fix(leetcode): remove solutions for problems 1365 and 1636

work/Baton-1/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js

@@ -1,69 +0,0 @@
-/*
- * 1365. How Many Numbers Are Smaller Than the Current Number
- * Difficulty: Easy
- * https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
- *
- * ──────────────────────────────────────────────────
- *
- * Given the array nums, for each nums[i] find out how many numbers in
- * the array are smaller than it. That is, for each nums[i] you have to
- * count the number of valid j's such that j != i and nums[j] < nums[i].
- *
- * Return the answer in an array.
- *
- *
- *
- * Example 1:
- *
- * Input: nums = [8,1,2,2,3]
- * Output: [4,0,1,1,3]
- * Explanation:
- * For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and
- * 3).
- * For nums[1]=1 does not exist any smaller number than it.
- * For nums[2]=2 there exist one smaller number than it (1).
- * For nums[3]=2 there exist one smaller number than it (1).
- * For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
- *
- * Example 2:
- *
- * Input: nums = [6,5,4,8]
- * Output: [2,1,0,3]
- *
- * Example 3:
- *
- * Input: nums = [7,7,7,7]
- * Output: [0,0,0,0]
- *
- *
- *
- * Constraints:
- *
- * 	• 2 <= nums.length <= 500
- *
- * 	• 0 <= nums[i] <= 100
- */
-
-/**
- * @param {number[]} nums
- * @return {number[]}
- */
-var smallerNumbersThanCurrent = function (nums) {
-  // Default for Hammer
-  const freq = {};
-  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
-  const sorted = Object.keys(freq).sort((a, b) => a - b);
-
-  // Problem Specific
-  let count = 0;
-  let smaller = {};
-
-  // Iterate over sorted.
-  for (let num of sorted) {
-    smaller[num] = count;
-    count += freq[num];
-  }
-
-  // Map output
-  return nums.map((n) => smaller[n]);
-};

work/Baton-1/Easy/Array/1636.sort-array-by-increasing-frequency.js

@@ -1,50 +0,0 @@
-/*
- * 1636. Sort Array by Increasing Frequency
- * Difficulty: Easy
- * https://leetcode.com/problems/sort-array-by-increasing-frequency/
- * 
- * ──────────────────────────────────────────────────
- * 
- * Given an array of integers nums, sort the array in increasing order
- * based on the frequency of the values. If multiple values have the same
- * frequency, sort them in decreasing order.
- * 
- * Return the sorted array.
- * 
- *  
- * 
- * Example 1:
- * 
- * Input: nums = [1,1,2,2,2,3]
- * Output: [3,1,1,2,2,2]
- * Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and
- * '2' has a frequency of 3.
- * 
- * Example 2:
- * 
- * Input: nums = [2,3,1,3,2]
- * Output: [1,3,3,2,2]
- * Explanation: '2' and '3' both have a frequency of 2, so they are
- * sorted in decreasing order.
- * 
- * Example 3:
- * 
- * Input: nums = [-1,1,-6,4,5,-6,1,4,1]
- * Output: [5,-1,4,4,-6,-6,1,1,1]
- * 
- *  
- * 
- * Constraints:
- * 
- * 	• 1 <= nums.length <= 100
- * 
- * 	• -100 <= nums[i] <= 100
-*/
-
-/**
- * @param {number[]} nums
- * @return {number[]}
- */
-var frequencySort = function(nums) {
-    
-};