Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 99.6 MB (52.11%)

Author: koralkulacoglu

LeetCode/0396-rotate-function/README.md

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+<p>You are given an integer array <code>nums</code> of length <code>n</code>.</p>
+
+<p>Assume <code>arr<sub>k</sub></code> to be an array obtained by rotating <code>nums</code> by <code>k</code> positions clock-wise. We define the <strong>rotation function</strong> <code>F</code> on <code>nums</code> as follow:</p>
+
+<ul>
+	<li><code>F(k) = 0 * arr<sub>k</sub>[0] + 1 * arr<sub>k</sub>[1] + ... + (n - 1) * arr<sub>k</sub>[n - 1].</code></li>
+</ul>
+
+<p>Return <em>the maximum value of</em> <code>F(0), F(1), ..., F(n-1)</code>.</p>
+
+<p>The test cases are generated so that the answer fits in a <strong>32-bit</strong> integer.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [4,3,2,6]
+<strong>Output:</strong> 26
+<strong>Explanation:</strong>
+F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
+F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
+F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
+F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
+So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [100]
+<strong>Output:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == nums.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
+</ul>

LeetCode/0396-rotate-function/solution.cpp

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+class Solution {
+public:
+    int maxRotateFunction(vector<int>& nums) {
+        int n = nums.size();
+        int cur = 0, sum = 0;
+        for (int i=0; i<n; i++) {
+            cur += i * nums[i];
+            sum += nums[i];
+        }
+        int ans = cur;
+        for (int i=1; i<n; i++) {
+            cur += sum - n * nums[n-i];
+            ans = max(ans, cur);
+        }
+        return ans;
+    }
+};