Sync LeetCode submission Runtime - 260 ms (14.25%), Memory - 30.3 MB (16.13%)

Author: jimit105

3152-maximum-value-of-an-ordered-triplet-ii/README.md

@@ -0,0 +1,40 @@
+<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>.</p>
+
+<p>Return <em><strong>the maximum value over all triplets of indices</strong></em> <code>(i, j, k)</code> <em>such that</em> <code>i &lt; j &lt; k</code><em>. </em>If all such triplets have a negative value, return <code>0</code>.</p>
+
+<p>The <strong>value of a triplet of indices</strong> <code>(i, j, k)</code> is equal to <code>(nums[i] - nums[j]) * nums[k]</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [12,6,1,2,7]
+<strong>Output:</strong> 77
+<strong>Explanation:</strong> The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
+It can be shown that there are no ordered triplets of indices with a value greater than 77. 
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,10,3,4,19]
+<strong>Output:</strong> 133
+<strong>Explanation:</strong> The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
+It can be shown that there are no ordered triplets of indices with a value greater than 133.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
+</ul>

3152-maximum-value-of-an-ordered-triplet-ii/solution.py

@@ -0,0 +1,21 @@
+# Approach 1: Greedy + Prefix Suffix Array
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def maximumTripletValue(self, nums: List[int]) -> int:
+        n = len(nums)
+        left_max = [0] * n
+        right_max = [0] * n
+
+        for i in range(1, n):
+            left_max[i] = max(left_max[i - 1], nums[i - 1])
+            right_max[n - i - 1] = max(right_max[n - i], nums[n - i])
+
+        res = 0
+        for j in range(1, n - 1):
+            res = max(res, (left_max[j] - nums[j]) * right_max[j])
+
+        return res
+