Related Articles
Find all possible subarrays having product less than or equal to K
• Last Updated : 11 Dec, 2020

Given an array arr[], the task is to print all possible subarrays having a product of its elements less than or equal to K.

Input: arr[] = {2, 1, 3, 4, 5, 6, 2}, K = 10
Output: [[2], [1], [2, 1], [3], [1, 3], [2, 1, 3], [4], [5], [6], [2]]
Explanation:
All possible subarrays having product ≤ K are {2}, {1}, {2, 1}, {3}, {1, 3}, {2, 1, 3}, {4}, {5}, {6}, {2}.

Input: arr[] = {2, 7, 1, 4}, K = 7
Output: [[2], [7], [1], [7, 1], [4], [1, 4]]

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays from the given array and for each subarray, check if its product is less than or equal to K or not and print accordingly.

Time Complexity: O(N3
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by observing that:

If the product of all the elements of a subarray is less than or equal to K, then all the subarrays possible from this subarray also has product less than or equal to K. Therefore, these subarrays need to be included in the answer as well.

Follow the steps below to solve the problem:

1. Initialize a pointer start pointing to the first index of the array.
2. Iterate over the array and keep calculating the product of the array elements and store it in a variable, say multi.
3. If multi exceeds K: keep dividing multi by arr[start] and keep incrementing start until multi reduces to ≤ K.
4. If multi ≤ K: Iterate from the current index to start, and store the subarrays in an Arraylist.
5. Finally, once all subarrays are generated, print the Arraylist which contains all the subarrays obtained.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include``using` `namespace` `std;` `// Function to return all possible``// subarrays having product less``// than or equal to K``vector> maxSubArray(``int` `arr[], ``int` `n,``                                ``int` `K)``{``    ` `    ``// Store the required subarrays``    ``vector> solution;``    ` `    ``// Stores the product of``    ``// current subarray``    ``int` `multi = 1;``    ` `    ``// Stores the starting index``    ``// of the current subarray``    ``int` `start = 0;``    ` `    ``// Check for empty array``    ``if` `(n <= 1 || K < 0)``    ``{``        ``return` `solution;``    ``}``    ` `    ``// Iterate over the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Calculate product``        ``multi = multi * arr[i];``    ` `        ``// If product exceeds K``        ``while` `(multi > K)``        ``{``            ` `            ``// Reduce product``            ``multi = multi / arr[start];``    ` `            ``// Increase starting index``            ``// of current subarray``            ``start++;``        ``}``    ` `        ``// Stores the subarray elements``        ``vector<``int``> list;``    ` `        ``// Store the subarray elements``        ``for``(``int` `j = i; j >= start; j--)``        ``{``            ``list.insert(list.begin(), arr[j]);``    ` `            ``// Add the subarrays``            ``// to the list``            ``solution.push_back(list);``        ``}``    ``}``    ` `    ``// Return the final``    ``// list of subarrays``    ``return` `solution;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 7, 1, 4 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 7;``    ` `    ``vector> v = maxSubArray(arr, n, K);``    ``cout << ``"["``;``    ` `    ``bool` `first = ``true``;``    ``for``(``auto` `x : v)``    ``{``        ``if` `(!first)``        ``{``            ``cout << ``", "``;``        ``}``        ``else``        ``{``            ``first = ``false``;``        ``}``        ``cout << ``"["``;``        ` `        ``bool` `ff = ``true``;``        ``for``(``int` `y : x)``        ``{``            ``if` `(!ff)``            ``{``                ``cout << ``", "``;``            ``}``            ``else``            ``{``                ``ff = ``false``;``            ``}``            ``cout << y;``        ``}``        ``cout << ``"]"``;``    ``}``    ``cout << ``"]"``;``    ` `    ``return` `0;``}` `// This code is contributed by rutvik_56`

## Java

 `// Java Program to implement``// the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``// Function to return all possible``    ``// subarrays having product less``    ``// than or equal to K``    ``public` `static` `List > maxSubArray(``        ``int``[] arr, ``int` `K)``    ``{` `        ``// Store the required subarrays``        ``List > solution``            ``= ``new` `ArrayList<>();` `        ``// Stores the product of``        ``// current subarray``        ``int` `multi = ``1``;` `        ``// Stores the starting index``        ``// of the current subarray``        ``int` `start = ``0``;` `        ``// Check for empty array``        ``if` `(arr.length <= ``1` `|| K < ``0``) {``            ``return` `new` `ArrayList<>();``        ``}` `        ``// Iterate over the array``        ``for` `(``int` `i = ``0``; i < arr.length; i++) {` `            ``// Calculate product``            ``multi = multi * arr[i];` `            ``// If product exceeds K``            ``while` `(multi > K) {` `                ``// Reduce product``                ``multi = multi / arr[start];` `                ``// Increase starting index``                ``// of current subarray``                ``start++;``            ``}` `            ``// Stores the subarray elements``            ``List list``                ``= ``new` `ArrayList<>();` `            ``// Store the subarray elements``            ``for` `(``int` `j = i; j >= start; j--) {` `                ``list.add(``0``, arr[j]);` `                ``// Add the subarrays``                ``// to the list``                ``solution.add(``                    ``new` `ArrayList<>(list));``            ``}``        ``}` `        ``// Return the final``        ``// list of subarrays``        ``return` `solution;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``2``, ``7``, ``1``, ``4` `};``        ``int` `K = ``7``;` `        ``System.out.println(maxSubArray(arr, K));``    ``}``}`

## Python3

 `# Python3 program to implement``# the above approach`` ` `# Function to return all possible``# subarrays having product less``# than or equal to K``def` `maxSubArray(arr, n, K):``     ` `    ``# Store the required subarrays``    ``solution ``=` `[]``     ` `    ``# Stores the product of``    ``# current subarray``    ``multi ``=` `1``     ` `    ``# Stores the starting index``    ``# of the current subarray``    ``start ``=` `0``     ` `    ``# Check for empty array``    ``if` `(n <``=` `1` `or` `K < ``0``):``        ``return` `solution``    ` `    ``# Iterate over the array``    ``for` `i ``in` `range``(n):``        ` `        ``# Calculate product``        ``multi ``=` `multi ``*` `arr[i]``     ` `        ``# If product exceeds K``        ``while` `(multi > K):``            ` `            ``# Reduce product``            ``multi ``=` `multi ``/``/` `arr[start]``     ` `            ``# Increase starting index``            ``# of current subarray``            ``start ``+``=` `1``     ` `        ``# Stores the subarray elements``        ``li ``=` `[]``        ` `        ``j ``=` `i``        ` `        ``# Store the subarray elements``        ``while``(j >``=` `start):       ``            ``li.insert(``0``, arr[j])``     ` `            ``# Add the subarrays``            ``# to the li``            ``solution.append(``list``(li))``            ``j ``-``=` `1``     ` `    ``# Return the final``    ``# li of subarrays``    ``return` `solution``    ` `# Driver Code``if` `__name__``=``=``'__main__'``:``    ` `    ``arr ``=` `[ ``2``, ``7``, ``1``, ``4` `]``    ``n ``=` `len``(arr)``    ``K ``=` `7``     ` `    ``v ``=` `maxSubArray(arr, n, K)``    ` `    ``print``(v)``     ` `# This code is contributed by pratham76`

## C#

 `// C# Program to implement``// the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG{` `// Function to return all possible``// subarrays having product less``// than or equal to K``public` `static` `List> maxSubArray(``int``[] arr,``                                          ``int` `K)``{``  ``// Store the required subarrays``  ``List > solution = ``new` `List>();` `  ``// Stores the product of``  ``// current subarray``  ``int` `multi = 1;` `  ``// Stores the starting index``  ``// of the current subarray``  ``int` `start = 0;` `  ``// Check for empty array``  ``if` `(arr.Length <= 1 || K < 0)``  ``{``    ``return` `new` `List>();``  ``}` `  ``// Iterate over the array``  ``for` `(``int` `i = 0; i < arr.Length; i++)``  ``{``    ``// Calculate product``    ``multi = multi * arr[i];` `    ``// If product exceeds K``    ``while` `(multi > K)``    ``{``      ``// Reduce product``      ``multi = multi / arr[start];` `      ``// Increase starting index``      ``// of current subarray``      ``start++;``    ``}` `    ``// Stores the subarray elements``    ``List<``int``> list = ``new` `List<``int``>();` `    ``// Store the subarray elements``    ``for` `(``int` `j = i; j >= start; j--)``    ``{``      ``list.Insert(0, arr[j]);` `      ``// Add the subarrays``      ``// to the list``      ``solution.Add(``new` `List<``int``>(list));``    ``}``  ``}` `  ``// Return the final``  ``// list of subarrays``  ``return` `solution;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``int``[] arr = {2, 7, 1, 4};``  ``int` `K = 7;``  ``List > list = maxSubArray(arr, K);``  ``foreach``(List<``int``> i ``in` `list)``  ``{``    ``Console.Write(``"["``);``    ``foreach``(``int` `j ``in` `i)``    ``{``      ``Console.Write(j);``      ``if``(i.Count > 1)``        ``Console.Write(``","``);``    ``}``    ``Console.Write(``"]"``);``  ``}``}``}` `// This code is contributed by 29AjayKumar`
Output:
`[[2], [7], [1], [7, 1], [4], [1, 4]]`

Time Complexity: O(N2)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up