refactor(work): restructure solution directories with bucket level and add .gitkeep placeholders

work/1/Easy/Array/1480.running-sum-of-1d-array.js

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+/*
+ * 1480. Running Sum of 1d Array
+ * Difficulty: Easy
+ * https://leetcode.com/problems/running-sum-of-1d-array/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given an array nums. We define a running sum of an array as
+ * runningSum[i] = sum(nums[0]…nums[i]).
+ * 
+ * Return the running sum of nums.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [1,2,3,4]
+ * Output: [1,3,6,10]
+ * Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3,
+ * 1+2+3+4].
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [1,1,1,1,1]
+ * Output: [1,2,3,4,5]
+ * Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1,
+ * 1+1+1+1, 1+1+1+1+1].
+ * 
+ * Example 3:
+ * 
+ * Input: nums = [3,1,2,10,1]
+ * Output: [3,4,6,16,17]
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= nums.length <= 1000
+ * 
+ * 	• -10^6 <= nums[i] <= 10^6
+*/
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var runningSum = function(nums) {
+  let prefix = [nums[0]];
+  for (let i = 1; i < nums.length; i++){
+    prefix.push(prefix[i - 1] + nums[i]);
+  }
+  return prefix;
+};