[LeetCode Sync] Runtime - 622 ms (55.99%), Memory - 18 MB (74.60%)

Author: NinePiece2

Solutions/0416-partition-equal-subset-sum/README.md

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+<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>

Solutions/0416-partition-equal-subset-sum/solution.py

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+class Solution:
+    def canPartition(self, nums: List[int]) -> bool:
+        # m, mod = divmod(sum(nums), 2)
+
+        # # if the sum is odd it can't be split into two subsets
+        # if mod:
+        #     return False
+
+        # n = len(nums)
+        # dp = [[False] * (m + 1) for _ in range(n + 1)]
+        # dp[0][0] = True
+
+        # for i, num in enumerate(nums, 1):
+        #     for j in range(m + 1):
+        #         dp[i][j] = dp[i - 1][j] or (j >= num and dp[i - 1][j - num])
+        
+        # return dp[-1][-1]
+
+        m, mod = divmod(sum(nums), 2)
+
+        # if the sum is odd it can't be split into two subsets
+        if mod:
+            return False
+
+        dp = [True] + [False] * m
+        for num in nums:
+            for j in range(m, num - 1, -1):
+                dp[j] = dp[j] or dp[j - num]
+        return dp[-1]