Given a positive array, find the number of subsequences having product smaller than or equal to K.
Examples:
Input : [1, 2, 3, 4]
k = 10
Output :11
Explanation: The subsequences are {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {1, 2, 3}, {1, 2, 4}Input : [4, 8, 7, 2]
k = 50
Output : 9
This problem can be solved using dynamic programming where dp[i][j] = number of subsequences having product less than i using first j terms of the array. Which can be obtained by : number of subsequences using first j-1 terms + number of subsequences that can be formed using j-th term.
Below is the implementation of the above approach:
C++
// CPP program to find number of subarrays having// product less than k.#include <bits/stdc++.h>using namespace std;// Function to count numbers of such subsequences// having product less than k.int productSubSeqCount(vector<int> &arr, int k){ int n = arr.size(); int dp[k + 1][n + 1]; memset(dp, 0, sizeof(dp)); for (int i = 1; i <= k; i++) { for (int j = 1; j <= n; j++) { // number of subsequence using j-1 terms dp[i][j] = dp[i][j - 1]; // if arr[j-1] > i it will surely make product greater // thus it won't contribute then if (arr[j - 1] <= i) // number of subsequence using 1 to j-1 terms // and j-th term dp[i][j] += dp[i/arr[j-1]][j-1] + 1; } } return dp[k][n];}// Driver codeint main(){ vector<int> A; A.push_back(1); A.push_back(2); A.push_back(3); A.push_back(4); int k = 10; cout << productSubSeqCount(A, k) << endl;} |
Java
// Java program to find number of subarrays // having product less than k.import java.util.*;class CountSubsequences{ // Function to count numbers of such // subsequences having product less than k. public static int productSubSeqCount(ArrayList<Integer> arr, int k) { int n = arr.size(); int dp[][]=new int[k + 1][n + 1]; for (int i = 1; i <= k; i++) { for (int j = 1; j <= n; j++) { // number of subsequence using j-1 terms dp[i][j] = dp[i][j - 1]; // if arr[j-1] > i it will surely make // product greater thus it won't contribute // then if (arr.get(j-1) <= i && arr.get(j-1) > 0) // number of subsequence using 1 to j-1 // terms and j-th term dp[i][j] += dp[i/arr.get(j - 1)][j - 1] + 1; } } return dp[k][n]; } // Driver code public static void main(String args[]) { ArrayList<Integer> A = new ArrayList<Integer>(); A.add(1); A.add(2); A.add(3); A.add(4); int k = 10; System.out.println(productSubSeqCount(A, k)); }}// This Code is contributed by Danish Kaleem |
Python3
# Python3 program to find # number of subarrays having # product less than k. def productSubSeqCount(arr, k): n = len(arr) dp = [[0 for i in range(n + 1)] for j in range(k + 1)] for i in range(1, k + 1): for j in range(1, n + 1): # number of subsequence # using j-1 terms dp[i][j] = dp[i][j - 1] # if arr[j-1] > i it will # surely make product greater # thus it won't contribute then if arr[j - 1] <= i and arr[j - 1] > 0: # number of subsequence # using 1 to j-1 terms # and j-th term dp[i][j] += dp[i // arr[j - 1]][j - 1] + 1 return dp[k][n]# Driver code A = [1,2,3,4]k = 10print(productSubSeqCount(A, k))# This code is contributed # by pk_tautolo |
C#
// C# program to find number of subarrays // having product less than k. using System ;using System.Collections ;class CountSubsequences { // Function to count numbers of such // subsequences having product less than k. public static int productSubSeqCount(ArrayList arr, int k) { int n = arr.Count ; int [,]dp = new int[k + 1,n + 1]; for (int i = 1; i <= k; i++) { for (int j = 1; j <= n; j++) { // number of subsequence using j-1 terms dp[i,j] = dp[i,j - 1]; // if arr[j-1] > i it will surely make // product greater thus it won't contribute // then if (Convert.ToInt32(arr[j-1]) <= i && Convert.ToInt32(arr[j-1]) > 0) // number of subsequence using 1 to j-1 // terms and j-th term dp[i,j] += dp[ i/Convert.ToInt32(arr[j - 1]),j - 1] + 1; } } return dp[k,n]; } // Driver code public static void Main() { ArrayList A = new ArrayList(); A.Add(1); A.Add(2); A.Add(3); A.Add(4); int k = 10; Console.WriteLine(productSubSeqCount(A, k)); } } // This Code is contributed Ryuga |
Javascript
<script> // Javascript program to find number of subarrays // having product less than k. // Function to count numbers of such // subsequences having product less than k. function productSubSeqCount(arr, k) { let n = arr.length; let dp = new Array(k + 1); for (let i = 0; i < k + 1; i++) { dp[i] = new Array(n + 1); for (let j = 0; j < n + 1; j++) { dp[i][j] = 0; } } for (let i = 1; i <= k; i++) { for (let j = 1; j <= n; j++) { // number of subsequence using j-1 terms dp[i][j] = dp[i][j - 1]; // if arr[j-1] > i it will surely make // product greater thus it won't contribute // then if (arr[j-1] <= i && arr[j-1] > 0) // number of subsequence using 1 to j-1 // terms and j-th term dp[i][j] += dp[parseInt(i/arr[j - 1], 10)][j - 1] + 1; } } return dp[k][n]; } let A = [1, 2, 3, 4]; let k = 10; document.write(productSubSeqCount(A, k)); // This code is contributed by suresh07.</script> |
Output
11
Time Complexity: O(K*N)
Auxiliary Space: O(K*N)
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