Grokking the Coding Interview: Patterns for Coding Questions
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Solution: Subset Sum
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Problem Statement

Basic Solution

Bottom-up Dynamic Programming

Code

Time and Space Complexity

Challenge

Problem Statement

Given a set of positive numbers, determine if a subset exists whose sum is equal to a given number ‘S’.

Example 1:

Input: {1, 2, 3, 7}, S=6
Output: True
The given set has a subset whose sum is '6': {1, 2, 3}

Example 2:

Input: {1, 2, 7, 1, 5}, S=10
Output: True
The given set has a subset whose sum is '10': {1, 2, 7}

Example 3:

Input: {1, 3, 4, 8}, S=6
Output: False
The given set does not have any subset whose sum is equal to '6'.

Constraints:

  • 1 <= num.length <= 200
  • 1 <= num[i] <= 100

Basic Solution

This problem follows the 0/1 Knapsack pattern and is quite similar to Equal Subset Sum Partition. A basic brute-force solution could be to try all subsets of the given numbers to see if any set has 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
  create a new set WITHOUT number 'i', and recursively process the remaining numbers 
return true if any of the above two sets has a sum equal to 'S', otherwise return false

Since this problem is quite similar to Equal Subset Sum Partition, let’s jump directly to the bottom-up dynamic programming solution.

Bottom-up Dynamic Programming

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

For every possible sum ‘s’ (where 0 <= s <= S), we have two options:

  1. Exclude the number. In this case, we will see if we can get the sum ‘s’ from the subset excluding this number => dp[index-1][s]
  2. Include the number if its value is not more than ‘s’. In this case, we will see if we can find a subset to get the remaining sum => dp[index-1][s-num[index]]

If either of the above two scenarios returns true, we can find a subset with a sum equal to ‘s’.

Let’s draw this visually, with the example input {1, 2, 3, 7}, and start with our base case of size zero:

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Code

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

Python3
Python3
. . . .

Time and Space Complexity

The above solution has the time and space complexity of O(N*S), where ‘N’ represents total numbers and ‘S’ is the required sum.

Challenge

Can we improve our bottom-up DP solution even further? Can you find an algorithm that has O(S) space complexity?

Python3
Python3
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On this page

Problem Statement

Basic Solution

Bottom-up Dynamic Programming

Code

Time and Space Complexity

Challenge