Solution Count of Subset Sum

Problem Statement

Given a set of positive numbers, find the total number of subsets whose sum is equal to a given number 'S'.

Example 1:

Input: {1, 1, 2, 3}, S=4
Output: 3
The given set has '3' subsets whose sum is '4': {1, 1, 2}, {1, 3}, {1, 3}
Note that we have two similar sets {1, 3}, because we have two '1' in our input.

Example 2:

Input: {1, 2, 7, 1, 5}, S=9
Output: 3
The given set has '3' subsets whose sum is '9': {2, 7}, {1, 7, 1}, {1, 2, 1, 5}

Basic Solution

This problem follows the 0/1 Knapsack pattern and is quite similar to "Subset Sum". The only difference in this problem is that we need to count the number of subsets, whereas in the "Subset Sum" we only wanted to know if there exists a subset with the given sum.

A basic brute-force solution could be to try all subsets of the given numbers to count the subsets that have a sum equal to 'S'. So our brute-force algorithm will look like:

for each number 'i' 
  create a new set which includes number 'i' if it does not exceed 'S', and recursively   
      process the remaining numbers and sum
  create a new set without number 'i', and recursively process the remaining numbers 
return the count of subsets who has a sum equal to 'S'

Code

Here is the code for the brute-force solution:

java

class Solution {

  public int countSubsets(int[] num, int sum) {
    return this.countSubsetsRecursive(num, sum, 0);
  }

  private int countSubsetsRecursive(int[] num, int sum, int currentIndex) {
    // base checks
    if (sum == 0) return 1;

    if (num.length == 0 || currentIndex >= num.length) return 0;

    // recursive call after selecting the number at the currentIndex
    // if the number at currentIndex exceeds the sum, we shouldn't process this
    int sum1 = 0;
    if (num[currentIndex] <= sum) sum1 = countSubsetsRecursive(
      num,
      sum - num[currentIndex],
      currentIndex + 1
    );

    // recursive call after excluding the number at the currentIndex
    int sum2 = countSubsetsRecursive(num, sum, currentIndex + 1);

    return sum1 + sum2;
  }

  public static void main(String[] args) {
    Solution ss = new Solution();
    int[] num = { 1, 1, 2, 3 };
    System.out.println(ss.countSubsets(num, 4));
    num = new int[] { 1, 2, 7, 1, 5 };
    System.out.println(ss.countSubsets(num, 9));
  }
}

The time complexity of the above algorithm is exponential , where ‘n’ represents the total number. The space complexity is , this memory is used to store the recursion stack.

Top-down Dynamic Programming with Memoization

We can use memoization to overcome the overlapping sub-problems. We will be using a two-dimensional array to store the results of solved sub-problems. As mentioned above, we need to store results for every subset and for every possible sum.

Code

Here is the code:

java

class Solution {

  public int countSubsets(int[] num, int sum) {
    Integer[][] dp = new Integer[num.length][sum + 1];
    return this.countSubsetsRecursive(dp, num, sum, 0);
  }

  private int countSubsetsRecursive(
    Integer[][] dp,
    int[] num,
    int sum,
    int currentIndex
  ) {
    // base checks
    if (sum == 0) return 1;

    if (num.length == 0 || currentIndex >= num.length) return 0;

    // check if we have not already processed a similar problem
    if (dp[currentIndex][sum] == null) {
      // recursive call after choosing the number at the currentIndex
      // if the number at currentIndex exceeds the sum, we shouldn't process this
      int sum1 = 0;
      if (num[currentIndex] <= sum) sum1 = countSubsetsRecursive(
        dp,
        num,
        sum - num[currentIndex],
        currentIndex + 1
      );

      // recursive call after excluding the number at the currentIndex
      int sum2 = countSubsetsRecursive(dp, num, sum, currentIndex + 1);

      dp[currentIndex][sum] = sum1 + sum2;
    }

    return dp[currentIndex][sum];
  }

  public static void main(String[] args) {
    Solution ss = new Solution();
    int[] num = { 1, 1, 2, 3 };
    System.out.println(ss.countSubsets(num, 4));
    num = new int[] { 1, 2, 7, 1, 5 };
    System.out.println(ss.countSubsets(num, 9));
  }
}

Bottom-up Dynamic Programming

We will try to find if we can make all possible sums with every subset to populate the array db[TotalNumbers][S+1].

So, at every step we have two options:

Exclude the number. Count all the subsets without the given number up to the given sum => dp[index-1][sum]
Include the number if its value is not more than the 'sum'. In this case, we will count all the subsets to get the remaining sum => dp[index-1][sum-num[index]]

To find the total sets, we will add both of the above two values:

    dp[index][sum] = dp[index-1][sum] + dp[index-1][sum-num[index]])

Let's start with our base case of size zero:

'0' sum can always be found through an empty set

With only one number, we can form a subset only when the required sum is equal to the number

sum: 1, index:1=> (dp[index-1][sum] + dp[index-1][sum - 1])

sum: 2, index:1=> (dp[index-1][sum] + dp[index-1][sum - 1])

sum: 4, index:3=> (dp[index-1][sum] + dp[index-1][sum - 3])

Code

Here is the code for our bottom-up dynamic programming approach:

java

class Solution {

  public int countSubsets(int[] num, int sum) {
    int n = num.length;
    int[][] dp = new int[n][sum + 1];

    // populate the sum=0 columns, as we will always have an empty set for zero sum
    for (int i = 0; i < n; i++) dp[i][0] = 1;

    // with only one number, we can form a subset only when the required sum is equal to its value
    for (int s = 1; s <= sum; s++) {
      dp[0][s] = (num[0] == s ? 1 : 0);
    }

    // process all subsets for all sums
    for (int i = 1; i < num.length; i++) {
      for (int s = 1; s <= sum; s++) {
        // exclude the number
        dp[i][s] = dp[i - 1][s];
        // include the number, if it does not exceed the sum
        if (s >= num[i]) dp[i][s] += dp[i - 1][s - num[i]];
      }
    }

    // the bottom-right corner will have our answer.
    return dp[num.length - 1][sum];
  }

  public static void main(String[] args) {
    Solution ss = new Solution();
    int[] num = { 1, 1, 2, 3 };
    System.out.println(ss.countSubsets(num, 4));
    num = new int[] { 1, 2, 7, 1, 5 };
    System.out.println(ss.countSubsets(num, 9));
  }
}

The above solution has time and space complexity of , where ‘N’ represents total numbers and ‘S’ is the desired sum.

Challenge

Can we further improve our bottom-up DP solution? Can you find an algorithm that has space complexity?

Similar to the space optimized solution for "0/1 Knapsack."

java

class Solution {

  static int countSubsets(int[] num, int sum) {
    int n = num.length;
    int[] dp = new int[sum + 1];
    dp[0] = 1;

    // with only one number, we can form a subset only when the required sum is equal to its value
    for (int s = 1; s <= sum; s++) {
      dp[s] = (num[0] == s ? 1 : 0);
    }

    // process all subsets for all sums
    for (int i = 1; i < num.length; i++) {
      for (int s = sum; s >= 0; s--) {
        if (s >= num[i]) dp[s] += dp[s - num[i]];
      }
    }

    return dp[sum];
  }

  public static void main(String[] args) {
    Solution ss = new Solution();
    int[] num = { 1, 1, 2, 3 };
    System.out.println(ss.countSubsets(num, 4));
    num = new int[] { 1, 2, 7, 1, 5 };
    System.out.println(ss.countSubsets(num, 9));
  }
}

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