Add solution #327

Author: JoshCrozier

README.md

@@ -263,6 +263,7 @@
 322|[Coin Change](./0322-coin-change.js)|Medium|
 324|[Wiggle Sort II](./0324-wiggle-sort-ii.js)|Medium|
 326|[Power of Three](./0326-power-of-three.js)|Easy|
+327|[Count of Range Sum](./0327-count-of-range-sum.js)|Hard|
 328|[Odd Even Linked List](./0328-odd-even-linked-list.js)|Medium|
 329|[Longest Increasing Path in a Matrix](./0329-longest-increasing-path-in-a-matrix.js)|Hard|
 331|[Verify Preorder Serialization of a Binary Tree](./0331-verify-preorder-serialization-of-a-binary-tree.js)|Medium|

solutions/0327-count-of-range-sum.js

@@ -0,0 +1,58 @@
+/**
+ * 327. Count of Range Sum
+ * https://leetcode.com/problems/count-of-range-sum/
+ * Difficulty: Hard
+ *
+ * Given an integer array nums and two integers lower and upper, return the number of range
+ * sums that lie in [lower, upper] inclusive.
+ *
+ * Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j
+ * inclusive, where i <= j.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} lower
+ * @param {number} upper
+ * @return {number}
+ */
+var countRangeSum = function(nums, lower, upper) {
+  const result = [0];
+  for (let i = 0; i < nums.length; i++) {
+    result.push(result[i] + nums[i]);
+  }
+
+  function mergeSort(input, left, right) {
+    if (left >= right) return 0;
+    const middle = Math.floor((left + right) / 2);
+    let count = mergeSort(input, left, middle) + mergeSort(input, middle + 1, right);
+    let i = left;
+    let j = middle + 1;
+    let k = middle + 1;
+
+    while (i <= middle) {
+      while (j <= right && input[j] - input[i] < lower) j++;
+      while (k <= right && input[k] - input[i] <= upper) k++;
+      count += k - j;
+      i++;
+    }
+
+    const sorted = [];
+    i = left;
+    j = middle + 1;
+    while (i <= middle || j <= right) {
+      if (i > middle) sorted.push(input[j++]);
+      else if (j > right) sorted.push(input[i++]);
+      else if (input[i] <= input[j]) sorted.push(input[i++]);
+      else sorted.push(input[j++]);
+    }
+
+    for (let i = 0; i < sorted.length; i++) {
+      input[left + i] = sorted[i];
+    }
+
+    return count;
+  }
+
+  return mergeSort(result, 0, result.length - 1);
+};