Solution Explanation for Partition Equal Subset Sum

The problem asks whether a given integer array can be partitioned into two subsets with equal sums. This can be efficiently solved using dynamic programming.

Core Idea:

The key insight is that if the sum of all numbers is odd, it's impossible to partition the array into two subsets with equal sums. Otherwise, we need to determine if there exists a subset whose sum is exactly half of the total sum.

Approach 1: 2D Dynamic Programming

1. Calculate Total Sum: Compute the sum of all numbers in the input array nums. If the sum is odd, return false.

2. Target Sum: Set the target sum m to half of the total sum.

3. DP Table: Create a 2D boolean array f of size (n+1) x (m+1), where n is the length of nums. f[i][j] will be true if there exists a subset of the first i numbers whose sum is j, otherwise false.

4. Base Case: f[0][0] = true (an empty subset sums to 0).

5. Iteration: Iterate through the array nums. For each number x and each possible sum j from 0 to m:

   • If j >= x, f[i][j] is true if either f[i-1][j] (without including x) or f[i-1][j-x] (including x) is true.
   • Otherwise, f[i][j] = f[i-1][j].

6. Result: f[n][m] will indicate whether a subset with sum m can be formed using the entire array.

Time Complexity: O(n*m), where n is the length of the array and m is half the sum of the array. This is because we iterate through the DP table.

Space Complexity: O(n*m) to store the DP table.

Approach 2: Optimized 1D Dynamic Programming

This approach optimizes space complexity by using a 1D DP array. We observe that the current row of the DP table only depends on the previous row.

1. Calculate Total Sum and Target: Same as Approach 1.

2. 1D DP Array: Create a 1D boolean array f of size (m+1). f[j] will be true if there is a subset summing to j using the numbers processed so far.

3. Base Case: f[0] = true.

4. Iteration: Iterate through nums. For each number x, iterate backward from m down to x:

   • Update f[j] to f[j] || f[j-x]. This means we can reach the sum j either by including x or not.

5. Result: f[m] will indicate if a subset with sum m is possible.

Time Complexity: O(n*m), same as Approach 1.

Space Complexity: O(m), significantly reduced compared to Approach 1.

Code Examples (Python - Approach 2):

class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        s = sum(nums)
        if s % 2 != 0:
            return False
        m = s // 2
        dp = [False] * (m + 1)
        dp[0] = True
        for num in nums:
            for i in range(m, num - 1, -1):
                dp[i] = dp[i] or dp[i - num]
        return dp[m]
 

Both approaches correctly solve the problem. The second approach is preferred for its improved space efficiency, especially beneficial for large input arrays. The provided code examples in various languages demonstrate both approaches. Remember to adapt the code to your specific language's syntax and libraries.