Add solution #1673

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,464 LeetCode solutions in JavaScript
+# 1,465 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1289,6 +1289,7 @@
 1670|[Design Front Middle Back Queue](./solutions/1670-design-front-middle-back-queue.js)|Medium|
 1671|[Minimum Number of Removals to Make Mountain Array](./solutions/1671-minimum-number-of-removals-to-make-mountain-array.js)|Hard|
 1672|[Richest Customer Wealth](./solutions/1672-richest-customer-wealth.js)|Easy|
+1673|[Find the Most Competitive Subsequence](./solutions/1673-find-the-most-competitive-subsequence.js)|Medium|
 1679|[Max Number of K-Sum Pairs](./solutions/1679-max-number-of-k-sum-pairs.js)|Medium|
 1716|[Calculate Money in Leetcode Bank](./solutions/1716-calculate-money-in-leetcode-bank.js)|Easy|
 1718|[Construct the Lexicographically Largest Valid Sequence](./solutions/1718-construct-the-lexicographically-largest-valid-sequence.js)|Medium|

solutions/1673-find-the-most-competitive-subsequence.js

@@ -0,0 +1,37 @@
+/**
+ * 1673. Find the Most Competitive Subsequence
+ * https://leetcode.com/problems/find-the-most-competitive-subsequence/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums and a positive integer k, return the most competitive subsequence
+ * of nums of size k.
+ *
+ * An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements
+ * from the array.
+ *
+ * We define that a subsequence a is more competitive than a subsequence b (of the same length) if
+ * in the first position where a and b differ, subsequence a has a number less than the
+ * corresponding number in b. For example, [1,3,4] is more competitive than [1,3,5] because the
+ * first position they differ is at the final number, and 4 is less than 5.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var mostCompetitive = function(nums, k) {
+  const result = [];
+  const n = nums.length;
+
+  for (let i = 0; i < n; i++) {
+    while (result.length && result[result.length - 1] > nums[i] && result.length + n - i > k) {
+      result.pop();
+    }
+    if (result.length < k) {
+      result.push(nums[i]);
+    }
+  }
+
+  return result;
+};