[LeetCode Sync] Runtime - 2986 ms (76.19%), Memory - 43.7 MB (95.24%)

Author: NinePiece2

Solutions/3592-find-x-sum-of-all-k-long-subarrays-ii/README.md

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+<p>You are given an array <code>nums</code> of <code>n</code> integers and two integers <code>k</code> and <code>x</code>.</p>
+
+<p>The <strong>x-sum</strong> of an array is calculated by the following procedure:</p>
+
+<ul>
+	<li>Count the occurrences of all elements in the array.</li>
+	<li>Keep only the occurrences of the top <code>x</code> most frequent elements. If two elements have the same number of occurrences, the element with the <strong>bigger</strong> value is considered more frequent.</li>
+	<li>Calculate the sum of the resulting array.</li>
+</ul>
+
+<p><strong>Note</strong> that if an array has less than <code>x</code> distinct elements, its <strong>x-sum</strong> is the sum of the array.</p>
+
+<p>Return an integer array <code>answer</code> of length <code>n - k + 1</code> where <code>answer[i]</code> is the <strong>x-sum</strong> of the <span data-keyword="subarray-nonempty">subarray</span> <code>nums[i..i + k - 1]</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,1,2,2,3,4,2,3], k = 6, x = 2</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[6,10,12]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>For subarray <code>[1, 1, 2, 2, 3, 4]</code>, only elements 1 and 2 will be kept in the resulting array. Hence, <code>answer[0] = 1 + 1 + 2 + 2</code>.</li>
+	<li>For subarray <code>[1, 2, 2, 3, 4, 2]</code>, only elements 2 and 4 will be kept in the resulting array. Hence, <code>answer[1] = 2 + 2 + 2 + 4</code>. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.</li>
+	<li>For subarray <code>[2, 2, 3, 4, 2, 3]</code>, only elements 2 and 3 are kept in the resulting array. Hence, <code>answer[2] = 2 + 2 + 2 + 3 + 3</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [3,8,7,8,7,5], k = 2, x = 2</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[11,15,15,15,12]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Since <code>k == x</code>, <code>answer[i]</code> is equal to the sum of the subarray <code>nums[i..i + k - 1]</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>nums.length == n</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= x &lt;= k &lt;= nums.length</code></li>
+</ul>

Solutions/3592-find-x-sum-of-all-k-long-subarrays-ii/solution.py

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+class Solution:
+    def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:
+        left = SortedList()
+        right = SortedList()
+        count = Counter()
+        sum_val = 0
+        n = len(nums)
+        result = [0] * (n - k + 1)
+
+        def add(val: int):
+            if count[val] == 0:
+                return
+            temp = (count[val], val)
+            if left and temp > left[0]:
+                nonlocal sum_val
+                sum_val += temp[0] * temp[1]
+                left.add(temp)
+            else:
+                right.add(temp)
+
+        def remove(val: int):
+            if count[val] == 0:
+                return
+            temp = (count[val], val)
+            if temp in left:
+                nonlocal sum_val
+                sum_val -= temp[0] * temp[1]
+                left.remove(temp)
+            else:
+                right.remove(temp)
+
+        for i, val in enumerate(nums):
+            remove(val)
+            count[val] += 1
+            add(val)
+            j = i - k + 1
+            if j < 0:
+                continue
+            while right and len(left) < x:
+                temp = right.pop()
+                left.add(temp)
+                sum_val += temp[0] * temp[1]
+
+            while len(left) > x:
+                temp = left.pop(0)
+                sum_val -= temp[0] * temp[1]
+                right.add(temp)
+            result[j] = sum_val
+
+            remove(nums[j])
+            count[nums[j]] -= 1
+            add(nums[j])
+
+        return result