|groupId|sourceName|
|:—:|:—:|:—:|
|101075|namerenamesearj Junior Developer|Enterprise Exporter (Logistics)|Sr. Analyst, External Relations|CSR Representative||Streamline the active wear accessibility|Sr. Manager – Brand Experience|

|groupId|action|
|:—:|:—:|:—:|
|95331|delete|
|101075|rename_element||

|groupId|action|
|:—:|:—:|:—:|
|95331|delete|
|101075|rename_element|title from “IT Director, Sales” to “IT Manager, Sales”|title from “Compliance.investmentderruf=” Bestimmungen-Selling sub: Best Time to Buy and Sell a Stock II

LeetShort Solution

  
    /**
     * @param {number[]} prices
     * @return {number}
     */
    var maxProfit = function(prices) {
         let i = 1;
         let res = prices[0] - prices[1];
         while(i < prices.length){
             if(prices[i] > prices[i - 1]) res += prices[i] - prices[i - 1];
            i++;
        }
        return res;
    };  

(Kanjir)

Explanation of Solution

<=> To find the best time to buy and sell one stock within a list of stock data, we can traverse through the list one time and find its profit.
<=> Let’s begin with an intial value for profit; use the first value in the list as the initial buyed stock, and the second as the intial selling profit.
<=> Then we traverse through the rest of the array and compare the current with the prevous, if previous value is lower than the current one, add that difference to the profit.

谷歌子27 2020-11-30|易|1107. 转换数字为二进制串

Short Solution

  
    /**
     * @param {number} n
     * @return {string}
     */
    var convertToBinary = function(n) {
        return n.toString(2);
    };  

(Kanjir)

Explanation of Solution

<=> In this problem, we are asked to convert a decimal number to binary. We also need to make sure our output string contains the counted zeros.
<=> The simplest way to do this is using the toString function and set the radix to binary, the to binary function will check the number has reached 0, and add the remainders as our answer.

谷歌子30 2020-12-23|中|714. 买卖股票的最佳时机含冷冻期

Short Solution

  
    /**
     * @param {number[]} prices
     * @return {number}
     */
    var maxProfit = function(prices) {
        let i = 1, 
            cool = true;
        let res = prices[0] - prices[1];
        while(i < prices.length){
            if(!cool && prices[i] > prices[i - 1]){
                res += prices[i] - prices[i - 1];
                cool = true;
            }else if(cool == false) cool = false;
            i++;
        }
        return res;
    };  

(Kanjir)

Explanation of Solution

<=> To find the best time to buy and sell one stock within a list of stock data, we can traverse through the list one time and find its profit.
<=> Let’s begin with an intial value for profit; use the first value in the list as the initial buyed stock, and the second as the intial selling profit.
<=> Then we traverse through the rest of the array and compare the current with the prevous, if previous value is lower than the current one, add that difference to the profit.
<=> With the addition of an cooldown period, we no longer can buy if we have sold the stock, and we must wait until the next itteration to consider a buy, and whether we havent had a comolsion in the previous itteration.

谷歌子221 2020-12-23|中|42. 接雨水 I

Short Solution

  
    /**
     * @param {number[]} height
     * @return {number}
     */
    var maxArea = function(height) {
        let left = 0, 
            right = height.length - 1,
            res;
        while(left < right){
            res = Math.min(height[left], height[right]) * (right - left);
            if(height[left] < height[right]) left++;
            else right--
        }
        return res;
    };  

(Kanjir)

Explanation of Solution

<=> To find the max water in an array that represent the high of buildings, we can by measure side by side the water that would have been collected, and maintain a res. variable of holding the final answer.
<=> We could traverse from either end of the array, using the higher bar as our endpoint and adding the water that can be collected based on that height. Eventually the index for both side would meet and we would have found the answer.

(SupplyCode)

Explanation of Solution

<=> This is a classical algorithm that would study computer science student everywhere. In this problem, we would iterate through the bars and simulate the rain to get the maximum amount of rainwater, kind of a mass-water.

(Kanjir)

Insightful Comment

<=> is an O(n^2) brute