Count of subsets with sum equal to X - GeeksforGeeks

29 Mar, 2022
Given an array arr[] of length N and an integer X, the task is to find the number of subsets with a sum equal to X.
Examples: 
Input: arr[] = {1, 2, 3, 3}, X = 6 Output: 3 All the possible subsets are {1, 2, 3}, {1, 2, 3} and {3, 3}
Input: arr[] = {1, 1, 1, 1}, X = 1 Output: 4 
Approach: A simple approach is to solve this problem by generating all the possible subsets and then checking whether the subset has the required sum. This approach will have exponential time complexity. However, for smaller values of X and array elements, this problem can be solved using dynamic programming. Let’s look at the recurrence relation first. 
This method is valid for all the integers.
dp[i][C] = dp[i – 1][C – arr[i]] + dp[i – 1][C] 
Let’s understand the states of the DP now. Here, dp[i][C] stores the number of subsets of the sub-array arr[i...N-1] such that their sum is equal to C. Thus, the recurrence is very trivial as there are only two choices i.e. either consider the ith element in the subset or don’t.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; #define maxN 20#define maxSum 50#define minSum 50#define base 50 // To store the states of DPint dp[maxN][maxSum + minSum];bool v[maxN][maxSum + minSum]; // Function to return the required countint findCnt(int* arr, int i, int required_sum, int n){    // Base case    if (i == n) {        if (required_sum == 0)            return 1;        else            return 0;    }     // If the state has been solved before    // return the value of the state    if (v[i][required_sum + base])        return dp[i][required_sum + base];     // Setting the state as solved    v[i][required_sum + base] = 1;     // Recurrence relation    dp[i][required_sum + base]        = findCnt(arr, i + 1, required_sum, n)          + findCnt(arr, i + 1, required_sum - arr[i], n);    return dp[i][required_sum + base];} // Driver codeint main(){    int arr[] = { 3, 3, 3, 3 };    int n = sizeof(arr) / sizeof(int);    int x = 6;     cout << findCnt(arr, 0, x, n);     return 0;}
// Java implementation of the approachimport java.util.*; class GFG{static int maxN = 20;static int maxSum = 50;static int minSum = 50;static int base = 50; // To store the states of DPstatic int [][]dp = new int[maxN][maxSum + minSum];static boolean [][]v = new boolean[maxN][maxSum + minSum]; // Function to return the required countstatic int findCnt(int []arr, int i,                   int required_sum, int n){    // Base case    if (i == n)    {        if (required_sum == 0)            return 1;        else            return 0;    }     // If the state has been solved before    // return the value of the state    if (v[i][required_sum + base])        return dp[i][required_sum + base];     // Setting the state as solved    v[i][required_sum + base] = true;     // Recurrence relation    dp[i][required_sum + base] =          findCnt(arr, i + 1, required_sum, n) +          findCnt(arr, i + 1, required_sum - arr[i], n);    return dp[i][required_sum + base];} // Driver codepublic static void main(String []args){    int arr[] = { 3, 3, 3, 3 };    int n = arr.length;    int x = 6;     System.out.println(findCnt(arr, 0, x, n));}} // This code is contributed by 29AjayKumar
# Python3 implementation of the approachimport numpy as np maxN = 20maxSum = 50minSum = 50base = 50 # To store the states of DPdp = np.zeros((maxN, maxSum + minSum));v = np.zeros((maxN, maxSum + minSum)); # Function to return the required countdef findCnt(arr, i, required_sum, n) :     # Base case    if (i == n) :        if (required_sum == 0) :            return 1;        else :            return 0;     # If the state has been solved before    # return the value of the state    if (v[i][required_sum + base]) :        return dp[i][required_sum + base];     # Setting the state as solved    v[i][required_sum + base] = 1;     # Recurrence relation    dp[i][required_sum + base] = findCnt(arr, i + 1,                                         required_sum, n) + \                                 findCnt(arr, i + 1,                                         required_sum - arr[i], n);         return dp[i][required_sum + base]; # Driver codeif __name__ == "__main__" :     arr = [ 3, 3, 3, 3 ];    n = len(arr);    x = 6;     print(findCnt(arr, 0, x, n)); # This code is contributed by AnkitRai01
// C# implementation of the approachusing System;     class GFG{ static int maxN = 20;static int maxSum = 50;static int minSum = 50;static int Base = 50; // To store the states of DPstatic int [,]dp = new int[maxN, maxSum + minSum];static Boolean [,]v = new Boolean[maxN, maxSum + minSum]; // Function to return the required countstatic int findCnt(int []arr, int i,                   int required_sum, int n){    // Base case    if (i == n)    {        if (required_sum == 0)            return 1;        else            return 0;    }     // If the state has been solved before    // return the value of the state    if (v[i, required_sum + Base])        return dp[i, required_sum + Base];     // Setting the state as solved    v[i, required_sum + Base] = true;     // Recurrence relation    dp[i, required_sum + Base] =          findCnt(arr, i + 1, required_sum, n) +          findCnt(arr, i + 1, required_sum - arr[i], n);    return dp[i,required_sum + Base];} // Driver codepublic static void Main(String []args){    int []arr = { 3, 3, 3, 3 };    int n = arr.Length;    int x = 6;     Console.WriteLine(findCnt(arr, 0, x, n));}} // This code is contributed by 29AjayKumar
<script> // Javascript implementation of the approach var maxN = 20var maxSum = 50var minSum = 50var base = 50 // To store the states of DPvar dp = Array.from(Array(maxN),()=>Array(maxSum+minSum));var v = Array.from(Array(maxN),()=>Array(maxSum+minSum)); // Function to return the required countfunction findCnt(arr, i, required_sum, n){    // Base case    if (i == n) {        if (required_sum == 0)            return 1;        else            return 0;    }     // If the state has been solved before    // return the value of the state    if (v[i][required_sum + base])        return dp[i][required_sum + base];     // Setting the state as solved    v[i][required_sum + base] = 1;     // Recurrence relation    dp[i][required_sum + base]        = findCnt(arr, i + 1,          required_sum, n)          + findCnt(arr, i + 1,          required_sum - arr[i], n);    return dp[i][required_sum + base];} // Driver codevar arr = [3, 3, 3, 3];var n = arr.length;var x = 6;document.write( findCnt(arr, 0, x, n)); </script>
6
This method is valid only for those arrays which contains positive elements.
In this method we use a 2D array of size (arr.size() + 1) * (target + 1) of type integer.
Initialization of Matrix:
mat[0][0] = 1 because If the size of sum is 
if (A[i] > j)
DP[i][j] = DP[i-1][j]
else 
DP[i][j] = DP[i-1][j] + DP[i-1][j-A[i]]
This means that if the current element has a value greater than the ‘current sum value’ we will copy the answer for previous cases
And if the current sum value is greater than the ‘ith’ element we will see if any of the previous states have already experienced the sum=’j’ and any previous states experienced a value ‘j – A[i]’ which will solve our purpose
C++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std; int subsetSum(int a[], int n, int sum){    // Initializing the matrix    int tab[n + 1][sum + 1];  // Initializing the first value of matrix    tab[0][0] = 1;    for (int i = 1; i <= sum; i++)        tab[0][i] = 0;            for (int i = 1; i <= n; i++)    {        for (int j = 0; j <= sum; j++)        {          // if the value is greater than the sum            if (a[i - 1] > j)                tab[i][j] = tab[i - 1][j];            else            {                tab[i][j] = tab[i - 1][j] + tab[i - 1][j - a[i - 1]];            }        }    }      return tab[n][sum];} int main(){    int n = 4;    int a[] = {3,3,3,3};    int sum = 6;     cout << (subsetSum(a, n, sum));}
import java.io.*;import java.lang.*;import java.util.*; class GFG{ static int subsetSum(int a[], int n, int sum){         // Initializing the matrix    int tab[][] = new int[n + 1][sum + 1];     // Initializing the first value of matrix    tab[0][0] = 1;     for(int i = 1; i <= sum; i++)        tab[0][i] = 0;      for(int i = 1; i <= n; i++)    {        for(int j = 0; j <= sum; j++)        {                         // If the value is greater than the sum            if (a[i - 1] > j)                tab[i][j] = tab[i - 1][j];             else            {                tab[i][j] = tab[i - 1][j] +                            tab[i - 1][j - a[i - 1]];            }        }    }     return tab[n][sum];} // Driver Codepublic static void main(String[] args){    int n = 4;    int a[] = { 3, 3, 3, 3 };    int sum = 6;     System.out.print(subsetSum(a, n, sum));}} // This code is contributed by Kingash
def subset_sum(a: list, n: int, sum: int):       # Initializing the matrix    tab = [[0] * (sum + 1) for i in range(n + 1)]    tab[0][0] = 1    for i in range(1, sum + 1):        tab[0][i] = 0         for i in range(1, n+1):        for j in range(sum + 1):            if a[i-1] <= j:                tab[i][j] = tab[i-1][j] + tab[i-1][j-a[i-1]]            else:                tab[i][j] = tab[i-1][j]    return tab[n][sum] if __name__ == '__main__':    a = [3, 3, 3, 3]    n = 4    sum = 6    print(subset_sum(a, n, sum))     # This code is contributed by Premansh2001.
using System; class GFG{ static int subsetSum(int []a, int n, int sum){         // Initializing the matrix    int [,]tab = new int[n + 1, sum + 1];     // Initializing the first value of matrix    tab[0, 0] = 1;     for(int i = 1; i <= sum; i++)        tab[0, i] = 0;      for(int i = 1; i <= n; i++)    {        for(int j = 0; j <= sum; j++)        {                         // If the value is greater than the sum            if (a[i - 1] > j)                tab[i, j] = tab[i - 1, j];            else            {                tab[i, j] = tab[i - 1, j] +                            tab[i - 1, j - a[i - 1]];            }        }    }    return tab[n, sum];} // Driver Codepublic static void Main(String[] args){    int n = 4;    int []a = { 3, 3, 3, 3 };    int sum = 6;     Console.Write(subsetSum(a, n, sum));}} // This code is contributed by shivanisinghss2110
<script>  function subsetSum( a,  n, sum){    // Initializing the matrix    var tab = new Array(n + 1);    for (let i = 0; i< n+1; i++)      tab[i] = new Array(sum + 1);  // Initializing the first value of matrix    tab[0][0] = 1;    for (let i = 1; i <= sum; i++)        tab[0][i] = 0;     for (let i = 1; i <= n; i++)    {        for (let j = 0; j <= sum; j++)        {          // if the value is greater than the sum            if (a[i - 1] > j)                tab[i][j] = tab[i - 1][j];            else            {                tab[i][j] = tab[i - 1][j] + tab[i - 1][j - a[i - 1]];            }        }    }      return tab[n][sum];} var n = 4;var a = new Array(3,3,3,3);var sum = 6; console.log(subsetSum(a, n, sum)); // This code is contributed by ukasp.</script>
6
Time Complexity: O(sum*n), where the sum is the ‘target sum’ and ‘n’ is the size of the array.Auxiliary Space: O(sum*n), as the size of the 2-D array, is sum*n. 
Method 3: Space Optimization:-
We can solve this problem by just taking care of last state and current state so we can wrap up this problem in O(target+1) space complexiy.
Example:-
vector<int> arr = { 3, 3, 3, 3 }; with targetSum of 6;
dp[0][arr[0]] — tells about what if at index 0 we need arr[0] to achieve the targetSum and fortunately we have that solve so mark them 1;
=====dp[0][3]=1
target
Index
at dp[2][6] --- tells tell me is at index 2 can count some subsets with sum=6, How can we achieve this?
so we can tell ok i have reached at index 2 by adding element of index 1 or not both case has been added ------ means dp[i-1] we need only bcoz we are need of last index decision only nothing more than that so this why we are using a huge 2D array
just store our running state and last state that's it.

1.Time Complexity:- O(N*val)
2.Space Compexity:- O(Val)
where val and n are targetSum and number of element.
C++
#include <bits/stdc++.h>using namespace std; int CountSubsetSum(vector<int>& arr, int val, int n){    int count = 0;    vector<int> PresentState(val + 1, 0),        LastState(val + 1, 0);    // consider only last and present state we dont need the    // (present-2)th state and above and we know for val to    // be 0 if we dont pick the current index element we can    // achieve    PresentState[0] = LastState[0] = 1;    if (arr[0] <= val)        LastState[arr[0]] = 1;    for (int i = 1; i < n; i++) {        for (int j = 0; j <= val; j++)            PresentState[j]                = ((j >= arr[i]) ? LastState[j - arr[i]]                                 : 0)                  + LastState[j];        // this we will need in the next iteration so just        // swap current and last state.        LastState = PresentState;    }    // Note after exit from loop we will having a present    // state which is nothing but the laststate itself;    return PresentState[val]; // or return                              // CurrentState[val];}int main(){    vector<int> arr = { 3, 3, 3, 3 };    cout << CountSubsetSum(arr, 6, arr.size());}
6
ankthon
29AjayKumar
ankitkumar774
architgwl2000
Kingash
noob2000
lakshya1st
premansh2001
shivanisinghss2110
ukasp
niyam239
abhishek2007
subset
Arrays
Combinatorial
Dynamic Programming
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Combinatorial
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