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 |
4
Time Complexity: O(N*X)
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Improved By : Rajput-Ji
