Add solution #1685

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,472 LeetCode solutions in JavaScript
+# 1,473 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1297,6 +1297,7 @@
 1680|[Concatenation of Consecutive Binary Numbers](./solutions/1680-concatenation-of-consecutive-binary-numbers.js)|Medium|
 1681|[Minimum Incompatibility](./solutions/1681-minimum-incompatibility.js)|Hard|
 1684|[Count the Number of Consistent Strings](./solutions/1684-count-the-number-of-consistent-strings.js)|Easy|
+1685|[Sum of Absolute Differences in a Sorted Array](./solutions/1685-sum-of-absolute-differences-in-a-sorted-array.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|

solutions/1685-sum-of-absolute-differences-in-a-sorted-array.js

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+/**
+ * 1685. Sum of Absolute Differences in a Sorted Array
+ * https://leetcode.com/problems/sum-of-absolute-differences-in-a-sorted-array/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums sorted in non-decreasing order.
+ *
+ * Build and return an integer array result with the same length as nums such that result[i] is
+ * equal to the summation of absolute differences between nums[i] and all the other elements
+ * in the array.
+ *
+ * In other words, result[i] is equal to sum(|nums[i]-nums[j]|) where 0 <= j < nums.length
+ * and j != i (0-indexed).
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var getSumAbsoluteDifferences = function(nums) {
+  const n = nums.length;
+  const result = new Array(n);
+  let prefixSum = 0;
+  let suffixSum = nums.reduce((sum, num) => sum + num, 0);
+
+  for (let i = 0; i < n; i++) {
+    const current = nums[i];
+    suffixSum -= current;
+    result[i] = (current * i - prefixSum) + (suffixSum - current * (n - 1 - i));
+    prefixSum += current;
+  }
+
+  return result;
+};