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Variety of subsequences of an array with given perform worth

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Given an array A[] of size N and integer F, the duty is to search out the quantity of subsequences the place the common of the sum of the sq. of components (of that exact subsequence) is the same as the worth F.


Enter: A[] = {1, 2, 1, 2}, F = 2
Output: 2
Clarification: Two subsets with worth F = 2 are {2} and {2}. The place 2^2/1 = 2

Enter: A[] = {1, 1, 1, 1}, F = 1
Output: 15
Clarification: All of the subsets will return the perform worth of 1 besides the empty subset. Therefore the overall variety of subsequences will probably be 2 ^ 4 – 1 = 15.

Strategy: This may be solved utilizing the next concept:

Utilizing Dynamic programming, we are able to cut back the overlapping of subsets which can be already computed.

Comply with the steps under to unravel the issue:

  • Initialize a 3D array, say dp[][][], the place dp[i][k][f] signifies the variety of methods to pick out okay integers from first i values such that their sum of squares or some other perform based on F is saved in dp[i][k][f]
  • Traverse the array, we nonetheless have 2 choices for every factor i.e. to embody or to exclude. The transition will probably be as proven on the finish:
  • If included then we can have okay + 1 components with purposeful sum worth equal to s+  sq the place sq is the sq. worth i.e. arr[i]^2.
  • Whether it is excluded we merely traverse to the following index storing the earlier state in it as it’s.
  • Lastly, the reply would be the sum of dp[N][j][F*j] because the purposeful worth is the squared sum common.

Transitions for DP:

  • Embody the ith factor: dp[i + 1][k + 1][s + sq] += dp[i][k][s]
  • Exclude the ith factor: dp[i + 1][k][s] += dp[i][k][s]

Beneath is the implementation of the code:


#embody <bits/stdc++.h>

utilizing namespace std;


int dp[101][101][1001];


int countValue(int n, int F, vector<int> v)




    dp[0][0][0] = 1;



    for (int i = 0; i < n; i++) {

        for (int okay = 0; okay < n; okay++) {

            for (int s = 0; s <= 1000; s++) {

                lengthy lengthy sq

                    = (lengthy lengthy)v[i] * (lengthy lengthy)v[i];





                dp[i + 1][k + 1][s + sq] += dp[i][k][s];



                dp[i + 1][k][s] += dp[i][k][s];





    int cnt = 0;



    for (int j = 1; j <= n; j++) {

        cnt += dp[n][j][F * j];




    return cnt;



int foremost()


    vector<int> v = { 1, 2, 1, 2 };

    int F = 2, N = v.measurement();



    cout << countValue(N, F, v);

    return 0;


Time Complexity: O(F*N2)
Auxilairy House: O(F*N2)



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