Add solution #870

Author: JoshCrozier

README.md

@@ -677,6 +677,7 @@
 867|[Transpose Matrix](./0867-transpose-matrix.js)|Easy|
 868|[Binary Gap](./0868-binary-gap.js)|Easy|
 869|[Reordered Power of 2](./0869-reordered-power-of-2.js)|Medium|
+870|[Advantage Shuffle](./0870-advantage-shuffle.js)|Medium|
 872|[Leaf-Similar Trees](./0872-leaf-similar-trees.js)|Easy|
 873|[Length of Longest Fibonacci Subsequence](./0873-length-of-longest-fibonacci-subsequence.js)|Medium|
 875|[Koko Eating Bananas](./0875-koko-eating-bananas.js)|Medium|

solutions/0870-advantage-shuffle.js

@@ -0,0 +1,36 @@
+/**
+ * 870. Advantage Shuffle
+ * https://leetcode.com/problems/advantage-shuffle/
+ * Difficulty: Medium
+ *
+ * You are given two integer arrays nums1 and nums2 both of the same length. The advantage of nums1
+ * with respect to nums2 is the number of indices i for which nums1[i] > nums2[i].
+ *
+ * Return any permutation of nums1 that maximizes its advantage with respect to nums2.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number[]}
+ */
+var advantageCount = function(nums1, nums2) {
+  const sortedNums1 = [...nums1].sort((a, b) => a - b);
+  const indexedNums2 = nums2.map((val, idx) => ({ val, idx }))
+    .sort((a, b) => a.val - b.val);
+
+  const result = new Array(nums1.length);
+  let left = 0;
+  let right = nums1.length - 1;
+
+  for (let i = nums1.length - 1; i >= 0; i--) {
+    const current = indexedNums2[i];
+    if (sortedNums1[right] > current.val) {
+      result[current.idx] = sortedNums1[right--];
+    } else {
+      result[current.idx] = sortedNums1[left++];
+    }
+  }
+
+  return result;
+};