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

work/1/Easy/Array/977.squares-of-a-sorted-array.js

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+/*
+ * 977. Squares of a Sorted Array
+ * Difficulty: Easy
+ * https://leetcode.com/problems/squares-of-a-sorted-array/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given an integer array nums sorted in non-decreasing order, return an
+ * array of the squares of each number sorted in non-decreasing order.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [-4,-1,0,3,10]
+ * Output: [0,1,9,16,100]
+ * Explanation: After squaring, the array becomes [16,1,0,9,100].
+ * After sorting, it becomes [0,1,9,16,100].
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [-7,-3,2,3,11]
+ * Output: [4,9,9,49,121]
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= nums.length <= 10^4
+ * 
+ * 	• -10^4 <= nums[i] <= 10^4
+ * 
+ * 	• nums is sorted in non-decreasing order.
+ * 
+ *  
+ * 
+ * Follow up: Squaring each element and sorting the new array is very
+ * trivial, could you find an O(n) solution using a different approach?
+*/
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var sortedSquares = function(nums) {
+  let n = nums.length;
+  let ans = new Array(nums.length);
+  let left = 0, right = nums.length - 1;
+
+  for (let i = n -1; i >= 0; i--){
+    let square;
+    if(Math.abs(nums[left]) < Math.abs(nums[right])){
+      square = nums[right];
+      right--;
+    }else{
+      square = nums[left];
+      left++;
+    }
+    ans[i] = square*square;
+  }
+  return ans;
+};