<p>Given an integer array <code>nums</code>, return <code>true</code> <em>if you can partition the array into two subsets such that the sum of the elements in both subsets is equal or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,11,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> The array can be partitioned as [1, 5, 5] and [11].
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        total_sum = sum(nums)
        if total_sum % 2:
            return False
        target_sum = total_sum // 2
        #return self.brute_force(0, nums, 0, sum(nums))
        #return self.memoization(0, nums, 0, sum(nums), {})
        return self.memoization_v2(0, nums, 0, target_sum, {})


    # TLE... time complexity = O(2^n), space complexity = O(n)
    def brute_force(self, i, nums, left_part_sum, right_part_sum):
        if i == len(nums):
            return left_part_sum == right_part_sum

        # skip this elem
        skip = self.brute_force(i + 1, nums, left_part_sum, right_part_sum)

        # consider this elem
        include = self.brute_force(i + 1, nums, left_part_sum + nums[i], right_part_sum - nums[i])

        return skip or include

    def memoization(self, i, nums, left_part_sum, right_part_sum, cache):
        if i == len(nums):
            return left_part_sum == right_part_sum

        cache_key = (i, left_part_sum)
        if cache_key in cache:
            return cache[cache_key]

        cache[cache_key] = self.memoization(i + 1, nums, left_part_sum, right_part_sum, cache) or self.memoization(i + 1, nums, left_part_sum + nums[i], right_part_sum - nums[i], cache)
        return cache[cache_key]

    def memoization_v2(self, i, nums, cur_sum, target, cache):
        if i == len(nums):
            return cur_sum == target
        cache_key = (i, cur_sum)
        if cache_key in cache:
            return cache[cache_key]

        if cur_sum == target:
            return True

        cache[cache_key] = self.memoization_v2(i + 1, nums, cur_sum, target, cache) or self.memoization_v2(i + 1, nums, cur_sum + nums[i], target, cache)
        return cache[cache_key]

    
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