[LeetCode Sync] Runtime - 390 ms (28.03%), Memory - 49.6 MB (5.07%)

Author: NinePiece2

Solutions/0698-partition-to-k-equal-sum-subsets/README.md

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+<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <code>true</code> if it is possible to divide this array into <code>k</code> non-empty subsets whose sums are all equal.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [4,3,2,3,5,2,1], k = 4
+<strong>Output:</strong> true
+<strong>Explanation:</strong> It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4], k = 3
+<strong>Output:</strong> false
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= nums.length &lt;= 16</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
+	<li>The frequency of each element is in the range <code>[1, 4]</code>.</li>
+</ul>

Solutions/0698-partition-to-k-equal-sum-subsets/solution.py

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+class Solution:
+    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
+        s, mod = divmod(sum(nums), k)
+        if mod:
+            return False
+        
+        nums.sort()
+        mask = (1 << len(nums)) - 1
+
+        @cache
+        def dfs(state, temp):
+            if state == mask:
+                return True
+            for i, num in enumerate(nums):
+                if (state >> i) & 1:
+                    continue
+                if temp + num > s:
+                    break
+                if dfs(state | 1 << i, (temp + num) % s):
+                    return True
+            return False
+        
+        return dfs(0, 0)