Add solution #1695

JoshCrozier · JoshCrozier · commit e9a9ab89fb04 · 2025-04-26T15:15:55.000-05:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,481 LeetCode solutions in JavaScript
+# 1,482 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1306,6 +1306,7 @@
 1690|[Stone Game VII](./solutions/1690-stone-game-vii.js)|Medium|
 1691|[Maximum Height by Stacking Cuboids](./solutions/1691-maximum-height-by-stacking-cuboids.js)|Hard|
 1694|[Reformat Phone Number](./solutions/1694-reformat-phone-number.js)|Easy|
+1695|[Maximum Erasure Value](./solutions/1695-maximum-erasure-value.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|
 1726|[Tuple with Same Product](./solutions/1726-tuple-with-same-product.js)|Medium|
diff --git a/solutions/1695-maximum-erasure-value.js b/solutions/1695-maximum-erasure-value.js
@@ -0,0 +1,37 @@
+/**
+ * 1695. Maximum Erasure Value
+ * https://leetcode.com/problems/maximum-erasure-value/
+ * Difficulty: Medium
+ *
+ * You are given an array of positive integers nums and want to erase a subarray containing unique
+ * elements. The score you get by erasing the subarray is equal to the sum of its elements.
+ *
+ * Return the maximum score you can get by erasing exactly one subarray.
+ *
+ * An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is,
+ * if it is equal to a[l],a[l+1],...,a[r] for some (l,r).
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maximumUniqueSubarray = function(nums) {
+  const seen = new Set();
+  let result = 0;
+  let currentSum = 0;
+  let start = 0;
+
+  for (let end = 0; end < nums.length; end++) {
+    while (seen.has(nums[end])) {
+      seen.delete(nums[start]);
+      currentSum -= nums[start];
+      start++;
+    }
+    seen.add(nums[end]);
+    currentSum += nums[end];
+    result = Math.max(result, currentSum);
+  }
+
+  return result;
+};
