Partition Equal Subset Sum Leetcode Problem 416 [Python Solution]

Welcome to another coding session!

Today, we'll tackle the Partition Equal Subset Sum problem, also known as LeetCode Problem 416. This problem falls under the category of 1-D Dynamic Programming and is of medium difficulty.

It's a great opportunity to hone your coding skills.

So, let's dive in!

Problem Overview

The problem statement is as follows: Given an integer array nums, your task is to determine whether it's possible to partition the array into two subsets such that the sum of elements in both subsets is equal.

In other words, can you divide the array into two equal-sum subsets?

If it's possible, return true; otherwise, return false.

Here's an example to illustrate the problem:

Example 1:

Input: nums = [1, 5, 11, 5]
Output: true
Explanation: The array can be partitioned into [1, 5, 5] and [11].

Example 2:

Input: nums = [1, 2, 3, 5]
Output: false
Explanation: The array cannot be partitioned into equal sum subsets.

Constraints

Before we delve into solving the problem, it's crucial to understand the constraints:

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

Now that we have a clear understanding of the problem, let's explore the solutions.

Understanding the Constraints

Before we dive into the solutions, let's take a moment to understand the constraints.

The array can have a maximum length of 200, with each element ranging from 1 to 100. These constraints give us an idea of the problem's scope and the need for an efficient solution.

Efficient Python Code Solution

We'll provide a Python solution that efficiently solves the problem.

We'll use dynamic programming to achieve this.

Here's the code:

def canPartition(nums):
    # Check if the sum of the array is odd, if so, it's impossible to partition
    if sum(nums) % 2:
        return False

    # Initialize a set to store possible sums
    dp = set()
    dp.add(0)  # We can always achieve a sum of 0 with no elements
    target = sum(nums) // 2

    # Iterate through the array in reverse order
    for i in range(len(nums) - 1, -1, -1):
        nextDP = set()  # Create a new set for the next iteration
        for t in dp:
            if (t + nums[i]) == target:
                return True  # We found a subset that sums up to half the total
            nextDP.add(t + nums[i])
            nextDP.add(t)
        dp = nextDP

    return False  # If we can't find a valid partition, return False

This Python function efficiently determines whether it's possible to partition the given array into two subsets with equal sums.

Time and Space Complexity

Let's analyze the time and space complexity of our efficient solution:

• Time Complexity: The time complexity is primarily influenced by the size of the input array, nums.

Since we iterate through the array once and perform set operations, the time complexity is O(n * sum(nums)), where 'n' is the length of the array and 'sum(nums)' is the sum of all elements in the array.

• Space Complexity: We use a set, dp, to store possible sums.

The space complexity depends on the number of unique sums we can generate.

In the worst case, it's also O(n * sum(nums)).

Reasoning Behind Our Approach

The core idea behind our approach is to use dynamic programming to keep track of the possible sums that can be achieved while iterating through the array in reverse order.

We maintain a set, dp, to store these sums.

1. We first check if the sum of the array is odd because, in such cases, it's impossible to divide the array into two equal-sum subsets.

If the sum is even, we proceed.

1. We initialize a set dp and add 0 to it because we can always achieve a sum of 0 with no elements.

2. We calculate the target sum, which is half of the total sum of the array.

This is the sum we aim to achieve in each of the two subsets.

1. We iterate through the array in reverse order.

For each element, we create a new set, nextDP, to store possible sums for the next iteration.

1. For each sum t in the dp set, we check if adding the current element nums[i] to t results in the target sum.

If it does, we return True because we've found a valid partition.

1. We add both t + nums[i] and t to the nextDP set to consider both cases: adding the current element to the sum and not adding it.

2. After iterating through all elements, we update the dp set to be the nextDP set, and the process continues.

3. If we can't find a valid partition during the entire iteration, we return False.

Related Interview Questions By Company:

• Reverse Integer LeetCode
• Spiral Matrix LeetCode
• Shortest Path In Binary Matrix LeetCode

Related Interview Questions By Difficulty:

• Word Break LeetCode
• Longest Increasing Subsequence LeetCode
• Min Cost Climbing Stairs LeetCode

Conclusion

In this blog post, we tackled the Partition Equal Subset Sum problem (LeetCode Problem 416) using an efficient Python solution.

We discussed the problem statement, constraints, and provided a detailed explanation of our dynamic programming approach.

This solution allows us to determine whether it's possible to partition an array into two equal-sum subsets.

We optimized our code for time and space efficiency, making it a practical solution for larger inputs.

Now it's your turn to put this knowledge into practice.

You can find the full Python code and more information about this problem on the LeetCode website.

If you have any questions, suggestions, or want to share your experiences with this problem, feel free to leave a comment below.

Happy coding!