Refactor: apply code review feedback

Author: HC-kang

kth-smallest-element-in-a-bst/HC-kang.ts

@@ -50,6 +50,7 @@ function kthSmallest(root: TreeNode | null, k: number): number {
 }
 
 /**
+ * NOT MY SOLUTION - NO NEED TO REVIEW
  * Awesome solution from leetcode
  */
 function kthSmallest(root: TreeNode | null, k: number): number {
@@ -58,29 +59,6 @@ function kthSmallest(root: TreeNode | null, k: number): number {
   assess. smallest value can be 0. At least k nodes, which can be at its smallest, 1.
   approach. DFS with backtracking. Traverse down the left edges until we hit null. if k is 1, return that value. Else, backtrack, go right, then try to go left again. Use a stack.
   */
-
-  // let currentRank = 0;
-
-  // let stack = [];
-
-  // let currentNode = root;
-
-  // while (currentNode || stack.length > 0) {
-  //     while (currentNode) {
-  //         stack.push(currentNode);
-
-  //         currentNode = currentNode.left;
-  //     }
-
-  //     currentNode = stack.pop();
-
-  //     currentRank++;
-
-  //     if (currentRank === k) return currentNode.val;
-
-  //     currentNode = currentNode.right;
-  // }
-
   const stack = [];
 
   let currentRank = 0;

top-k-frequent-elements/HC-kang.ts

@@ -13,11 +13,11 @@ Output: [1]
 // Time complexity: O(nlogn)
 // Space complexity: O(n)
 function topKFrequent(nums: number[], k: number): number[] {
-  const acc = new Map<number, number>();
+  const frequentMap = new Map<number, number>();
   for (const num of nums) { // s.c. O(n)
-    acc.set(num, (acc.get(num) || 0) + 1);
+    frequentMap.set(num, (frequentMap.get(num) || 0) + 1);
   }
-  return Array.from(acc.entries())
+  return Array.from(frequentMap.entries())
     .sort((a, b) => b[1] - a[1]) // this will cost t.c. O(nlogn)
     .slice(0, k)
     .map((v) => v[0]);