Add solution #1499

JoshCrozier · JoshCrozier · commit edf768d5b931 · 2025-04-16T12:20:58.000-05:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,331 LeetCode solutions in JavaScript
+# 1,332 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1146,6 +1146,7 @@
 1496|[Path Crossing](./solutions/1496-path-crossing.js)|Easy|
 1497|[Check If Array Pairs Are Divisible by k](./solutions/1497-check-if-array-pairs-are-divisible-by-k.js)|Medium|
 1498|[Number of Subsequences That Satisfy the Given Sum Condition](./solutions/1498-number-of-subsequences-that-satisfy-the-given-sum-condition.js)|Medium|
+1499|[Max Value of Equation](./solutions/1499-max-value-of-equation.js)|Hard|
 1502|[Can Make Arithmetic Progression From Sequence](./solutions/1502-can-make-arithmetic-progression-from-sequence.js)|Easy|
 1507|[Reformat Date](./solutions/1507-reformat-date.js)|Easy|
 1512|[Number of Good Pairs](./solutions/1512-number-of-good-pairs.js)|Easy|
diff --git a/solutions/1499-max-value-of-equation.js b/solutions/1499-max-value-of-equation.js
@@ -0,0 +1,46 @@
+/**
+ * 1499. Max Value of Equation
+ * https://leetcode.com/problems/max-value-of-equation/
+ * Difficulty: Hard
+ *
+ * You are given an array points containing the coordinates of points on a 2D plane,
+ * sorted by the x-values, where points[i] = [xi, yi] such that xi < xj for all
+ * 1 <= i < j <= points.length. You are also given an integer k.
+ *
+ * Return the maximum value of the equation yi + yj + |xi - xj| where |xi - xj| <= k
+ * and 1 <= i < j <= points.length.
+ *
+ * It is guaranteed that there exists at least one pair of points that satisfy the
+ * constraint |xi - xj| <= k.
+ */
+
+/**
+ * @param {number[][]} points
+ * @param {number} k
+ * @return {number}
+ */
+var findMaxValueOfEquation = function(points, k) {
+  let result = -Infinity;
+  const q = [];
+
+  for (let j = 0; j < points.length; j++) {
+    const [xj, yj] = points[j];
+
+    while (q.length && xj - points[q[0]][0] > k) {
+      q.shift();
+    }
+
+    if (q.length) {
+      const [xi, yi] = points[q[0]];
+      result = Math.max(result, yi + yj + xj - xi);
+    }
+
+    while (q.length && yj - xj >= points[q[q.length - 1]][1] - points[q[q.length - 1]][0]) {
+      q.pop();
+    }
+
+    q.push(j);
+  }
+
+  return result;
+};
