# Count all subsequences having product less than K

Given a positive array, find the number of subsequences having product smaller than or equal to K.**Examples:**

Input :[1, 2, 3, 4]

k = 10Output :11Explanation: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 = 50Output :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 code` `int` `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 ` `=` `10` `print` `(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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