Add solution #1458

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,312 LeetCode solutions in JavaScript
+# 1,313 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1114,6 +1114,7 @@
 1455|[Check If a Word Occurs As a Prefix of Any Word in a Sentence](./solutions/1455-check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence.js)|Easy|
 1456|[Maximum Number of Vowels in a Substring of Given Length](./solutions/1456-maximum-number-of-vowels-in-a-substring-of-given-length.js)|Medium|
 1457|[Pseudo-Palindromic Paths in a Binary Tree](./solutions/1457-pseudo-palindromic-paths-in-a-binary-tree.js)|Medium|
+1458|[Max Dot Product of Two Subsequences](./solutions/1458-max-dot-product-of-two-subsequences.js)|Hard|
 1460|[Make Two Arrays Equal by Reversing Sub-arrays](./solutions/1460-make-two-arrays-equal-by-reversing-sub-arrays.js)|Easy|
 1462|[Course Schedule IV](./solutions/1462-course-schedule-iv.js)|Medium|
 1464|[Maximum Product of Two Elements in an Array](./solutions/1464-maximum-product-of-two-elements-in-an-array.js)|Easy|

solutions/1458-max-dot-product-of-two-subsequences.js

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+/**
+ * 1458. Max Dot Product of Two Subsequences
+ * https://leetcode.com/problems/max-dot-product-of-two-subsequences/
+ * Difficulty: Hard
+ *
+ * Given two arrays nums1 and nums2.
+ *
+ * Return the maximum dot product between non-empty subsequences of nums1 and nums2 with the
+ * same length.
+ *
+ * A subsequence of a array is a new array which is formed from the original array by deleting
+ * some (can be none) of the characters without disturbing the relative positions of the remaining
+ * characters. (ie, [2,3,5] is a subsequence of [1,2,3,4,5] while [1,5,3] is not).
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+var maxDotProduct = function(nums1, nums2) {
+  const m = nums1.length;
+  const n = nums2.length;
+  const dp = new Array(m + 1).fill().map(() => new Array(n + 1).fill(-Infinity));
+
+  let result = -Infinity;
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      result = Math.max(result, computeMax(i, j));
+    }
+  }
+
+  return result;
+
+  function computeMax(i, j) {
+    if (i === 0 || j === 0) return -Infinity;
+    if (dp[i][j] !== -Infinity) return dp[i][j];
+
+    dp[i][j] = nums1[i - 1] * nums2[j - 1];
+    dp[i][j] = Math.max(
+      dp[i][j],
+      dp[i][j] + computeMax(i - 1, j - 1),
+      computeMax(i - 1, j),
+      computeMax(i, j - 1)
+    );
+
+    return dp[i][j];
+  }
+};