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 Xint 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 Xint 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 Codeint 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 Xclass GFG{// Utility function to return the count// of subsequence in an array with sum// less than or equal to Xstatic 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 Xstatic 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 Codepublic 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 Xdef 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 Xdef 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 codeif __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 Xusing System;class GFG{// Utility function to return the count// of subsequence in an array with sum// less than or equal to Xstatic 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 Xstatic 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 Codepublic 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 Xfunction 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 Xfunction 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 Codelet arr = [ 25, 13, 40 ], X = 50;let N = arr.length;document.write(countSubsequence(arr, N, X));// This code is contributed by susmitakundugoaldanga</script> |
Output
4
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 Codeint 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 Codearr = [25, 13, 40]X = 50N = len(arr)# function callprint(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 Codeconst arr = [25, 13, 40];const X = 50;const N = arr.length;// function callconsole.log(countSubsequence(arr, N, X));// This code is contributed by Samim Hossain Mondal. |
Output
4
Time Complexity: O(N*X)
Auxiliary Space: O(N*X)
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