[LeetCode Sync] Runtime - 1787 ms (51.15%), Memory - 55.7 MB (7.86%)

Author: NinePiece2

Solutions/3851-find-sum-of-array-product-of-magical-sequences/README.md

@@ -0,0 +1,62 @@
+<p>You are given two integers, <code>m</code> and <code>k</code>, and an integer array <code>nums</code>.</p>
+A sequence of integers <code>seq</code> is called <strong>magical</strong> if:
+
+<ul>
+	<li><code>seq</code> has a size of <code>m</code>.</li>
+	<li><code>0 &lt;= seq[i] &lt; nums.length</code></li>
+	<li>The <strong>binary representation</strong> of <code>2<sup>seq[0]</sup> + 2<sup>seq[1]</sup> + ... + 2<sup>seq[m - 1]</sup></code> has <code>k</code> <strong>set bits</strong>.</li>
+</ul>
+
+<p>The <strong>array product</strong> of this sequence is defined as <code>prod(seq) = (nums[seq[0]] * nums[seq[1]] * ... * nums[seq[m - 1]])</code>.</p>
+
+<p>Return the <strong>sum</strong> of the <strong>array products</strong> for all valid <strong>magical</strong> sequences.</p>
+
+<p>Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>A <strong>set bit</strong> refers to a bit in the binary representation of a number that has a value of 1.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">m = 5, k = 5, nums = [1,10,100,10000,1000000]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">991600007</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>All permutations of <code>[0, 1, 2, 3, 4]</code> are magical sequences, each with an array product of 10<sup>13</sup>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">m = 2, k = 2, nums = [5,4,3,2,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">170</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The magical sequences are <code>[0, 1]</code>, <code>[0, 2]</code>, <code>[0, 3]</code>, <code>[0, 4]</code>, <code>[1, 0]</code>, <code>[1, 2]</code>, <code>[1, 3]</code>, <code>[1, 4]</code>, <code>[2, 0]</code>, <code>[2, 1]</code>, <code>[2, 3]</code>, <code>[2, 4]</code>, <code>[3, 0]</code>, <code>[3, 1]</code>, <code>[3, 2]</code>, <code>[3, 4]</code>, <code>[4, 0]</code>, <code>[4, 1]</code>, <code>[4, 2]</code>, and <code>[4, 3]</code>.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">m = 1, k = 1, nums = [28]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">28</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The only magical sequence is <code>[0]</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= k &lt;= m &lt;= 30</code></li>
+	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>8</sup></code></li>
+</ul>

Solutions/3851-find-sum-of-array-product-of-magical-sequences/solution.py

@@ -0,0 +1,60 @@
+class Solution:
+    def mul_with_mod(self, x: int, y: int, mod: int) -> int:
+        result, current = 1, x % mod
+        while y:
+            if y & 1:
+                result = result * current % mod
+            y >>= 1
+            current = current * current % mod
+        return result
+
+    def magicalSum(self, m: int, k: int, nums: List[int]) -> int:
+        n = len(nums)
+        mod = 10**9 + 7
+
+        fac = [1] * (m + 1)
+        for i in range(1, m + 1):
+            fac[i] = fac[i - 1] * i % mod
+        
+        ifac = [1] * (m + 1)
+        for i in range(2, m + 1):
+            ifac[i] = self.mul_with_mod(i, mod - 2, mod)
+        for i in range(2, m + 1):
+            ifac[i] = ifac[i - 1] * ifac[i] % mod
+        
+        numsPower = [[1] * (m + 1) for _ in range(n)]
+        for i in range(n):
+            for j in range(1, m + 1):
+                numsPower[i][j] = numsPower[i][j - 1] * nums[i] % mod
+
+        dp = [[[[0] * (k + 1) for _ in range(m * 2 + 1)] for _ in range(m + 1)] for _ in range(n)]
+
+        for j in range(m + 1):
+            dp[0][j][j][0] = numsPower[0][j] * ifac[j] % mod
+        
+        for i in range(n - 1):
+            for j in range(m + 1):
+                for p in range(m * 2 + 1):
+                    for q in range(k + 1):
+                        if dp[i][j][p][q] == 0:
+                            continue
+                        q2 = (p % 2) + q
+
+                        if q2 > k:
+                            break
+
+                        for r in range(m - j + 1):
+                            p2 = (p // 2) + r
+                            if p2 > m * 2:
+                                continue
+                            dp[i + 1][j + r][p2][q2] = (dp[i + 1][j + r][p2][q2] + dp[i][j][p][q] 
+                                                    * numsPower[i + 1][r] % mod * ifac[r] % mod
+                                                    ) % mod
+
+        result = 0
+        for p in range(m * 2 + 1):
+            for q in range(k + 1):
+                if bin(p).count("1") + q == k:
+                    result = (result + dp[n - 1][m][p][q] * fac[m] % mod) % mod
+        
+        return result