Given a non negative array, find the number of subsequences having product smaller than K.

**Examples:**

Input : [1, 2, 3, 4] k = 10 Output :11 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.

## 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 && 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 ` `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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

11

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