Count of subsequences in an array with sum less than or equal to X
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(2N)
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 |
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> |
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)) |
4
Time Complexity: O(N*X)
Auxiliary Space: O(N*X)
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