try this for both part of array.
- Sort both new vector which contains products of elements for each possible subsets.
- Traverse any one vector and find upper-bound of element k/vector[i] to find how many subsets are there for vector[i] whose product of elements is less than k.

**Some key points to improve complexity :**

- Ignore elements from array if greater than k.
- Ignore product of elements to push into vector (subset1 or subset2) if greater than k.

## C++

`// CPP to find the count subset having product ` `// less than k ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `findSubset(` `long` `long` `int` `arr[], ` `int` `n, ` ` ` `long` `long` `int` `k) ` `{ ` ` ` `// declare four vector for dividing array into ` ` ` `// two halves and storing product value of ` ` ` `// possible subsets for them ` ` ` `vector<` `long` `long` `int` `> vect1, vect2, subset1, subset2; ` ` ` ` ` `// ignore element greater than k and divide ` ` ` `// array into 2 halves ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// ignore element if greater than k ` ` ` `if` `(arr[i] > k) ` ` ` `continue` `; ` ` ` `if` `(i <= n / 2) ` ` ` `vect1.push_back(arr[i]); ` ` ` `else` ` ` `vect2.push_back(arr[i]); ` ` ` `} ` ` ` ` ` `// generate all subsets for 1st half (vect1) ` ` ` `for` `(` `int` `i = 0; i < (1 << vect1.size()); i++) { ` ` ` `long` `long` `value = 1; ` ` ` `for` `(` `int` `j = 0; j < vect1.size(); j++) { ` ` ` `if` `(i & (1 << j)) ` ` ` `value *= vect1[j]; ` ` ` `} ` ` ` ` ` `// push only in case subset product is less ` ` ` `// than equal to k ` ` ` `if` `(value <= k) ` ` ` `subset1.push_back(value); ` ` ` `} ` ` ` ` ` `// generate all subsets for 2nd half (vect2) ` ` ` `for` `(` `int` `i = 0; i < (1 << vect2.size()); i++) { ` ` ` `long` `long` `value = 1; ` ` ` `for` `(` `int` `j = 0; j < vect2.size(); j++) { ` ` ` `if` `(i & (1 << j)) ` ` ` `value *= vect2[j]; ` ` ` `} ` ` ` ` ` `// push only in case subset product is ` ` ` `// less than equal to k ` ` ` `if` `(value <= k) ` ` ` `subset2.push_back(value); ` ` ` `} ` ` ` ` ` `// sort subset2 ` ` ` `sort(subset2.begin(), subset2.end()); ` ` ` ` ` `long` `long` `count = 0; ` ` ` `for` `(` `int` `i = 0; i < subset1.size(); i++) ` ` ` `count += upper_bound(subset2.begin(), subset2.end(), ` ` ` `(k / subset1[i])) - subset2.begin(); ` ` ` ` ` `// for null subset decrement the value of count ` ` ` `count--; ` ` ` ` ` `// return count ` ` ` `return` `count; ` `} ` ` ` `// driver program ` `int` `main() ` `{ ` ` ` `long` `long` `int` `arr[] = { 4, 2, 3, 6, 5 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `long` `long` `int` `k = 25; ` ` ` `cout << findSubset(arr, n, k); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 to find the count subset ` `# having product less than k ` `import` `bisect ` ` ` `def` `findSubset(arr, n, k): ` ` ` ` ` `# declare four vector for dividing ` ` ` `# array into two halves and storing ` ` ` `# product value of possible subsets ` ` ` `# for them ` ` ` `vect1, vect2, subset1, subset2 ` `=` `[], [], [], [] ` ` ` ` ` `# ignore element greater than k and ` ` ` `# divide array into 2 halves ` ` ` `for` `i ` `in` `range` `(` `0` `, n): ` ` ` ` ` `# ignore element if greater than k ` ` ` `if` `arr[i] > k: ` ` ` `continue` ` ` `if` `i <` `=` `n ` `/` `/` `2` `: ` ` ` `vect1.append(arr[i]) ` ` ` `else` `: ` ` ` `vect2.append(arr[i]) ` ` ` ` ` `# generate all subsets for 1st half (vect1) ` ` ` `for` `i ` `in` `range` `(` `0` `, (` `1` `<< ` `len` `(vect1))): ` ` ` `value ` `=` `1` ` ` `for` `j ` `in` `range` `(` `0` `, ` `len` `(vect1)): ` ` ` `if` `i & (` `1` `<< j): ` ` ` `value ` `*` `=` `vect1[j] ` ` ` ` ` `# push only in case subset product ` ` ` `# is less than equal to k ` ` ` `if` `value <` `=` `k: ` ` ` `subset1.append(value) ` ` ` ` ` `# generate all subsets for 2nd half (vect2) ` ` ` `for` `i ` `in` `range` `(` `0` `, (` `1` `<< ` `len` `(vect2))): ` ` ` `value ` `=` `1` ` ` `for` `j ` `in` `range` `(` `0` `, ` `len` `(vect2)): ` ` ` `if` `i & (` `1` `<< j): ` ` ` `value ` `*` `=` `vect2[j] ` ` ` ` ` `# push only in case subset product ` ` ` `# is less than equal to k ` ` ` `if` `value <` `=` `k: ` ` ` `subset2.append(value) ` ` ` ` ` `# sort subset2 ` ` ` `subset2.sort() ` ` ` ` ` `count ` `=` `0` ` ` `for` `i ` `in` `range` `(` `0` `, ` `len` `(subset1)): ` ` ` `count ` `+` `=` `bisect.bisect(subset2, (k ` `/` `/` `subset1[i])) ` ` ` ` ` `# for null subset decrement the ` ` ` `# value of count ` ` ` `count ` `-` `=` `1` ` ` ` ` `# return count ` ` ` `return` `count ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `arr ` `=` `[` `4` `, ` `2` `, ` `3` `, ` `6` `, ` `5` `] ` ` ` `n ` `=` `len` `(arr) ` ` ` `k ` `=` `25` ` ` `print` `(findSubset(arr, n, k)) ` ` ` `# This code is contributed by Rituraj Jain ` |

*chevron_right*

*filter_none*

**Output :**

15

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Maximum product from array such that frequency sum of all repeating elements in product is less than or equal to 2 * k
- Count of subsets having sum of min and max element less than K
- Number of subarrays having product less than K
- Count the number of words having sum of ASCII values less than and greater than k
- Count all subsequences having product less than K
- Count pairs in a sorted array whose product is less than k
- Numbers less than N which are product of exactly two distinct prime numbers
- Sum and Product of all the nodes which are less than K in the linked list
- Find all possible subarrays having product less than or equal to K
- Partition an array of non-negative integers into two subsets such that average of both the subsets is equal
- Sum of subsets of all the subsets of an array | O(3^N)
- Sum of subsets of all the subsets of an array | O(2^N)
- Sum of subsets of all the subsets of an array | O(N)
- Divide array in two Subsets such that sum of square of sum of both subsets is maximum
- Count of alphabets having ASCII value less than and greater than k
- Sum of all array elements less than X and greater than Y for Q queries
- Number of elements less than or equal to a number in a subarray : MO's Algorithm
- Total number of subsets in which the product of the elements is even
- Split Array into min number of subsets with difference between each pair greater than 1
- Length of longest subarray in which elements greater than K are more than elements not greater than K

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.