Given an array A[] of length N and integer F, the task is to find the number of subsequences where the average of the sum of the square of elements (of that particular subsequence) is equal to the value F.
Examples:
Input: A[] = {1, 2, 1, 2}, F = 2
Output: 2
Explanation: Two subsets with value F = 2 are {2} and {2}. Where 2^2/1 = 2Input: A[] = {1, 1, 1, 1}, F = 1
Output: 15
Explanation: All the subsets will return the function value of 1 except the empty subset. Hence the total number of subsequences will be 2 ^ 4 – 1 = 15.
Approach: This can be solved using the following idea:
Using Dynamic programming, we can reduce the overlapping of subsets that are already computed.
Follow the steps below to solve the problem:
- Initialize a 3D array, say dp[][][], where dp[i][k][f] indicates the number of ways to select k integers from first i values such that their sum of squares or any other function according to F is stored in dp[i][k][f]
- Traverse the array, we still have 2 options for each element i.e. to include or to exclude. The transition will be as shown at the end:
- If included then we will have k + 1 elements with functional sum value equal to s+ sq where sq is the square value i.e. arr[i]^2.
- If it is excluded we simply traverse to the next index storing the previous state in it as it is.
- Finally, the answer will be the sum of dp[N][j][F*j] as the functional value is the squared sum average.
Transitions for DP:
- Include the ith element: dp[i + 1][k + 1][s + sq] += dp[i][k][s]
- Exclude the ith element: dp[i + 1][k][s] += dp[i][k][s]
Below is the implementation of the code:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Storing the DP states// dp[i][k][f] indicates that the number// of ways to select k integers from first// i values such that their sum of squares// or any other function according to F// is stored in dp[i][k][F]int dp[101][101][1001];// Calculate the number of subsequences// with given function valueint countValue(int n, int F, vector<int> v){ // Base condition dp[0][0][0] = 1; // Three loops for three states for (int i = 0; i < n; i++) { for (int k = 0; k < n; k++) { for (int s = 0; s <= 1000; s++) { long long sq = (long long)v[i] * (long long)v[i]; // Recurrence relation // Include the ith element dp[i + 1][k + 1][s + sq] += dp[i][k][s]; // Exclude the ith element dp[i + 1][k][s] += dp[i][k][s]; } } } int cnt = 0; // Iterate over the range [1, N] for (int j = 1; j <= n; j++) { cnt += dp[n][j][F * j]; } // Return the final count return cnt;}// Driver Codeint main(){ vector<int> v = { 1, 2, 1, 2 }; int F = 2, N = v.size(); // Function call cout << countValue(N, F, v); return 0;} |
Java
import java.util.*;public class Main { // Storing the DP states // dp[i][k][f] indicates that the number // of ways to select k integers from first // i values such that their sum of squares // or any other function according to F // is stored in dp[i][k][F] static long[][][] dp = new long[101][101][1001]; // Calculate the number of subsequences // with given function value static long countValue(int n, int F, List<Integer> v) { // Base condition dp[0][0][0] = 1; // Three loops for three states for (int i = 0; i < n; i++) { for (int k = 0; k < n; k++) { for (int s = 0; s <= 1000; s++) { long sq = (long) v.get(i) * (long) v.get(i); // Recurrence relation // Include the ith element if (i + 1 <= n && k + 1 <= n && s + sq <= 1000) { dp[i + 1][k + 1][s + (int) sq] += dp[i][k][s]; } // Exclude the ith element if (i + 1 <= n && k <= n && s <= 1000) { dp[i + 1][k][s] += dp[i][k][s]; } } } } long cnt = 0; // Iterate over the range [1, N] for (int j = 1; j <= n; j++) { if (n <= 100 && j <= 100 && F * j <= 1000) { cnt += dp[n][j][F * j]; } } // Return the final count return cnt; } // Driver code public static void main(String[] args) { List<Integer> v = Arrays.asList(1, 2, 1, 2); int F = 2, N = v.size(); // Function call System.out.println(countValue(N, F, v)); }} |
Python3
# Python program for the above approach# Storing the DP states# dp[i][k][f] indicates that the number# of ways to select k integers from first# i values such that their sum of squares# or any other function according to F# is stored in dp[i][k][F]dp = [[[0 for i in range(1005)] for j in range(101)] for k in range(101)]# Calculate the number of subsequences# with given function valuedef countValue(n, F, v): # Base condition dp[0][0][0] = 1 # Three loops for three states for i in range(n): for k in range(n): for s in range(1001): sq = v[i] * v[i] # Recurrence relation # Include the ith element dp[i + 1][k + 1][s + sq] += dp[i][k][s] # Exclude the ith element dp[i + 1][k][s] += dp[i][k][s] cnt = 0 # Iterate over the range [1, N] for j in range(1, n + 1): cnt += dp[n][j][F * j] # Return the final count return cnt# Driver Codeif __name__ == '__main__': v = [1, 2, 1, 2] F, N = 2, len(v) # Function call print(countValue(N, F, v))# This code is contributed by Susobhan Akhuil |
C#
// C# program for the above approachusing System;public class GFG { // Storing the DP states // dp[i][k][f] indicates that the number // of ways to select k integers from first // i values such that their sum of squares // or any other function according to F // is stored in dp[i][k][F] static int[,,] dp = new int[101,101,1005]; // Calculate the number of subsequences // with given function value static int countValue(int n, int F, int[] v) { // Base condition dp[0,0,0] = 1; // Three loops for three states for (int i = 0; i < n; i++) { for (int k = 0; k < n; k++) { for (int s = 0; s <= 1000; s++) { long sq = (long)v[i] * (long)v[i]; // Recurrence relation // Include the ith element dp[i + 1,k + 1,s + sq] += dp[i,k,s]; // Exclude the ith element dp[i + 1,k,s] += dp[i,k,s]; } } } int cnt = 0; // Iterate over the range [1, N] for (int j = 1; j <= n; j++) { cnt += dp[n,j,F * j]; } // Return the final count return cnt; } // Driver Code public static void Main() { int[] v = { 1, 2, 1, 2 }; int F = 2, N = v.Length; // Function call Console.WriteLine(countValue(N, F, v)); }} |
Javascript
// Storing the DP states// dp[i][k][f] indicates that the number// of ways to select k integers from first// i values such that their sum of squares// or any other function according to F// is stored in dp[i][k][F]const dp = Array.from(Array(101), () => Array.from(Array(101), () => new Array(1005).fill(0)));// Calculate the number of subsequences// with given function valuefunction countValue(n, F, v) { // Base condition dp[0][0][0] = 1; // Three loops for three states for (let i = 0; i < n; i++) { for (let k = 0; k < n; k++) { for (let s = 0; s <= 1000; s++) { const sq = v[i] * v[i]; // Recurrence relation // Include the ith element dp[i + 1][k + 1][s + sq] += dp[i][k][s]; // Exclude the ith element dp[i + 1][k][s] += dp[i][k][s]; } } } let cnt = 0; // Iterate over the range [1, N] for (let j = 1; j <= n; j++) { cnt += dp[n][j][F * j]; } // Return the final count return cnt;}// Driver Codeconst v = [1, 2, 1, 2];const F = 2;const N = v.length;// Function callconsole.log(countValue(N, F, v)); |
Output
2
Time Complexity: O(F*N2)
Auxiliary Space: O(F*N2)
Efficient approach : Space optimization
In previous approach the current value dp[i][j][s] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 2D vector to store the computations.
Implementation:
C++
#include <bits/stdc++.h>using namespace std;// Calculate the number of subsequences// with given function valueint countValue(int n, int F, vector<int> v){ // Initialize the DP table vector<vector<int> > dp(n + 1, vector<int>(1001, 0)); dp[0][0] = 1; // Compute the DP table for (int i = 0; i < n; i++) { for (int k = i + 1; k >= 1; k--) { for (int s = 0; s <= 1000; s++) { int sq = v[i] * v[i]; // Recurrence relation if (s + sq <= 1000) { dp[k][s + sq] += dp[k - 1][s]; } } } } // Compute the final count int cnt = 0; for (int k = 1; k <= n; k++) { cnt += dp[k][k * F]; } return cnt;}// Driver Codeint main(){ vector<int> v = { 1, 2, 1, 2 }; int F = 2, N = v.size(); // Function call cout << countValue(N, F, v); return 0;} |
Java
import java.util.*;public class GFG { // Calculate the number of subsequences // with the given function value public static int countValue(int n, int F, ArrayList<Integer> v) { // Initialize the DP table int[][] dp = new int[n + 1][1001]; dp[0][0] = 1; // Compute the DP table for (int i = 0; i < n; i++) { for (int k = i + 1; k >= 1; k--) { for (int s = 0; s <= 1000; s++) { int sq = v.get(i) * v.get(i); // Recurrence relation if (s + sq <= 1000) { dp[k][s + sq] += dp[k - 1][s]; } } } } // Compute the final count int cnt = 0; for (int k = 1; k <= n; k++) { cnt += dp[k][k * F]; } return cnt; } // Driver Code public static void main(String[] args) { ArrayList<Integer> v = new ArrayList<>(Arrays.asList(1, 2, 1, 2)); int F = 2; int N = v.size(); // Function call System.out.println(countValue(N, F, v)); }} |
Python
# Calculate the number of subsequences# with given function valuedef countValue(n, F, v): # Initialize the DP table dp = [[0] * 1001 for _ in range(n + 1)] dp[0][0] = 1 # Compute the DP table for i in range(n): for k in range(i + 1, 0, -1): for s in range(1001): sq = v[i] * v[i] # Recurrence relation if s + sq <= 1000: dp[k][s + sq] += dp[k - 1][s] # Compute the final count cnt = 0 for k in range(1, n + 1): cnt += dp[k][k * F] return cnt# Driver Codedef main(): v = [1, 2, 1, 2] F = 2 N = len(v) # Function call result = countValue(N, F, v) print(result)if __name__ == "__main__": main() |
C#
using System;using System.Collections.Generic;public class GFG{ // Calculate the number of subsequences with a given // function value public static int CountValue(int n, int F, List<int> v) { // Initialize the DP table int[,] dp = new int[n + 1, 1001]; dp[0, 0] = 1; // Compute the DP table for (int i = 0; i < n; i++) { for (int k = i + 1; k >= 1; k--) { for (int s = 0; s <= 1000; s++) { int sq = v[i] * v[i]; // Recurrence relation if (s + sq <= 1000) { dp[k, s + sq] += dp[k - 1, s]; } } } } // Compute the final count int cnt = 0; for (int k = 1; k <= n; k++) { cnt += dp[k, k * F]; } return cnt; } public static void Main(string[] args) { List<int> v = new List<int> { 1, 2, 1, 2 }; int F = 2; int N = v.Count; // Function call int result = CountValue(N, F, v); Console.WriteLine(result); }} |
Javascript
function countValue(n, F, v) { const dp = new Array(n + 1).fill(null).map(() => new Array(1001).fill(0)); dp[0][0] = 1; // Compute the DP table for (let i = 0; i < n; i++) { for (let k = i + 1; k >= 1; k--) { for (let s = 0; s <= 1000; s++) { const sq = v[i] * v[i]; // Recurrence relation if (s + sq <= 1000) { dp[k][s + sq] += dp[k - 1][s]; } } } } // Compute the final count let cnt = 0; for (let k = 1; k <= n; k++) { cnt += dp[k][k * F]; } return cnt;}// Driver Code const v = [1, 2, 1, 2]; const F = 2; const N = v.length; // Function call console.log(countValue(N, F, v)); |
Output
2
Time Complexity: O(F*N^2)
Auxiliary Space: O(N^2)
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