Count of subsequences in an array with sum less than or equal to X

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Given an integer array arr[] of size N and an integer X, the task is to count the number of subsequences in that array such that its sum is less than or equal to X.
Note: 1 <= N <= 1000 and 1 <= X <= 1000, where N is the size of the array.

Examples:

Input : arr[] = {84, 87, 73}, X = 100
Output : 3
Explanation: The three subsequences with sum less than or equal to 100 are {84}, {87} and {73}.

Input : arr[] = {25, 13, 40}, X = 50
Output : 4
Explanation: The four subsequences with sum less than or equal to 50 are {25}, {13}, {40} and {25, 13}.

Naive Approach: Generate all the subsequences of the array and check if the sum is less than or equal to X.
Time complexity:O(2^N)

Efficient Approach: Generate the count of subsequences using Dynamic Programming. In order to solve the problem, follow the steps below:

For any index ind, if arr[ind] ? X then, the count of subsequences including as well as excluding the element at the current index:

countSubsequence(ind, X) = countSubsequence(ind + 1, X) (excluding) + countSubsequence(ind + 1, X – arr[ind]) (including)

Else, count subsequences excluding the current index:

countSubsequence(ind, X) = countSubsequence(ind + 1, X) (excluding)

Finally, subtract 1 from the final count returned by the function as it also counts an empty subsequence.

Below is the implementation of the above approach:

C++

// C++ Program to count number
// of subsequences in an array
// with sum less than or equal to X
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to return the count
// of subsequence in an array with sum
// less than or equal to X
int countSubsequenceUtil(
    int ind, int sum,
    int* A, int N,
    vector<vector<int> >& dp)
{
    // Base condition
    if (ind == N)
        return 1;
 
    // Return if the sub-problem
    // is already calculated
    if (dp[ind][sum] != -1)
        return dp[ind][sum];
 
    // Check if the current element is
    // less than or equal to sum
    if (A[ind] <= sum) {
        // Count subsequences excluding
        // the current element
        dp[ind][sum]
            = countSubsequenceUtil(
                  ind + 1,
                  sum, A, N, dp)
              +
 
              // Count subsequences including
              // the current element
              countSubsequenceUtil(
                  ind + 1,
                  sum - A[ind],
                  A, N, dp);
    }
 
    else {
        // Exclude current element
        dp[ind][sum]
            = countSubsequenceUtil(
                ind + 1,
                sum, A,
                N, dp);
    }
 
    // Return the result
    return dp[ind][sum];
}
 
// Function to return the count of subsequence
// in an array with sum less than or equal to X
int countSubsequence(int* A, int N, int X)
{
    // Initialize a DP array
    vector<vector<int> > dp(
        N,
        vector<int>(X + 1, -1));
 
    // Return the result
    return countSubsequenceUtil(0, X, A,
                                N, dp)
           - 1;
}
 
// Driver Code
int main()
{
    int arr[] = { 25, 13, 40 }, X = 50;
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << countSubsequence(arr, N, X);
 
    return 0;
}

Java

// Java program to count number
// of subsequences in an array
// with sum less than or equal to X
class GFG{
 
// Utility function to return the count
// of subsequence in an array with sum
// less than or equal to X
static int countSubsequenceUtil(int ind, int sum,
                                int []A, int N,
                                int [][]dp)
{
     
    // Base condition
    if (ind == N)
        return 1;
 
    // Return if the sub-problem
    // is already calculated
    if (dp[ind][sum] != -1)
        return dp[ind][sum];
 
    // Check if the current element is
    // less than or equal to sum
    if (A[ind] <= sum) 
    {
         
        // Count subsequences excluding
        // the current element
        dp[ind][sum] = countSubsequenceUtil(
                           ind + 1, sum, 
                           A, N, dp) +
                        
                       // Count subsequences 
                       // including the current
                       // element
                       countSubsequenceUtil(
                           ind + 1, 
                           sum - A[ind], 
                           A, N, dp);
    }
    else
    {
         
        // Exclude current element
        dp[ind][sum] = countSubsequenceUtil(
                           ind + 1, sum,
                           A, N, dp);
    }
 
    // Return the result
    return dp[ind][sum];
}
 
// Function to return the count of subsequence
// in an array with sum less than or equal to X
static int countSubsequence(int[] A, int N, int X)
{
     
    // Initialize a DP array
    int [][]dp = new int[N][X + 1];
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < X + 1; j++)
        {
            dp[i][j] = -1;
        }
    }
     
    // Return the result
    return countSubsequenceUtil(0, X, A,
                                N, dp) - 1;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 25, 13, 40 }, X = 50;
    int N = arr.length;
 
    System.out.print(countSubsequence(arr, N, X));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python program for the above approach:
 
## Utility function to return the count
## of subsequence in an array with sum
## less than or equal to X
def countSubsequenceUtil(ind, s, A, N, dp):
 
    ## Base condition
    if (ind == N):
        return 1
 
    ## Return if the sub-problem
    ## is already calculated
    if (dp[ind][s] != -1):
        return dp[ind][s]
 
    ## Check if the current element is
    ## less than or equal to sum
    if (A[ind] <= s):
         
        ## Count subsequences excluding
        ## the current element
        ## Also, Count subsequences including
        ## the current element
        dp[ind][s] = countSubsequenceUtil(ind + 1, s, A, N, dp) + countSubsequenceUtil(ind + 1, s - A[ind], A, N, dp)
 
             
 
    else:
        ## Exclude current element
        dp[ind][s] = countSubsequenceUtil(ind + 1, s, A, N, dp)
 
    ## Return the result
    return dp[ind][s]
 
## Function to return the count of subsequence
## in an array with sum less than or equal to X
def countSubsequence(A, N, X):
 
    ## Initialize a DP array
    dp = [[-1 for _ in range(X + 1)] for i in range(N)]
 
    ## Return the result
    return countSubsequenceUtil(0, X, A, N, dp) - 1
 
 
## Driver code
if __name__=='__main__':
 
    arr = [25, 13, 40]
    X = 50
 
    N = len(arr)
 
    print(countSubsequence(arr, N, X))

C#

// C# program to count number
// of subsequences in an array
// with sum less than or equal to X
using System;
 
class GFG{
 
// Utility function to return the count
// of subsequence in an array with sum
// less than or equal to X
static int countSubsequenceUtil(int ind, int sum,
                                int []A, int N,
                                int [,]dp)
{
     
    // Base condition
    if (ind == N)
        return 1;
 
    // Return if the sub-problem
    // is already calculated
    if (dp[ind, sum] != -1)
        return dp[ind, sum];
 
    // Check if the current element is
    // less than or equal to sum
    if (A[ind] <= sum) 
    {
         
        // Count subsequences excluding
        // the current element
        dp[ind, sum] = countSubsequenceUtil(
                           ind + 1, sum, 
                           A, N, dp) +
                            
                       // Count subsequences 
                       // including the current
                       // element
                       countSubsequenceUtil(
                           ind + 1, 
                           sum - A[ind], 
                           A, N, dp);
    }
    else
    {
         
        // Exclude current element
        dp[ind, sum] = countSubsequenceUtil(
                           ind + 1, sum,
                           A, N, dp);
    }
 
    // Return the result
    return dp[ind, sum];
}
 
// Function to return the count of subsequence
// in an array with sum less than or equal to X
static int countSubsequence(int[] A, int N, int X)
{
     
    // Initialize a DP array
    int [,]dp = new int[N, X + 1];
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < X + 1; j++)
        {
            dp[i, j] = -1;
        }
    }
     
    // Return the result
    return countSubsequenceUtil(0, X, A,
                                N, dp) - 1;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 25, 13, 40 };
    int X = 50;
    int N = arr.Length;
 
    Console.Write(countSubsequence(arr, N, X));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// JavaScript program to count number
// of subsequences in an array
// with sum less than or equal to X
 
// Utility function to return the count
// of subsequence in an array with sum
// less than or equal to X
function countSubsequenceUtil(ind, sum, A, N, dp)
{
     
    // Base condition
    if (ind == N)
        return 1;
 
    // Return if the sub-problem
    // is already calculated
    if (dp[ind][sum] != -1)
        return dp[ind][sum];
 
    // Check if the current element is
    // less than or equal to sum
    if (A[ind] <= sum) 
    {
         
        // Count subsequences excluding
        // the current element
        dp[ind][sum] = countSubsequenceUtil(
                           ind + 1, sum, 
                           A, N, dp) +
                        
                       // Count subsequences 
                       // including the current
                       // element
                       countSubsequenceUtil(
                           ind + 1, 
                           sum - A[ind], 
                           A, N, dp);
    }
    else
    {
         
        // Exclude current element
        dp[ind][sum] = countSubsequenceUtil(
                           ind + 1, sum,
                           A, N, dp);
    }
 
    // Return the result
    return dp[ind][sum];
}
 
// Function to return the count of subsequence
// in an array with sum less than or equal to X
function countSubsequence(A, N, X)
{
     
    // Initialize a DP array
    let dp = new Array(N);
    for(var i = 0; i < dp.length; i++)
    {
        dp[i] = new Array(2);
    }
 
    for(let i = 0; i < N; i++)
    {
        for(let j = 0; j < X + 1; j++)
        {
            dp[i][j] = -1;
        }
    }
     
    // Return the result
    return countSubsequenceUtil(0, X, A,
                                N, dp) - 1;
}
   
// Driver Code
let arr = [ 25, 13, 40 ], X = 50;
let N = arr.length;
 
document.write(countSubsequence(arr, N, X));
 
// This code is contributed by susmitakundugoaldanga
 
</script>

Output

Time Complexity: O(N*X)
Auxiliary Space: O(N*X)

Efficient approach: Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

Create a DP to store the solution of the subproblems.
Initialize the DP with base cases when index = n then dp[i][j] = 1.
Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
Return the final solution stored in dp[0][x] – 1.

Implementation :

C++

// C++ program for above approach
 
#include <bits/stdc++.h>
using namespace std;
 
int countSubsequence(int* A, int N, int X)
{
    // Initialize a DP array
    int dp[N+1][X+1];
    memset(dp, 0, sizeof(dp));
     
    // Set Base Case
    for(int i=0 ; i<=N ;i++){
        for(int j=0 ;j<=X ; j++){
            if(i==N){
                dp[i][j] = 1;
            }
        }
    }
 
    // Fill the DP table
    // iterate over subproblems and get the current 
    // solution for previous computations
    for(int i=N-1; i>=0; i--) {
        for(int j=1; j<=X; j++) {
             
            // update current value
            if(A[i] <= j) { // Fixed index here
                dp[i][j] = dp[i+1][j] + dp[i+1][j-A[i]];
            } else {
                dp[i][j] = dp[i+1][j];
            }
        }
    }
 
    // Return the result
    return dp[0][X] -1;
}
 
// Driver Code
int main()
{
    int arr[] = { 25, 13, 40 }, X = 50;
    int N = sizeof(arr) / sizeof(arr[0]);
     
    // function call
    cout << countSubsequence(arr, N, X);
    return 0;
}
 
// This code is contributed by bhardwajji.

Java

import java.util.*;
 
public class Main {
  public static void main(String[] args) {
    int[] arr = {25, 13, 40};
    int X = 50;
    int N = arr.length;
    System.out.println(countSubsequence(arr, N, X));
  }
 
  public static int countSubsequence(int[] A, int N, int X) {
    // Initialize a DP array
    int[][] dp = new int[N+1][X+1];
    for (int[] row : dp) {
      Arrays.fill(row, 0);
    }
 
    // Set Base Case
    for (int i = 0; i <= N; i++) {
      for (int j = 0; j <= X; j++) {
        if (i == N) {
          dp[i][j] = 1;
        }
      }
    }
 
    // Fill the DP table
    // iterate over subproblems and get the current 
    // solution for previous computations
    for (int i = N-1; i >= 0; i--) {
      for (int j = 1; j <= X; j++) {
        // update current value
        if (A[i] <= j) {
          dp[i][j] = dp[i+1][j] + dp[i+1][j-A[i]];
        } else {
          dp[i][j] = dp[i+1][j];
        }
      }
    }
 
    // Return the result
    return dp[0][X] - 1;
  }
}

Python3

def countSubsequence(A, N, X):
    # Initialize a DP array
    dp = [[0 for j in range(X+1)] for i in range(N+1)]
 
    # Set Base Case
    for i in range(N+1):
        for j in range(X+1):
            if i == N:
                dp[i][j] = 1
 
    # Fill the DP table
    # iterate over subproblems and get the current
    # solution for previous computations
    for i in range(N-1, -1, -1):
        for j in range(1, X+1):
 
            # update current value
            if A[i] <= j:
                dp[i][j] = dp[i+1][j] + dp[i+1][j-A[i]]
            else:
                dp[i][j] = dp[i+1][j]
 
    # Return the result
    return dp[0][X] - 1
 
 
# Driver Code
arr = [25, 13, 40]
X = 50
N = len(arr)
 
# function call
print(countSubsequence(arr, N, X))

C#

using System;
 
class Program {
    // Function to count subsequences of an array with sum X
    static int CountSubsequence(int[] A, int N, int X)
    {
        // Initialize a DP array
        int[, ] dp = new int[N + 1, X + 1];
        for (int i = 0; i <= N; i++) {
            for (int j = 0; j <= X; j++) {
                dp[i, j] = 0;
            }
        }
 
        // Set Base Case
        for (int i = 0; i <= N; i++) {
            for (int j = 0; j <= X; j++) {
                if (i == N) {
                    dp[i, j] = 1;
                }
            }
        }
 
        // Fill the DP table
        // iterate over subproblems and get the current
        // solution for previous computations
        for (int i = N - 1; i >= 0; i--) {
            for (int j = 1; j <= X; j++) {
 
                // update current value
                if (A[i] <= j) // Fixed index here
                {
                    dp[i, j] = dp[i + 1, j]
                               + dp[i + 1, j - A[i]];
                }
                else {
                    dp[i, j] = dp[i + 1, j];
                }
            }
        }
 
        // Return the result
        return dp[0, X] - 1;
    }
 
    static void Main(string[] args)
    {
        int[] arr = { 25, 13, 40 };
        int X = 50;
        int N = arr.Length;
 
        // function call
        Console.WriteLine(CountSubsequence(arr, N, X));
    }
}

Javascript

function countSubsequence(A, N, X) {
    // Initialize a DP array
    const dp = Array.from({ length: N + 1 }, () => Array(X + 1).fill(0));
 
    // Set Base Case
    for (let i = 0; i <= N; i++) {
        for (let j = 0; j <= X; j++) {
            if (i === N) {
                dp[i][j] = 1;
            }
        }
    }
 
    // Fill the DP table
    // iterate over subproblems and get the current
    // solution for previous computations
    for (let i = N - 1; i >= 0; i--) {
        for (let j = 1; j <= X; j++) {
 
            // update current value
            if (A[i] <= j) {
                dp[i][j] = dp[i + 1][j] + dp[i + 1][j - A[i]];
            } else {
                dp[i][j] = dp[i + 1][j];
            }
        }
    }
 
    // Return the result
    return dp[0][X] - 1;
}
 
// Driver Code
const arr = [25, 13, 40];
const X = 50;
const N = arr.length;
 
// function call
console.log(countSubsequence(arr, N, X));
 
// This code is contributed by Samim Hossain Mondal.

Output

Time Complexity: O(N*X)
Auxiliary Space: O(N*X)

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