Number of subsets with product less than k

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 2ⁿ subset and for each of them 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*2ⁿ). 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.

Below is the implementation of the above approach:

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;
}

Java

// Java to find the count subset having product less than k
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    static int upper_bound(List<Integer> arr, int key)
    {
        int upperBound = 0;
        while (upperBound < arr.size()) {
            if (arr.get(upperBound) <= key) {
                upperBound++;
            }
            else {
                return arr.get(upperBound);
            }
        }
        return -1;
    }
 
    static int findSubset(int[] arr, int n, int k)
    {
        // declare four vector for dividing array into two
        // halves and storing product value of possible
        // subsets for them
        List<Integer> vect1 = new ArrayList<>();
        List<Integer> vect2 = new ArrayList<>();
        List<Integer> subset1 = new ArrayList<>();
        List<Integer> subset2 = new ArrayList<>();
 
        // 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.add(arr[i]);
            }
            else {
                vect2.add(arr[i]);
            }
        }
 
        // generate all subsets for 1st half (vect1)
        for (int i = 0; i < (1 << vect1.size()); i++) {
            int value = 1;
            for (int j = 0; j < vect1.size(); j++) {
                if ((i & (1 << j)) != 0) {
                    value *= vect1.get(j);
                }
            }
            // push only in case subset product is less than
            // equal to k
            if (value <= k) {
                subset1.add(value);
            }
        }
 
        // generate all subsets for 2nd half (vect2)
        for (int i = 0; i < (1 << vect2.size()); i++) {
            int value = 1;
            for (int j = 0; j < vect2.size(); j++) {
                if ((i & (1 << j)) != 0) {
                    value *= vect2.get(j);
                }
            }
            // push only in case subset product is less than
            // equal to k
            if (value <= k) {
                subset2.add(value);
            }
        }
 
        // sort subset2
        Collections.sort(subset2);
 
        int count = 0;
        for (int i = 0; i < subset1.size(); i++) {
            count += upper_bound(subset2,
                                 (k / subset1.get(i)));
        }
        // for null subset decrement the value of count
        count--;
 
        // return count
        return count;
    }
 
    public static void main(String[] args)
    {
        int[] arr = { 4, 2, 3, 6, 5 };
        int n = arr.length;
        int k = 25;
        System.out.print(findSubset(arr, n, k));
    }
}
 
// This code is contributed by lokeshmvs21.

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

C#

// C# to find the count subset having product less than k
 
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG {
 
  static int upper_bound(List<int> arr, int key)
  {
    int upperBound = 0;
    while (upperBound < arr.Count) {
      if (arr[upperBound] <= key) {
        upperBound++;
      }
      else {
        return arr[upperBound];
      }
    }
    return -1;
  }
 
  static int findSubset(int[] arr, int n, int k)
  {
    // declare four vector for dividing array into two
    // halves and storing product value of possible
    // subsets for them
    List<int> vect1 = new List<int>();
    List<int> vect2 = new List<int>();
    List<int> subset1 = new List<int>();
    List<int> subset2 = new List<int>();
 
    // 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.Add(arr[i]);
      }
      else {
        vect2.Add(arr[i]);
      }
    }
 
    // generate all subsets for 1st half (vect1)
    for (int i = 0; i < (1 << vect1.Count); i++) {
      int value = 1;
      for (int j = 0; j < vect1.Count; j++) {
        if ((i & (1 << j)) != 0) {
          value *= vect1[j];
        }
      }
      // push only in case subset product is less than
      // equal to k
      if (value <= k) {
        subset1.Add(value);
      }
    }
 
    // generate all subsets for 2nd half (vect2)
    for (int i = 0; i < (1 << vect2.Count); i++) {
      int value = 1;
      for (int j = 0; j < vect2.Count; j++) {
        if ((i & (1 << j)) != 0) {
          value *= vect2[j];
        }
      }
      // push only in case subset product is less than
      // equal to k
      if (value <= k) {
        subset2.Add(value);
      }
    }
 
    // sort subset2
    subset2.Sort();
 
    int count = 0;
    for (int i = 0; i < subset1.Count; i++) {
      count += upper_bound(subset2, (k / subset1[i]));
    }
    // for null subset decrement the value of count
    count--;
 
    // return count
    return count;
  }
 
  public static void Main(string[] args)
  {
    int[] arr = { 4, 2, 3, 6, 5 };
    int n = arr.Length;
    int k = 25;
    Console.Write(findSubset(arr, n, k));
  }
}
 
// This code is contributed by karandeep1234.

Javascript

// JavaScript to find the count subset
// having product less than k
function upper_bound(arr, key) {
    let upperBound = 0;
    while (upperBound < arr.length) {
        if (arr[upperBound] <= key) {
        upperBound++;
        } else {
            return arr[upperBound];
        }
    }
    return -1;
}
 
function findSubset(arr, n, k)
{
 
    // declare four vector for dividing
    // array into two halves and storing
    // product value of possible subsets
    // for them
    let vect1 = [];
    let vect2 = [];
    let subset1 = [];
    let subset2 = [];
 
    // ignore element greater than k
    // and divide array into 2 halves
    for (let i = 0; i < n; i++)
    {
     
        // ignore element if greater than k
        if (arr[i] > k) {
            continue;
        }
        if (i <= n / 2) {
            vect1.push(arr[i]);
        } else {
            vect2.push(arr[i]);
        }
    }
 
    // generate all subsets for 1st half (vect1)
    for (let i = 0; i < (1 << vect1.length); i++) {
        let value = 1;
        for (let j = 0; j < vect1.length; j++) {
            if ((i & (1 << j)) !== 0) {
                value *= vect1[j];
            }
        }
         
        // push only in case subset product
        // is less than equal to k
        if (value <= k) {
            subset1.push(value);
        }
    }
 
    // generate all subsets for 2nd half (vect2)
    for (let i = 0; i < (1 << vect2.length); i++) {
        let value = 1;
        for (let j = 0; j < vect2.length; j++) {
            if ((i & (1 << j)) !== 0) {
                value *= vect2[j];
            }
        }
        // push only in case subset product
        // is less than equal to k
        if (value <= k) {
            subset2.push(value);
        }
    }
 
    // sort subset2
    subset2.sort(function(a, b) {
    return a - b;
    });
 
    let count = 0;
    for (let i = 0; i < subset1.length; i++) {
        count += upper_bound(subset2, k / subset1[i]);
    }
     
    // for null subset decrement the value of count
    count--;
 
    // return count
    return count;
}
 
let arr = [4, 2, 3, 6, 5];
let n = arr.length;
let k = 25;
console.log(findSubset(arr, n, k));
 
// This code is contributed by lokesh.

Output

Last Updated : 21 Mar, 2023

Number of subsets with product less than k

C++

Java

Python3

C#

Javascript

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