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Commit 199efae

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feat: add No.739,2966
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# 2966. Divide Array Into Arrays With Max Difference
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- Difficulty: Medium.
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- Related Topics: Array, Greedy, Sorting.
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- Similar Questions: .
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## Problem
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You are given an integer array `nums` of size `n` and a positive integer `k`.
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Divide the array into one or more arrays of size `3` satisfying the following conditions:
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- **Each** element of `nums` should be in **exactly** one array.
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- The difference between **any** two elements in one array is less than or equal to `k`.
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Return **a ****2D**** array containing all the arrays. If it is impossible to satisfy the conditions, return an empty array. And if there are multiple answers, return **any** of them.**
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Example 1:
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```
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Input: nums = [1,3,4,8,7,9,3,5,1], k = 2
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Output: [[1,1,3],[3,4,5],[7,8,9]]
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Explanation: We can divide the array into the following arrays: [1,1,3], [3,4,5] and [7,8,9].
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The difference between any two elements in each array is less than or equal to 2.
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Note that the order of elements is not important.
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```
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Example 2:
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```
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Input: nums = [1,3,3,2,7,3], k = 3
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Output: []
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Explanation: It is not possible to divide the array satisfying all the conditions.
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```
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**Constraints:**
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- `n == nums.length`
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- `1 <= n <= 105`
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- `n` is a multiple of `3`.
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- `1 <= nums[i] <= 105`
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- `1 <= k <= 105`
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## Solution
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```javascript
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/**
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* @param {number[]} nums
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* @param {number} k
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* @return {number[][]}
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*/
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var divideArray = function(nums, k) {
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nums.sort((a, b) => a - b);
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var res = [];
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for (var i = 0; i < nums.length; i += 3) {
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if (nums[i + 2] - nums[i] <= k) {
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res.push([nums[i], nums[i + 1], nums[i + 2]]);
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} else {
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return [];
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}
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}
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return res;
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};
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```
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**Explain:**
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nope.
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**Complexity:**
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* Time complexity : O(n * log(n)).
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* Space complexity : O(n).

‎701-800/739. Daily Temperatures.md‎

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# 739. Daily Temperatures
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- Difficulty: Medium.
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- Related Topics: Array, Stack, Monotonic Stack.
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- Similar Questions: Next Greater Element I, Online Stock Span.
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## Problem
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Given an array of integers `temperatures` represents the daily temperatures, return **an array** `answer` **such that** `answer[i]` **is the number of days you have to wait after the** `ith` **day to get a warmer temperature**. If there is no future day for which this is possible, keep `answer[i] == 0` instead.
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Example 1:
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```
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Input: temperatures = [73,74,75,71,69,72,76,73]
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Output: [1,1,4,2,1,1,0,0]
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```Example 2:
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```
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Input: temperatures = [30,40,50,60]
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Output: [1,1,1,0]
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```Example 3:
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```
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Input: temperatures = [30,60,90]
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Output: [1,1,0]
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```
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**Constraints:**
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- `1 <= temperatures.length <= 105`
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- `30 <= temperatures[i] <= 100`
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## Solution
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```javascript
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/**
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* @param {number[]} temperatures
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* @return {number[]}
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*/
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var dailyTemperatures = function(temperatures) {
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var stack = [];
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var res = Array(temperatures.length).fill(0);
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for (var i = 0; i < temperatures.length; i++) {
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while (stack.length && temperatures[i] > temperatures[stack[stack.length - 1]]) {
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var index = stack.pop();
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res[index] = i - index;
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}
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stack.push(i);
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}
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return res;
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};
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```
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**Explain:**
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nope.
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**Complexity:**
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* Time complexity : O(n).
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* Space complexity : O(n).

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