Add solution #315

Author: JoshCrozier

Diff for: README.md

@@ -242,6 +242,7 @@
 303|[Range Sum Query - Immutable](./0303-range-sum-query-immutable.js)|Easy|
 306|[Additive Number](./0306-additive-number.js)|Medium|
 309|[Best Time to Buy and Sell Stock with Cooldown](./0309-best-time-to-buy-and-sell-stock-with-cooldown.js)|Medium|
+315|[Count of Smaller Numbers After Self](./0315-count-of-smaller-numbers-after-self.js)|Hard|
 316|[Remove Duplicate Letters](./0316-remove-duplicate-letters.js)|Medium|
 318|[Maximum Product of Word Lengths](./0318-maximum-product-of-word-lengths.js)|Medium|
 319|[Bulb Switcher](./0319-bulb-switcher.js)|Medium|

Diff for: solutions/0315-count-of-smaller-numbers-after-self.js

@@ -0,0 +1,31 @@
+/**
+ * 315. Count of Smaller Numbers After Self
+ * https://leetcode.com/problems/count-of-smaller-numbers-after-self/
+ * Difficulty: Hard
+ *
+ * Given an integer array nums, return an integer array counts where counts[i] is the
+ * number of smaller elements to the right of nums[i].
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var countSmaller = function(nums) {
+  const result = new Array(nums.length).fill(0);
+  const rank = new Map([...nums].sort((a, b) => a - b).map((n, i) => [n, i]));
+  const group = new Array(nums.length + 1).fill(0);
+
+  for (let i = nums.length - 1; i >= 0; i--) {
+    const rankIndex = rank.get(nums[i]) + 1;
+    for (let j = rankIndex - 1, sum = 0; j > 0; j -= j & -j) {
+      sum += group[j];
+      result[i] = sum;
+    }
+    for (let j = rankIndex; j < group.length; j += j & -j) {
+      group[j]++;
+    }
+  }
+
+  return result;
+};