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Count all sub-sequences having product <= K – Recursive approach

  • Difficulty Level : Medium
  • Last Updated : 26 May, 2021

Given an integer K and a non-negative array arr[], the task is to find the number of sub-sequences having the product ≤ K
This problem already has a dynamic programming solution. This solution aims to provide an optimized recursive strategy to the problem.

Examples: 

Input: arr[] = { 1, 2, 3, 4 }, K = 10 
Output: 11 
The sub-sequences are {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {1, 2, 3}, {1, 2, 4}

Input: arr[] = { 4, 8, 7, 2 }, K = 50 
Output: 9

Approach: Convert the product problem to a sum problem by performing the conversions arr[i] = log(arr[i]) and K = log(K). Generate all subsets and store the sum of elements that have been taken in the sub-sequence. If at any point, the sum becomes larger than K, then we know that if we add another element to the sub-sequence, its sum will also be larger than K. Therefore, we discard all such sub-sequences that have sum larger than K without making a recursive call for them. Also, if we currently have sum less than K then we check if there are any chances to further discard any sub-sequences. If any further sub-sequences can’t be discarded, then no recursive calls are made.
Below is the implementation of the above approach: 

C++




// C++ implementation of the above approach.
#include <bits/stdc++.h>
 
#define ll long long
 
using namespace std;
 
// This variable counts discarded subsequences
ll discard_count = 0;
 
// Function to return a^n
ll power(ll a, ll n)
{
    if (n == 0)
        return 1;
    ll p = power(a, n / 2);
    p = p * p;
    if (n & 1)
        p = p * a;
    return p;
}
 
// Recursive function that counts discarded
// subsequences
void solve(int i, int n, float sum, float k,
                   float* a, float* prefix)
{
 
    // If at any stage, sum > k
    // discard further subsequences
    if (sum > k) {
        discard_count += power(2, n - i);
 
        // Recursive call terminated
        // No further calls
        return;
    }
 
    if (i == n)
        return;
 
    // rem = Sum of array[i+1...n-1]
    float rem = prefix[n - 1] - prefix[i];
 
    // If there are chances of discarding
    // further subsequences then make a
    // recursive call, otherwise not
    // Including a[i]
    if (sum + a[i] + rem > k)
        solve(i + 1, n, sum + a[i], k,
                          a, prefix);
 
    // Excluding a[i]
    if (sum + rem > k)
        solve(i + 1, n, sum, k, a, prefix);
}
 
// Function to return count of non-empty
// subsequences whose product doesn't
// exceed k
int countSubsequences(const int* arr,
                         int n, ll K)
{
    float sum = 0.0;
 
    // Converting k to log(k)
    float k = log2(K);
 
    // Prefix sum array and array to
    // store log values.
    float prefix[n], a[n];
 
    // a[] is the array obtained
    // after converting numbers to
    // logarithms
    for (int i = 0; i < n; i++) {
        a[i] = log2(arr[i]);
        sum += a[i];
    }
 
    // Computing prefix sums
    prefix[0] = a[0];
    for (int i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1] + a[i];
    }
 
    // Calculate non-empty subsequences
    // hence 1 is subtracted
    ll total = power(2, n) - 1;
 
    // If total sum is <= k, then
    // answer = 2^n - 1
    if (sum <= k) {
        return total;
    }
 
    solve(0, n, 0.0, k, a, prefix);
    return total - discard_count;
}
 
// Driver code
int main()
{
    int arr[] = { 4, 8, 7, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    ll k = 50;
    cout << countSubsequences(arr, n, k);
    return 0;
}

Java




// Java implementation of the above approach.
class GFG
{
 
// This variable counts discarded subsequences
static long discard_count = 0;
 
// Function to return a^n
static long power(long a, long n)
{
    if (n == 0)
        return 1;
    long p = power(a, n / 2);
    p = p * p;
    if (n % 2 == 1)
        p = p * a;
    return p;
}
 
// Recursive function that counts discarded
// subsequences
static void solve(int i, int n, float sum, float k,
                float []a, float []prefix)
{
 
    // If at any stage, sum > k
    // discard further subsequences
    if (sum > k)
    {
        discard_count += power(2, n - i);
 
        // Recursive calong terminated
        // No further calongs
        return;
    }
 
    if (i == n)
        return;
 
    // rem = Sum of array[i+1...n-1]
    float rem = prefix[n - 1] - prefix[i];
 
    // If there are chances of discarding
    // further subsequences then make a
    // recursive calong, otherwise not
    // Including a[i]
    if (sum + a[i] + rem > k)
        solve(i + 1, n, sum + a[i], k,
                        a, prefix);
 
    // Excluding a[i]
    if (sum + rem > k)
        solve(i + 1, n, sum, k, a, prefix);
}
 
// Function to return count of non-empty
// subsequences whose product doesn't
// exceed k
static int countSubsequences(int []arr,
                        int n, long K)
{
    float sum = 0.0f;
 
    // Converting k to log(k)
    float k = (float) Math.log(K);
 
    // Prefix sum array and array to
    // store log values.
    float []prefix = new float[n];
    float []a = new float[n];
 
    // a[] is the array obtained
    // after converting numbers to
    // logarithms
    for (int i = 0; i < n; i++)
    {
        a[i] = (float) Math.log(arr[i]);
        sum += a[i];
    }
 
    // Computing prefix sums
    prefix[0] = a[0];
    for (int i = 1; i < n; i++)
    {
        prefix[i] = prefix[i - 1] + a[i];
    }
 
    // Calculate non-empty subsequences
    // hence 1 is subtracted
    long total = power(2, n) - 1;
 
    // If total sum is <= k, then
    // answer = 2^n - 1
    if (sum <= k)
    {
        return (int) total;
    }
 
    solve(0, n, 0.0f, k, a, prefix);
    return (int) (total - discard_count);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 8, 7, 2 };
    int n = arr.length;
    long k = 50;
    System.out.print(countSubsequences(arr, n, k));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the
# above approach.
 
# From math lib import log2
from math import log2
 
# This variable counts discarded
# subsequences
discard_count = 0
 
# Function to return a^n
def power(a, n) :
     
    if (n == 0) :
        return 1
         
    p = power(a, n // 2)
    p = p * p
    if (n & 1) :
        p = p * a
    return p
 
# Recursive function that counts
# discarded subsequences
def solve(i, n, sum, k, a, prefix) :
    global discard_count
     
    # If at any stage, sum > k
    # discard further subsequences
    if (sum > k) :
        discard_count += power(2, n - i)
         
        # Recursive call terminated
        # No further calls
        return;
     
    if (i == n) :
        return
     
    # rem = Sum of array[i+1...n-1]
    rem = prefix[n - 1] - prefix[i]
     
    # If there are chances of discarding
    # further subsequences then make a
    # recursive call, otherwise not
    # Including a[i]
    if (sum + a[i] + rem > k) :
        solve(i + 1, n, sum + a[i], k, a, prefix)
     
    # Excluding a[i]
    if (sum + rem > k) :
        solve(i + 1, n, sum, k, a, prefix)
 
# Function to return count of non-empty
# subsequences whose product doesn't
# exceed k
def countSubsequences(arr, n, K) :
     
    sum = 0.0
 
    # Converting k to log(k)
    k = log2(K)
 
    # Prefix sum array and array to
    # store log values.
    prefix = [0] * n
    a = [0] * n
 
    # a[] is the array obtained after
    # converting numbers to logarithms
    for i in range(n) :
        a[i] = log2(arr[i])
        sum += a[i]
     
    # Computing prefix sums
    prefix[0] = a[0]
     
    for i in range(1, n) :
        prefix[i] = prefix[i - 1] + a[i]
 
    # Calculate non-empty subsequences
    # hence 1 is subtracted
    total = power(2, n) - 1
 
    # If total sum is <= k, then
    # answer = 2^n - 1
    if (sum <= k) :
        return total
 
    solve(0, n, 0.0, k, a, prefix)
    return total - discard_count
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 4, 8, 7, 2 ]
    n = len(arr)
    k = 50;
    print(countSubsequences(arr, n, k))
 
# This code is contributed by Ryuga

C#




// C# implementation of the above approach.
using System;
 
class GFG
{
 
// This variable counts discarded subsequences
static long discard_count = 0;
 
// Function to return a^n
static long power(long a, long n)
{
    if (n == 0)
        return 1;
    long p = power(a, n / 2);
    p = p * p;
    if (n % 2 == 1)
        p = p * a;
    return p;
}
 
// Recursive function that counts discarded
// subsequences
static void solve(int i, int n, float sum, float k,
                     float []a, float []prefix)
{
 
    // If at any stage, sum > k
    // discard further subsequences
    if (sum > k)
    {
        discard_count += power(2, n - i);
 
        // Recursive calong terminated
        // No further calongs
        return;
    }
 
    if (i == n)
        return;
 
    // rem = Sum of array[i+1...n-1]
    float rem = prefix[n - 1] - prefix[i];
 
    // If there are chances of discarding
    // further subsequences then make a
    // recursive calong, otherwise not
    // Including a[i]
    if (sum + a[i] + rem > k)
        solve(i + 1, n, sum + a[i], k,
                           a, prefix);
 
    // Excluding a[i]
    if (sum + rem > k)
        solve(i + 1, n, sum, k, a, prefix);
}
 
// Function to return count of non-empty
// subsequences whose product doesn't
// exceed k
static int countSubsequences(int []arr,
                             int n, long K)
{
    float sum = 0.0f;
 
    // Converting k to log(k)
    float k = (float) Math.Log(K);
 
    // Prefix sum array and array to
    // store log values.
    float []prefix = new float[n];
    float []a = new float[n];
 
    // []a is the array obtained
    // after converting numbers to
    // logarithms
    for (int i = 0; i < n; i++)
    {
        a[i] = (float) Math.Log(arr[i]);
        sum += a[i];
    }
 
    // Computing prefix sums
    prefix[0] = a[0];
    for (int i = 1; i < n; i++)
    {
        prefix[i] = prefix[i - 1] + a[i];
    }
 
    // Calculate non-empty subsequences
    // hence 1 is subtracted
    long total = power(2, n) - 1;
 
    // If total sum is <= k, then
    // answer = 2^n - 1
    if (sum <= k)
    {
        return (int) total;
    }
 
    solve(0, n, 0.0f, k, a, prefix);
    return (int) (total - discard_count);
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 8, 7, 2 };
    int n = arr.Length;
    long k = 50;
    Console.Write(countSubsequences(arr, n, k));
}
}
 
// This code is contributed by Rajput-Ji

PHP




<?php
// PHP implementation of the above approach.
 
// This variable counts discarded subsequences
$discard_count = 0;
 
// Function to return a^n
function power($a, $n)
{
    if ($n == 0)
        return 1;
    $p = power($a, $n / 2);
    $p = $p * $p;
    if ($n & 1)
        $p = $p * $a;
    return $p;
}
 
// Recursive function that counts discarded
// subsequences
function solve($i, $n, $sum, $k, &$a, &$prefix)
{
    global $discard_count;
 
    // If at any stage, sum > k
    // discard further subsequences
    if ($sum > $k)
    {
        $discard_count += power(2, $n - $i);
 
        // Recursive call terminated
        // No further calls
        return;
    }
 
    if ($i == $n)
        return;
 
    // rem = Sum of array[i+1...n-1]
    $rem = $prefix[$n - 1] - $prefix[$i];
 
    // If there are chances of discarding
    // further subsequences then make a
    // recursive call, otherwise not
    // Including a[i]
    if ($sum + $a[$i] + $rem > $k)
        solve($i + 1, $n, $sum + $a[$i], $k,
                               $a, $prefix);
 
    // Excluding a[i]
    if ($sum + $rem > $k)
        solve($i + 1, $n, $sum, $k, $a, $prefix);
}
 
// Function to return count of non-empty
// subsequences whose product doesn't
// exceed k
function countSubsequences(&$arr, $n, $K)
{
    global $discard_count;
    $sum = 0.0;
 
    // Converting k to log(k)
    $k = log($K, 2);
 
    // Prefix sum array and array to
    // store log values.
    $prefix = array_fill(0, $n, NULL);
    $a = array_fill(0, $n, NULL);
 
    // a[] is the array obtained after
    // converting numbers to logarithms
    for ($i = 0; $i < $n; $i++)
    {
        $a[$i] = log($arr[$i], 2);
        $sum += $a[$i];
    }
 
    // Computing prefix sums
    $prefix[0] = $a[0];
    for ($i = 1; $i < $n; $i++)
    {
        $prefix[$i] = $prefix[$i - 1] + $a[$i];
    }
 
    // Calculate non-empty subsequences
    // hence 1 is subtracted
    $total = power(2, $n) - 1;
 
    // If total sum is <= k, then
    // answer = 2^n - 1
    if ($sum <= $k)
    {
        return $total;
    }
 
    solve(0, $n, 0.0, $k, $a, $prefix);
    return $total - $discard_count;
}
 
// Driver code
$arr = array(4, 8, 7, 2 );
$n = sizeof($arr);
$k = 50;
echo countSubsequences($arr, $n, $k);
 
// This code is contributed by ita_c
?>

Javascript




<script>
// Javascript implementation of the above approach.
     
// This variable counts discarded subsequences
let discard_count = 0;
 
// Function to return a^n
function power(a,n)
{
       if (n == 0)
        return 1;
    let p = power(a, Math.floor(n / 2));
    p = p * p;
    if (n % 2 == 1)
        p = p * a;
    return p;
}
 
// Recursive function that counts discarded
// subsequences
function solve(i,n,sum,k,a,prefix)
{
    // If at any stage, sum > k
    // discard further subsequences
    if (sum > k)
    {
        discard_count += power(2, n - i);
   
        // Recursive calong terminated
        // No further calongs
        return;
    }
   
    if (i == n)
        return;
   
    // rem = Sum of array[i+1...n-1]
    let rem = prefix[n - 1] - prefix[i];
   
    // If there are chances of discarding
    // further subsequences then make a
    // recursive calong, otherwise not
    // Including a[i]
    if (sum + a[i] + rem > k)
        solve(i + 1, n, sum + a[i], k,
                        a, prefix);
   
    // Excluding a[i]
    if (sum + rem > k)
        solve(i + 1, n, sum, k, a, prefix);
}
 
// Function to return count of non-empty
// subsequences whose product doesn't
// exceed k
function countSubsequences(arr,n,K)
{
    let sum = 0.0;
   
    // Converting k to log(k)
    let k =  Math.log(K);
   
    // Prefix sum array and array to
    // store log values.
    let prefix = new Array(n);
    let a = new Array(n);
   
    // a[] is the array obtained
    // after converting numbers to
    // logarithms
    for (let i = 0; i < n; i++)
    {
        a[i] =  Math.log(arr[i]);
        sum += a[i];
    }
   
    // Computing prefix sums
    prefix[0] = a[0];
    for (let i = 1; i < n; i++)
    {
        prefix[i] = prefix[i - 1] + a[i];
    }
   
    // Calculate non-empty subsequences
    // hence 1 is subtracted
    let total = power(2, n) - 1;
   
    // If total sum is <= k, then
    // answer = 2^n - 1
    if (sum <= k)
    {
        return total;
    }
   
    solve(0, n, 0.0, k, a, prefix);
    return (total - discard_count);
}
 
// Driver code
let arr=[4, 8, 7, 2];
let n = arr.length;
let k = 50;
document.write(countSubsequences(arr, n, k));
 
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
9

 


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