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sushant-sinhasushant-sinha
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Merge branch 'master' into master
2 parents 29412be + a816d0d commit d379ea0

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‎README.md

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‎database/_1113.sql

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select extra as report_reason, count(distinct(post_id)) as report_count from Actions where action_date = '2019-07-04' and action = 'report' group by extra;

‎database/_1142.sql

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--# Write your MySQL query statement below
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select ifnull(round(count(distinct session_id)/count(distinct user_id), 2), 0.00)
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as average_sessions_per_user
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from Activity
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where activity_date between '2019-06-28' and '2019-07-27';

‎database/_1371.sql

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--# Write your MySQL query statement below
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select b.employee_id, b.name, count(*) as reports_count, round(avg(a.age)) as average_age
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from Employees a join Employees b on a.reports_to = b.employee_id
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group by b.employee_id
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order by b.employee_id

‎database/_1393.sql

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select stock_name, sum(
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case
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when operation = 'Buy' then -price
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else price
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end
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) as capital_gain_loss from Stocks group by stock_name;

‎database/_1596.sql

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--credit: https://leetcode.com/problems/the-most-frequently-ordered-products-for-each-customer/discuss/861257/simple-and-easy-solution-using-window-function
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select customer_id, product_id, product_name from
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(
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select o.customer_id, o.product_id, p.product_name,
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rank() over (partition by customer_id order by count(o.product_id) desc) as ranking
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from Orders o
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join Products p
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on o.product_id = p.product_id
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group by customer_id, product_id
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) tmp
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where ranking = 1
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order by customer_id, product_id

‎database/_1729.sql

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--# Write your MySQL query statement below
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select user_id, count(follower_id) as followers_count from followers group by user_id order by user_id;

‎src/main/java/com/fishercoder/solutions/_127.java

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import java.util.List;
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import java.util.Set;
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/**
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* 127. Word Ladder
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*
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* Given two words (beginWord and endWord), and a dictionary's word list,
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* find the length of shortest transformation sequence from beginWord to endWord, such that:
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* Only one letter can be changed at a time.
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* Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
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For example,
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Given:
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beginWord = "hit"
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endWord = "cog"
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wordList = ["hot","dot","dog","lot","log","cog"]
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As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", return its length 5.
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Note:
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Return 0 if there is no such transformation sequence.
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All words have the same length.
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All words contain only lowercase alphabetic characters.
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You may assume no duplicates in the word list.
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You may assume beginWord and endWord are non-empty and are not the same.
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*/
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public class _127 {
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public static class Solution1 {
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‎src/main/java/com/fishercoder/solutions/_1437.java

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package com.fishercoder.solutions;
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import java.util.ArrayList;
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import java.util.List;
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public class _1437 {
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public static class Solution1 {
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public boolean kLengthApart(int[] nums, int k) {
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List<Integer> indexes = new ArrayList<>();
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for (int i = 0; i < nums.length; i++) {
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int lastOneIndex = nums[0] == 1 ? 0 : -1;
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for (int i = 1; i < nums.length; i++) {
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if (nums[i] == 1) {
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indexes.add(i);
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}
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}
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for (int i = 0; i < indexes.size() - 1; i++) {
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if (indexes.get(i + 1) - indexes.get(i) < k + 1) {
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return false;
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if (i - lastOneIndex <= k) {
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return false;
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} else {
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lastOneIndex = i;
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}
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}
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}
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return true;

‎src/main/java/com/fishercoder/solutions/_151.java

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return stringBuilder.substring(0, stringBuilder.length() - 1).toString();
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}
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}
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public static class Solution2 {
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public String reverseWords(String s) {
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int len = s.length();
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int i = 0;
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int j = 0;
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String result = "";
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while (i < len) {
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// index i keeps track of the spaces and ignores them if found
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while (i < len && s.charAt(i) == ' ') {
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i++;
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}
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if (i == len) {
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break;
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}
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j = i + 1;
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// index j keeps track of non-space characters and gives index of the first occurrence of space after a non-space character
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while (j < len && s.charAt(j) != ' ') {
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j++;
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}
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// word found
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String word = s.substring(i, j);
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if (result.length() == 0) {
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result = word;
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} else {
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result = word + " " + result;
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}
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i = j + 1;
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}
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return result;
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}
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}
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}

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